13 April 2010

kinematics

Problem 1
A train covers 60 miles between 2 p.m. and 4 p.m. How fast was it going at 3 p.m.?
Solution:
The speed is traveled distance (60 miles) divided by traveled time (4pm – 2pm = 2hours):


Problem 2
Is it possible that the car could have accelerated to 55mph within 268 meters if the car can only accelerate from 0 to 60 mph in 15 seconds?
Solution:
Let us find the maximum acceleration of the car:
The car can accelerate from 0 to in 15 seconds. Then maximum acceleration is
If the car needs to accelerate to within 268 meters then its acceleration should be
This acceleration is less than the maximum possible acceleration, so the car can reach the speed 55 mph within 268 meters.
Problem 3
A car travels up a hill at a constant speed of 37 km/h and returns down the hill at a constant speed of 66 km/h. Calculate the average speed for the whole trip.
Solution:
By definition the average speed is the ration of the total traveled distance and the total traveled time. Let us introduce the total traveled distance of the car as L. Then the time of the travel up the hill is
The time of the travel down the hill is
The total traveled time is
Then the average velocity is


Problem 4
An archer shoots an arrow with a velocity of 30 m/s at an angle of 20 degrees with respect to the horizontal. An assistant standing on the level ground 30 m downrange from the launch point throws an apple straight up with the minimum initial speed necessary to meet the path of the arrow. What is the initial speed of the apple and at what time after the arrow is launched should the apple be thrown so that the arrow hits the apple?


Solution:
The motion of the arrow is a projectile motion. Then the motion of the arrow along horizontal direction is a motion with constant velocity:
Where the initial position is 0 and . Then
Motion along vertical axis is a motion with constant acceleration, then
Where , initial position is 0, and . Then
Then we need to find the time when the arrow will be exactly above the assistant. At this moment of time x(t) =100. Then we can find the time:
Then we can find the height of the arrow at this moment of time:
The assistant should through the apple with minimal velocity so it will reach point 5.4 m and at this height the velocity should be 0. From this condition we can find the minimal velocity:
The time of the motion of the apple to this point is
Then the apple should be thrown after


Problem 5
A box sits on a horizontal wooden board. The coefficient of static friction between the box and the board is 0.5. You grab one end of the board and lift it up, keeping the other end of the board on the ground. What is the angle between the board and the horizontal direction when the box begins to slide down the board?


Solution:
The critical angle is determined by the condition:
From this equation we can find an angle .


Problem 6
A 8 kg block is at rest on a horizontal floor. If you push horizontally on the 8 kg block with a force of 20 N, it just starts to move.


(a) What is the coefficient of static friction?


(b) A 10.0 kg block is stacked on top of the 8 kg block. What is the magnitude F of the force, acting horizontally on the 8 kg block as before, that is required to make the two blocks start to move?


Solution:
The magnitude of horizontal force should be equal to the magnitude of the maximal static friction force, which is equal to the product of the coefficient of static friction and the normal force (gravitation force in the present problem).
(a) The gravitation force is mg=8*9.8 = 78.4 N. Then the coefficient of static friction is
(b) Now we know the coefficient of static friction and we know the normal force: 18*9.8 = 176.4 N. Then we can find the magnitude of force F:
Problem 7
A car is accelerating at . Find its acceleration in .


Solution:
To find an acceleration in we need to use the relations:
,
Then we can write:


Problem 8.
We drive a distance of 1 km at 16 km/h. Then we drive an additional distance of 1 km at 32 km/h. What is our average speed?


Solution:
By definition the average speed is the ratio of traveled distance and traveled time.
The traveled distance is 2 km.
The traveled time is the sum of two contributions:
• time of the motion a distance 1 km with speed 16 km/h. It is
• time of the motion a distance 1 km with speed 32 km/h. It is
Then the average speed is





Problem 9.
An airliner reaches its takeoff speed of 163 mph in 36.2 s. What is the magnitude of its average acceleration.


Solution:
By definition the acceleration is the ratio of the change of velocity and the traveled time. In the present problem the change of velocity is 163 mph = 0.447*163 m/s = 72.9 m/s. The traveled time is 36.2 s.
Then the average acceleration is


Problem 10.
A car is initially traveling due north at 23 m/s.
(a) Find the velocity of the car after 4 s if its acceleration is due north.
(b) Find the velocity of the car after 4 s if its acceleration is instead due south.


Solution:
This is the motion with constant acceleration. The dependence of velocity on time is given by the equation:
Where .
(a) In this case the direction of acceleration is the same as the direction of initial velocity. Then and we have
(b) In this case the direction of acceleration is opposite to the direction of initial velocity. Then and we have


Problem 11.
From the dimensional analysis find the time t it takes for a ball to fall from a height h.


Solution:
We know that the time should depend on the height, h, and free fall acceleration, g. So we can write:
Where a is a dimensionless constant (we cannot find this constant) and x, y – are constant, which can be found from dimensional analysis.
The units of t is seconds. The units of h is meter, the unit of g is . Then we have:
From this equation we have:
Then
Then


Problem 12.
You are driving along the street at the speed limit (35mph) and 50 meters before reaching a traffic light you notice it becoming yellow. You accelerate to make the traffic light within the 3 seconds it takes for it to turn red. What is your speed as you cross the intersection? Assume that the acceleration is constant and that there is no air resistance.


Solution:
This is the motion with constant acceleration. If the acceleration of the car is a then we can write the expression for traveled distance:
We know the initial velocity , we also know that after 3 seconds the car travels distance 50 meters. Then we can find acceleration
Now we know acceleration, then we can find the final velocity:


Problem 13.
How high can a human throw a ball if he can throw it with initial velocity 90 mph.


Solution:
The height is given by the expression:
where . Then


Problem 14.
Mr. Letourneau is flying his broom stick parallel to the ground. He undergoes two consecutive displacements. The first is 100 km 10 degrees west of north, and the second is 120 km 50 degrees east of north. What is the magnitude of the broom stick's displacement?


Solution:
The displacement are shown schematically in the figure. From this figure we can see that the angle between the first and the second displacements is . Then from the triangle based on the first, second, and net displacements we can find the magnitude of the net displacement (cosine formula):








Problem 15.
In reaching her destination, a backpacker walks with an average velocity of 1 m/s, due west. This average velocity results, because she hikes for 6 km with an average velocity of 3 m/s due west, turns around, and hikes with an average velocity of 0.3 m/s due east.
How far east did she walk (in kilometers)?
Solution:
We will define the average velocity as the ratio of total traveled distance and total traveled time (the average velocity can be also defined as the ratio of displacement and traveled time).
The backpacker walks 6km=6000m due to west. The average velocity was 3 m/s. Then the traveled time for this motion is
Then she travels east. Let assume that her traveled time due to east is . Then the traveled distance due to east is
Then the total traveled distance is and the total traveled time is . Then the final average velocity is
From this equation we can find :
Then we can find :


Problem 16.
It takes you 9.5 minutes to walk with an average velocity of 1.2 m/s to the north from the bus stop to museum entrance. What is your displacement?
Solution:
If you travel along the straight line than the displacement is equal to the traveled distance.
Then the displacement is
where and .
Then
Problem 17.
What is 3.3 slug in kilograms?
Solution:
Since 1 slug = 14.5939 kg then we get:


Problem 18.
An athlete swims the length of a 50.0-m pool in 20.0s and makes the return trip to the starting position in 22.0s. Determine her average velocities in
(a) the first half of the swim,
(b) the second half of the swim, and
(c) the round trip.


Solution:
The average velocity is the ratio of traveled distance and traveled time.
Then
(a)
(b)
(c)


Problem 19.
A speedboat increases its speed uniformly from 20 m/s to 30 m/s in a distance of 200m. Find
(a) the magnitude of its acceleration and
(b) the time it takes the boat to travel the 200-m distance.


Solution:
(a) This is the motion with constant acceleration. We can use the following equation to find the magnitude of acceleration
where , , . Then
(b) We can find the time of the travel from the following equation:


Problem 20.
The car drives straight off the edge of a cliff that is 57 m high. The investigator at the scene of the accident notes that the point of impact is 130 m from the base of the cliff. How fast was the car traveling when it went over the cliff?


Solution:
Motion along axis x (horizontal axis) is the motion with constant velocity. So we can write down the dependence of x-coordinate as a function of time:
Where is the initial velocity (the initial velocity has only x-component, its direction is along axis x).
We know the final x-coordinate of the car – it is 130 m. But we do not know the traveled time and the initial velocity.
We can find the traveled time from the motion along axis y. This is the motion with constant acceleration (free fall acceleration). We know the height of the cliff – this is the traveled distance in y-direction. We know that initial velocity (in y-direction) is 0. Then we can write the following equation:
From this equation we can find the traveled time:
Then from the motion along axis x we have:
From this equation we can find initial velocity:


Problem 21.
An aircraft has a lift-off of 120km/h.
(a) What minimum constant acceleration does the aircraft require if it is to be airborne after a takeoff run of 240m?
(b) How long does it take the aircraft to become airborne?


Solution:
(a) This is the motion with constant acceleration. We know that the initial velocity is 0. And we know that after run of 240 m the velocity of aircraft becomes 120 km/h = 120 *1000/3600 m/s = 33.3 m/s.
Then we need to use the following equation:
From this equation we can find acceleration:
(b) Now we know acceleration and we know the final velocity so we can find the traveled time:


Problem 22.
A ball is thrown vertically upward with a speed of 25.0m/s.
(a) How high does it rise?
(b) How long does it take to reach its highest point?
(c) How long does the ball take to hit the ground after it reaches its highest point?
(d) What is its velocity when it returns to the level from which it started?


Solution:
This is the motion with constant acceleration (free fall acceleration).
(a) We need to use the following equation (the initial velocity is 25 m/s and the final velocity is 0):
(b) We need to use the following equation (the initial velocity is 25 m/s and the final velocity is 0):
(c) At this moment we do not know the final velocity, but we know the initial velocity (it is 0) and we know the height (the traveled distance) – it is 31.9 m. Then we can use the following equation:
From this equation we can find time:
It is the same time as in part (b).
(d) Now we can find the final velocity (the magnitude):


Problem 23.
A tortoise and a hare are in a road race to defend the honor of their breed. The tortoise crawls the entire 1000 meters at a speed of 0.2 m/s. The rabbit runs the first 200 meters at 2 m/s, stops to take a nap for 1.3 hours, and awakens to finish the last 800 meters with an average speed of 3 m/s. Who wins the race and by how much time?


Solution:
At first let us calculate the traveled time of tortoise. We know the speed and the distance, so we can easily find the time:
Now let us calculate the traveled time of rabbit. The traveled time consists of three parts:
1. He runs the first 200 m at 2 m/s. The time of this motion is
2. Then he take a nap for 1.3 hours:
3. Then he run the last 800 m with speed 3 m/s. The time of this motion is
Then the total traveled time is
Since then tortoise wins the race by 47 s.


Problem 24.
The slowest animal ever discovered is a crab found in the Red Sea that travels an average speed of 5.7 km/year. How long will it take this crab to travel 1 meter?


Solution:
The speed of the crab is
Then the time of the motion is


Problem 25.
A "moving sidewalk" in a busy airport terminal moves 1 m/s and is 200 m long. A passenger steps onto one end and walks, in the same direction as the sidewalk is moving, at a rate of 2.0 m/s relative to the moving sidewalk. How much time does it take the passenger to reach the opposite end of the walkway?


Solution:
The speed of the passenger relative to the ground is 2m/s+1m/s =3 m/s (since the he is working in the same direction as the direction of the motion of the sidewalk.
Then the passenger reaches the end of the sidewalk after


Problem 26.
Assume it takes 8 minutes to fill a 35.0 gal gasoline tank. (1 U.S. gal = 231 cubic inches)
(a) Calculate the rate at which the tank is filled in gallons per second.
(b) Calculate the rate at which the tank is filled in cubic meters per second.
(c) Determine the time interval, in hours, required to fill a volume at the same rate.


Solution:
The rate of filling is 35galons/8 minutes = 4.375 galons/minutes.
(a) Since 1 minute = 60 seconds then
(b) Since 1 minute = 60 seconds and then
(c) From part (b) - is filled within


Problem 27.
A plane flies 955 km due east, then turns due north and flies another 469 km. Draw an x-y axis at the starting point with the positive x-axis pointing east, and determine the polar coordinates of the plane's finale position.


Solution:
The positive x-axis is pointing east, while the positive y-axis is pointing north. Then the final position of the plane has coordinate:
x-coordinate is 955 km,
y-coordinate is 469 km.
Then the polar angle can be found from the equation:
Then .


Problem 28.
A car travels east at 89 km/h for 1 h. It then travels 26° east of north at 141 km/h for 1 h.
(a) What is the average speed for the trip?
(b) What is the average velocity for the trip?


Solution:
(a) The average speed is defined as the ratio of the total traveled distance and the traveled time. The total traveled distance is the sum of length AB (=89 km) and length BC (=141 km). The total traveled time is 2 hours. Then the average speed is
(b) The average velocity is defined as the ratio of the net displacement (the magnitude of the vector, connecting points A and C) and the traveled time (= 2 h)
We can find the length AC from the triangle ABC:
(cosine theorem).
Then the average velocity is






Problem 29.
(a) If a particle's position is given by (where t is in seconds, and x is in meters), what is it's velocity at t=1s?
(b) what is it's speed at t=1s?
(c) Is there ever an instant when the velocity is 0? If so, give the time.


Solution:
The velocity is the derivative of x(t) with respect to time. Then
(a) at t=1 s we get:
What we calculate here is an x-component of velocity. The negative sign means that the direction of velocity is opposite to the direction of axis x.
(b) the speed is the magnitude of velocity. Then at t=1 s the speed is 4 m/s
(c) to find time at which the velocity is 0 we just need to solve an equation:
From this equation we find time:
At this moment of time the velocity is 0.
Problem 30.
The captain of a plane wishes to proceed due west. The cruising speed of the plane is 245 m/s relative to the air. A weather report indicates that a 38-m/s wind is blowing from the south to the north. In what direction, measured to due west, should the pilot head the plane relative to the air?
Solution:
This is problem on the relative motion. The velocity of a plane is the vector sum of the velocity of an air and the relative velocity of the plane (relative to the air) as shown in the figure.
We know that the direction of the final velocity of the plane is from east to west (as shown in the figure).
Then from the triangle based shown in the figure we can find angle . This angle characterizes the direction of the relative velocity of the plane.
Then










Problem 31.
A missile is launched into the air at an initial velocity of 80 m/s. It is moving with constant velocity until it reaches 1000m, when the engine fails.
(a) How long does it take it to reach 1000m?
(b) How high does the missile go?
(c) How long does it take for it to fall back to the earth?
(d) How long does it stay in the air?
(e) How fast is it going when it hits the ground?


Solution:

(a) Since initial we have the motion with constant velocity we can easily find the time of the motion of the missile till it reaches the height 1000 m. The time is given by the expression:


After this point we have free fall motion – there is only one force acting on the object (it is gravitational force) – this force provide free fall acceleration.
The initial velocity is 80 m/s pointing upward. The acceleration is pointing downward. The initial height of the missile is 1000 m. Then the equations which describe this motion are the following:
 
(b) To find the maximum height of the missile we can use the last equation. The velocity at the maximum height is 0. Then

 
(c) To find the time when the missile hits the ground we need to use the first equation:
When the missile hits the ground h=0. Then
From this equation we can find time: 24.6 s.
 
(d) Then we can find the time when the missile is in the air: it is the sum of the time when it reaches 1000 m and the time when it hits the ground:
 
 
(e) To find the speed of the missile when it hits the ground we need to use the last equation:
When the missile hits the ground h=0. Then
 
 
Problem 32.
The highest barrier that a projectile can clear is 14 m, when the projectile is launched at an angle of 30.0 degrees above the horizontal. What is the projectile's launch speed?
The maximum height of the projectile is given by the equation:
where is the launch angle. Then we can find initial velocity:
 
Problem 33.
A rocket is fired vertically upwards with initial velocity 80 m/s at the ground level. Its engines then fire and it is accelerated at until it reaches an altitude of 1000 m. At that point the engines fail and the rocket goes into free-fall. Disregard air resistance.
(a) How long was the rocket above the ground?
(b) What is the maximum altitude?
(c) What is the velocity just before it collides with the ground?
Solution: 

The first part of the motion is the motion with constant acceleration at . The initial velocity for this motion is 80 m/s. Then we can write the equation, which describe the dependence of height of the rocket on time:
From this equation we can find the time when the rocket reach the height 1000 m = h:
The solution of this equation is 10 s. So after 10 seconds the engine fails. The velocity at this moment of time is
 
 
After this moment of time we have free fall motion – there is only one force acting on the object (it is gravitational force) – this force provide free fall acceleration.
The initial velocity is 120 m/s pointing upward. The acceleration is pointing downward. The initial height of the rocket is 1000 m. Then the equations which describe this motion are the following:
 
To find the maximum height of the rocket we can use the last equation. The velocity at the maximum height is 0. Then
This is the answer to part (b).


To find the time when the rocket hits the ground we need to use the first equation:
When the rocket hits the ground h=0. Then
From this equation we can find time: 31 s.
 
Then we can find the time when the rocket is in the air: it is the sum of the time when it reaches 1000 m and the time when it hits the ground:
This is the answer to part (a).

 
To find the speed of the rocket when it hits the ground we need to use the last equation:
When the missile hits the ground h=0. Then
Problem 34.
An object is fired straight up at a speed of 9.8m/s. Compute its maximum altitude and the time it takes to reach that height. Ignore air resistance.
Solution: 

This is the free fall motion. The initial velocity is 9.8 m/s pointing upward. The acceleration is pointing downward. The initial height of the object is 0 m. Then the equations which describe this motion are the following:
 
The maximum altitude corresponds to the condition that the velocity at this point is 0. Then from the second equation we can find the time:
Then
From the first equation we can find height:
Problem 35.
A rock is dropped from rest into a well.
(a) The sound of the splash is heard 4 s after the rock is released from rest.
How far below to top of the well is the surface of the water? (the speed of sound in air at ambient temperature is 336m/s).
(b) If the travel time for the sound is neglected, what % error is introduced when the depth of the well is calculated?


Solution:
(a) If h is the depth of the well then to find the time of the rock's fall we need to use the following equation:
From this equation we can find time:
After the rock hits the water the sound of splash will propagate the distance h with speed 336 m/s. After time the sound reaches the ground. Then total time is
This time is equal to 4 s. Then
From this equation we can find h:

 
(b) If we neglect the sound velocity then t=4 s and we can find the height from the equation:
If we compare this result with the result from part (a) then we can find an error:
Problem 36.
A ball is thrown upward with initial velocity v. What is the change in speed on the way up? What is the change in speed on the way down? Disregard air resistance.
Solution :
he speed of the ball is the magnitude of velocity.
On the way up: initial speed is v, the final speed is 0. Then the change of the speed is 0-v=-v (it is negative).

On the way down: initial speed is 0, the final speed is v. Then the change of the speed is v-0=v (it is positive).
 
Problem 37.
A displacement, s, of an object as a function of time, t, is given by
a) Find an expression for the acceleration of the object
b) Explain why this expression indicates that the acceleration is not constant
  
Solution :
(a) Acceleration is the second derivative of displacement with respect to time.
The first derivative of displacement with respect to time is a velocity:
The derivative of the velocity is an acceleration:

 
(b) From this expression we can see that the acceleration depends on time (at different moments of time we have different acceleration) – it means that acceleration is not constant.
Problem 38.
Snow is falling vertically at a constant speed of 3 m/s. At what angle from the vertical do the snowflakes appear to be falling as viewed by the driver of a car traveling on a straight, level road with a speed of 60 km/h?
Solution :


At first we need to have the same units, so we need to convert the speed of the car in m/s:

Then we need to find the direction of the velocity of the snowflakes relative to the car:

 
Then from the triangle shown in the picture we can find angle :
 
Then

Problem 39.
A dart is thrown horizontally with an initial speed of 10 m/s toward point P, the bull's-eye on a dart board. It hits at point Q on the rim, vertically below P, 0.2 s later. What is the distance PQ? How far away from the dart board is the dart released?

 Solution :


The initial velocity is horizontal, the magnitude is 10 m/s. Then the equation of motion along horizontal axis has the following form:
After 0.2 seconds the dart hits the target. From this condition and from the above equation we can find the distance between the point where the dart is released and the board:

 
The equations which describe the motion along the vertical axis are the following:
If we substitute into the first equation the traveled time then we can find the distance between points P and Q:
The minus sign means that point Q is below point P.
Problem 40.
A boy walks to the store using the following path: 0.4 miles west, 0.2 miles north, 0.3 miles east. What is his total displacement? That is, what is the length and direction of the vector that points from her house directly to the store?


Solution :



 
The net displacement is the vector sum of three displacements as shown in the figure. It is more easier to find the net displacement in terms of components: x component (along east direction), y component (along north direction).
x component = x component of the first displacement + x component of the second displacement + x component of the third displacement = -0.4 + 0 + 0.3 = -0.1 miles.
y component = y component of the first displacement + y component of the second displacement + y component of the third displacement = 0 + 0.2 + 0 = 0.2 miles.
 
Then the magnitude of the net displacement is

The angle between the direction of the net displacement and west direction is
 
Then
Problem 41.
A motor scooter travels east at a speed of 13 m/s. The driver then reverses direction and heads west at 17 m/s. What was the change in velocity of the scooter?


Solution :


By definition the change in the velocity is equal to final velocity minus initial velocity.
The final velocity is 17 m/s pointing west. The initial velocity is 13 m/s pointing east. Since the velocity is a vector we need to describe the change of the velocity in terms of the components.
We choose the coordinate axis pointing to the west. Then the component of the final velocity is +17 m/s, and the component of the initial velocity is -13 m/s. Then the change in the velocity is
17 – (-13) = 30 m/s
Problem 42.
A bicycle travels 3.2 km due east in 0.1 h, the 3.2 km at 15.0 degrees east of north in 0.21 h, and finally another 3.2km due east in 0.1 h to reach its destination. The time lost in turning is negligible. What is the average velocity for the entire trip?
Solution :


By definition the average velocity is equal to the ratio of the magnitude of the net displacement and the traveled time. The net displacement is shown in the figure.
 
 
We can find the x and y components of the vector of net displacement.
The x-component is
The y-component is

Then the magnitude of the net displacement is

The traveled time is 0.1 h + 0.21 h + 0.1 h = 0.41 h. Then the average velocity is

Problem 43.
A swimmer dived off a cliff with a running horizontal leap. What must his minimum speed be just as he leaves the top of the cliff so that he will miss the ledge at the bottom which is 2 m wide and 9 m below the top of the cliff?
Solution: 

This is the projectile motion with horizontal initial velocity. In this case the motion in horizontal direction is described by the following equation:
Where v is the initial velocity.
 
The motion along vertical axis y is described by the following equations (y-component of initial velocity is 0):
 
We know that the final point of the motion is the end of the ledge. The coordinates of the final point are: x=2 m and y = -9m.
We can substitute these numbers in the above equations and obtain:
 
From the seconds equation we can find time:
Then we substitute this time in the first equation and obtain the initial velocity:
Problem 44.
A ferries wheel with radius 20 m which rotates counterclockwise, is just starting up. At a given moment, a passenger on the rim of the wheel and passing through the lowest point of his circular motion is moving at 3.00m/s and in gaining speed at a rate of .
Find the magnitude of the passenger's acceleration at the instant. 


Solution: 

The acceleration at this point has two components: the tangential acceleration, which is equal to , and centripetal acceleration, which is given by the expression:

These components are orthogonal. Then the net acceleration is

Problem 45.
In an action film hero is supposed to throw a grenade from his car, which is going 90.km/h, to his enemy's car, which is going 110 km/h. The enemy's car is 15.8 m in front of the hero's when he lets go of the grenade. If the hero throws the grenade so its initial velocity relative to him is at an angle of 45(degree) above the horizontal, what should the magnitude of the initial velocity be? The cars are both traveling in the same direction on a level road. Ignore air resistance.
Find the magnitude of the velocity both relative to the hero and relative to the earth.
Solution: 

At first we need to convert all the velocities into m/s:
 
We will describe the motion in the hero's reference frame. In this reference frame the velocity of the enemy's car is
And the grenade is released with the launch angle . We introduce the magnitude of initial velocity of grenade (relative to the hero's car) as v.

After the grenade is released it is moving according to the equations of projectile motion. Along axis x we have motion with constant velocity:
Along axis y we have motion with constant acceleration:
In these equations at the initial moment of time the grenade has zero coordinates.
 
 
We can also write down the equation of enemy's car motion:
Now we need to write down the condition that the grenade hits the enemy's car:
At some moment of time the x coordinate of grenade should be equation to the x-coordinate of enemy's car and at the same moment of time the y-coordinate of grenade should be equal 0.
Then we have:
 
Then we just need to solve the system of two equations with two unknown variables:
Then
And
Then
 
This is the velocity of the grenade relative to the hero.

 
We can also find the magnitude of velocity relative to the earth. The x-component of this velocity is
The y-component of the velocity is
Then the magnitude is
This is the magnitude of the velocity of grenade relative to the earth.

Problem 46.
A flowerpot falls from a windowsill 25.0 m above the sidewalk.
(a) how fast is the flowerpot moving when it strikes the ground?
(b) how much time does a passerby on the sidewalk below have to move out of the
way before the flowerpot hits the ground?


Solution :


This is the free fall motion with constant acceleration. The equations which describe this motion are the following:
where we introduced an axis y with upward direction. The origin of axis y is the initial position of the flowerpot. The initial velocity of flowerpot is 0.
 
We know the final position of the flowerpot: the coordinate of the final point is -25 m.

(a) We substitute this coordinate in the third equation an obtain the velocity:
The negative sign means that the direction of velocity is downward. The magnitude of velocity is 22 m/s.

(b) The traveled time can be found from the first equation in the above system. We substitute y(t)=-25 and obtain time:



Problem 47.
A tennis ball is thrown vertically upward with an initial velocity of +8.0 m/s.
(a) what will the ball\'s speed be when it returns to its starting point?
(b) how long will the ball take to reach its starting point?
Solution :


This is the free fall motion with constant acceleration. The equations which describe this motion are the following:


where we introduced an axis y with upward direction. The origin of axis y is the initial position of the ball. The initial velocity of the ball is 8 m/s.
 
(a) The final point is the starting point. It means that the coordinate of the final point is 0. We substitute this coordinate in the third equation and obtain:
It means that the magnitude of the final velocity is 8 m/s. The direction of the final velocity is downward. So we should write
Minus sign means that the direction is downward.

(b) To find the traveled time we substitute the final position (0 m) in the first equation of the system and obtain:
Then
Problem 48.
Problem 4.14 of Serway & Jewett: A firefighter, a distance d from a burning building,
directs a stream of water from a fire hose at angle above the horizontal. If the initial
speed of the stream is , at what height h does the water strike the building?
Solution :


This is the projectile motion. The equation which describe the motion along horizontal axis x is the following
where is the x-component of initial velocity.
 
Along axis y we have motion with constant acceleration:
 
In these equations at the initial moment of time the water has zero coordinate.
 
What do we know about the final point: we know only the x-coordinate of the final point – it is x=d. We can substitute this coordinate in the first equation (which describe the motion along axis x):
From this equation we can find the traveled time:
Then we substitute this time into the equation which describe the y-coordinate of the water:
 
This expression gives as the height where the water strikes the building.



Problem 49.
If it takes a player 3 seconds to run from the batter's box to the first base at an average speed of 6.5 m/s, what is the distance she covers in that time?
Solution :


In this problem we just need to use the relation between the traveled time, traveled distance, and the average velocity:
We know the traveled time and the average velocity, then the distance is
Problem 50.
 
A car goes down a certain road at an average speed of 40 km/h and returns along the same road at an average speed of 60 km/h. Calculate the average speed for the round trip.
Solution :


Solution:
By definition the average speed is the traveled distance divided by the traveled time. Let introduce L as the traveled distance of the car in one direction. Then the total traveled distance is 2L. Let us find the traveled time. It is
 
Then the average speed is
 

Problem 51. 
(Inquiry into Physics-5th ed.,Ostdiek,Bord) On a day the wind is blowing toward the south at 3m/s, a runner jogs west at 4m/s. What is the velocity of the air relative to the runner?
Solution :


By definition the relative velocity is the vector difference of the velocity of the air minus velocity of the runner (see figure).
From the triangle we can find the magnitude of relative velocity:
 
 



Problem 52. 
(Inquiry into Physics-5th ed.,Ostdiek,Bord) A runner has an average speed of 4 m/s during a race. How far does the runner travel in 20 minutes?
Solution :


By definition the traveled distance is equal to the product of traveled time and the average velocity. In this problem we need to convert time into seconds: 20 minutes = 20*60 seconds = 1200 seconds. Then

Problem 53.
A car starts from the rest and accelerates uniformly for t=5 seconds over a distance of 100m. Find the acceleration of the car.
Solution :


This is the motion with constant acceleration. The equations which describe this motion are the following:
 
We know the traveled time and we know the traveled distance, then from the first equation we can find acceleration:
Problem 54. 
On the moon, an object is dropped from a height of 2 m. The acceleration of gravity on the moon is . Determine the time it takes for the object to fall to the surface of the moon.
Solution :


This is the motion with constant acceleration (free fall motion). The equations which describe this motion are the following:
 
We know the traveled distance (y=2 m), we know acceleration, then we can find time:
Problem 55.
A coin is dropped from a 400 m high tower. What is the coin's velocity when it hits the ground? How long does it take to get there?
Solution:


This is the free fall motion – motion with constant acceleration, . The equations, which describe this motion, are the following:
 
where the axis y has upward direction.

The final position of the coin is the ground. The coordinate of the final point is -400m. We can substitute this value in the first equation and obtain the traveled time:
Then
Then from the second equation we can find the velocity of the coin at the ground:

The minus sign means that the direction of velocity is downward.


Problem 56. 
Does the odometer of a car measure a scalar or a vector quantity?
Solution :


The odometer of the car shows the magnitude of the vector quantity.
The odometer shows the speed of the car.
The speed of the car is the magnitude of the velocity (which is the vector) of the car.
So the odometer gives us only the number, it does not show us the direction of velocity. The odometer measure the scalar quantity – the speed of the car. 

Problem 57.
The speed of light is about . Convert it into miles per hour (mph).
Solution :


Solution:
In this problem we need to know the following relations (see unit conversion):
1 mile = 1609 meter
1 hour = 3600 s

Then
Problem 58. 
You are driving towards a traffic signal when it turns yellow. Your speed is 55 km/h, and your best deceleration has the magnitude of . Your best reaction time before breaking is . To avoid having the front of your car enter the intersection after the light turns red, should you break to a stop or continue to move at 55 km/h if the distance to the intersection and the duration of the yellow light are
(a) 40m and 2.8s, and
(b) 32m and 1.8s?
Solution :


The first step – we need to convert the speed to m/s:
We have two possible types of motion:
1. motion with constant speed till the driver reaches the intersection. Time of this motion is
2. motion with deceleration. The driver starts deceleration after . During this time the driver travels the distance:
 
The equations which describe these motion are the following:
 

where the initial position of the car has zero coordinate.
At the final point the velocity is 0. Then
During this time the position of the car is
Now we can analyze the problem:

 
(a) 40m and 2.8s. Then for motion 1 we have:
Since 2.6<2.8 then the driver can cross the intersection if he will move with constant speed.

 
(b) 32m and 1.8s. In this case:
Since 2.1>1,8 the driver cannot cross the intersection.
Therefore in this part we need to consider case 2. But even in this case the distance he will travel before he stops is 34.2 m, which is greater than 32 m. Therefore he cannot stop before the intersection.
Problem 59. 
The tips of the blades in a food blender are moving with a speed of 20 m/s in a circle that has a radius of 0.06 m. How much time does it take for the blades to make one revolution?
Solution :


The traveled distance of a tip is
where R=0.06 m.

Then the time is equal to the ratio of the traveled time and the average velocity:
Problem 60. 
While you are traveling in a car on a straight road at 90 km/hr, another car passes you in the same direction; its speedometer reads 120km/hr. What is your velocity relative to the other driver? What is the other car's velocity relative to you?
Solution :


The relative velocity is equal to the difference of velocities.
So if you need to find your relative velocity then you need to subtract from your velocity the velocity of another car:

The minus sign here gives us the direction of relative velocity: the direction is opposite to the directions of motions of the cars.

 
Similarly we can find:
The plus sign here gives us the direction of relative velocity: the direction is the same as the directions of motions of the cars.
Problem 61.
 
A ball is thrown from a point 1 m above the ground. The initial velocity is 20 m/s at an angle of 40 degrees above the horizontal.
(a) Find the maximum height of the ball above the ground.
(b) Calculate the speed of the ball at the highest point in the trajectory.
Solution :


This is the projectile motion. The equation which describe the motion along horizontal axis x is the following
where is the x-component of initial velocity.
 
Along axis y we have motion with constant acceleration:
 

(a) The point of the maximum height corresponds to the condition that . Then
and
Then
This is the maximum height.


(b) The y-component of velocity at maximum height is 0, then the speed is equal to x-component of velocity:
Problem 62. 
A tortoise can run with a speed of 10.0 cm/s, and a hare can run exactly 10 times as fast. In a race, they both start at the same time, but the hare stops to rest for 3.00 min. The tortoise wins by 10 cm.
(a) How long does the race take?
(b) What is the length of the race?
Solution :


(a) If L is the length of the race then the tortoise run this race the time:

When the tortoise crosses the finish line the hare is 20 cm=0.2 m behind tortoise. It means that during time it traveled the distance L-0.2. Since he stopped for rest for 3 minutes =3*60 s=180 s and his speed is , then we can write the equation:

Taking into account the first equation we have:
Then

(b) Then we can find the length of the race:

Problem 63.


Emily takes a trip, driving with a constant velocity of 90 km/h to the north except for a 30 min rest stop. If Emily's average velocity is 75 km/h to the north, how long does the trip take?
Solution :


The average velocity is the ratio of traveled distance and the traveled time:
If the traveled time is t, then Emily traveled time (t-0.5)h with the speed 90 km/h. Then the traveled distance is

We substitute this relation in the first equation and obtain:
Then

Then we can find L (length of the trip):

Problem 64.
To qualify for the finals in a racing event, a race car must achieve an average speed of 250 km/h on a track with a total length of 2000 m. If a particular car covers the first half of the track at an average speed of 230 km/h, what minimum average speed must it have in the second half of the event to qualify?
Solution :


The average speed is the ratio of traveled distance and traveled time:

For the whole track we have: and . Then the traveled time is
 
The car travels the first half of the track (1 km) with speed 230 km/h. The time of this motion is

Since the total traveled time should be 0.008 h, then the second part of the track the car should travel 0.008-0.0043 = 0.0037 h. Since the distance of this motion is 1 km, then the average speed is

Problem 65. 
A skier is accelerating down a 30 degree hill at .
(a) What is the vertical component of her acceleration?
(b) How long will it take her to reach the bottom of the hill, assuming she starts from rest and accelerates uniformly, if the elevation is 300 m?
Solution :


(a) If we assume that the direction of the vertical axis is downward then the vertical component is positive and is equal to


(b) This is the motion with constant acceleration then the equation which describe this motion is the following:

The traveled distance (L) is related to the elevation (h) by the equation:

Now we know acceleration and the traveled distance then we can find the traveled time:
Problem 66.
You are trying to cross a river that flows due south with a strong current. You start out in you motorboat on the west bank desiring to reach the east bank directly across from your starting point. Which direction should head your motorboat? Dram a picture of the river, the banks, and your motorboat, and include the relevant velocity vectors. What information would you need in order to determine the actual direction you need to head?
Solution :


To find the actual direction of the boat velocity we need to know the velocity of the river and the speed of the boat relative to the river (relative velocity). Then the actual velocity (this is the vector) is

The diagram is shown below.

Problem 67.
(a) One liter ( ) of oil is spilled onto a smooth lake. If the oil spreads out uniformly until it makes an oil slick just one molecule thick, with adjacent molecules just touching, estimate the diameter of the (roughly circular) oil slick. Assume the oil molecules have a diameter of .
(b) Recalculate for the Exxon Valdez oil spill (March 1989), in which 11 million gallons (42 million L) of crude oil coated Prince William Sound in Alaska.
Solution :


(a) To find the diameter of the oil slick we just need to write down the equation that the volume of the oil slick is equal to . The height of the oil slick is equal to the diameter of the molecule: .
Then the diameter of the oil slick can be found from the equation:
Then

(b) Now the volume is not but

Then using the same equations as in part (a) we obtain
Problem 68.
The acceleration of an object as a function of time is . Determine the
(a) velocity and
(b) the position of the object as a function of time
if it is located at x = 2 m and has a velocity of 3 m/s at time t = 0 s.
Solution :


(a) By definition the velocity of the object is the integral of acceleration with respect to time:

where is the initial velocity (at t=0).

(b) The position can found as an integral of velocity with respect to time:
where is the initial position (at t=0) of the object.
Problem 69. 
An automobile traveling 90 km/h overtakes a 1.5-km-long train traveling in the same direction on a track parallel to the road. If the train's speed is 70 km/h,
(a) how long does it take the car to pass it, and
(b) how far will the car have traveled in the time?
Solution :


It is easier to solve this problem if we introduce the relative velocity of the car (relative to the train). The velocity of the car relative to the train is (90-70) km/h=20 km/h.
In the relative description the train is not moving and the car is moving with constant speed 20 km/h.

(a) Then we can easily find the time the car needs to pass the train. It is

(b) To find the actual traveled distance of the car we just need to multiply the traveled time (0.075 h) by the actual speed of the car:

Problem 70. 
Two cannonballs, A and B, are fired from the ground with identical initial speeds, but with launch angle of cannonball A larger than the launch angle of cannonball B.
(a) Which cannonball reaches a higher elevation?
(b) Which cannonball stays longer in the air?
(c) Which cannonball travels farther?
Solution :


(a) The vertical component of the initial velocity of the projectile is proportional to sinus of the launch angle. It means that the vertical component of initial velocity of cannonball A is larger than the vertical component of cannonball B. Then cannonball A reaches a higher elevation.

(b) The time in the air depends only on the vertical component of initial velocity of projectile. The larger the vertical component of the velocity the larger the time in the air. Then the cannonball A will stay longer in the air.

(c) From the data of the problem we cannot tell which one will travel farther. The horizontal length is proportional to . It has maximum at angle . Therefore we can have both possibilities:

cannonball A travels further than cannonball B (for example, if the launch angle of cannonball A is and the launch angler of cannonball B is ;

cannonball B travels further than cannonball A (for example, if the launch angle of cannonball B is and launch angler of cannonball A is ;
Problem 71. 
A stone falls freely from rest for 10 s. What is the stones displacement during this time.
Solution : 
This is the free fall motion. The displacement (axis y is pointing downward) is given by the equation
Then at t=10 s we have:


Problem 72.
A typical atom has a diameter of . What is this in inches?
Solution : 
In this problem we need to use the relation: 1m = 39.37 inch. Then




Problem 73. 
A rock is shot up vertically upward from the edge of the top of the building. The rock reaches its maximum height 2 s after being shot. Them, after barely missing the edge of the building as it falls downward, the rock strikes the ground 8 s after it was launched. Find
(a) upward velocity the rock was shot at;
(b) the maximum height above the building the rock reaches; and
(c) how tall is the building?
Disregard air resistance.
Solution : 
We have free fall motion of the rock. The equations, which describe the motion of the rock are the following:
Where – initial velocity, - initial height (height of the building).
(a) At the maximum height the velocity of the rock is zero. Then from the second equation we can find the initial velocity (since t= 2s)
(b) Then we can substitute the initial velocity, the traveled time (2 s) into the first equation and find the maximum height above the building:
(c) We know that after 8 s the rock hits the ground. It means that the position of the rock at this moment of time is 0. Then we substitute this time and this position into the first equation and find the height of the building:
Problem 74.
A car going at 10 m/s undergoes an acceleration of for 6 seconds. How far did it go when it was accelerating?
Solution :
 
In this problem we just need to use the equation:
Then at t = 6s we have:
Problem 75. 
What is the displacement of a car accelerating from 5 m/s (right) to 10 m/s (right) in 2.0s?
Solution:
In this problem we need to use two equations which describe the motion with constant acceleration:
Where – acceleration, - initial velocity.
Then from the second equation we can find acceleration:
Then from the first equation we can find displacement


Problem 76.
A plane flies 50 degrees east of south for 100 km then 400 km north and then 20 degrees north of west for 250 km.
Solution: 
This problem is easier to solve in terms of components.
The net displacement is the vector sum of three displacements: the first displacement (, the second displacement ( ), and the third displacement ( ):
Then we can find the x and y components of the displacement vectors and finally the net displacement:
Then the magnitude of the net displacement is









Problem 77. 
Calculate the hang time of an athlete who jumps a vertical distance of .9 meters considering the acceleration of gravity is


Solution: 
Without air resistance we have free fall motion. Then the maximum height can be related to the traveled distance (from the ground to the maximum height) by the equation:
where is the initial velocity.
At the maximum height the velocity is 0. Then
Then
And
The total hang time is the twice of this time: 2*0.42=0.84 s.
Problem 78. 
A man is driving at the speed 40 mph when he see an obstacle at distance 300 ft ahead of his position. The driver applies the brakes and decelerates at How long does it take him to stop the vehicle? How far from the obstacle will the driver be when he finally stops?
Solution: 
At first we need to convert all the variables into correct units (SI units):
Then we have motion with constant deceleration. Then
The car stops when then
And
The traveled distance is
Then the distance between the car and the obstacle is
Problem 79. 
The highest barrier that a projectile can clear is 20 m, when the projectile is launched at an angle of 40.0 degrees above the horizontal. What is the projectile's launch speed?
Solution: 
The meaning that the highest barrier the projectile can clear is 20 m is that the maximum height of the projectile is 20 m.
Since we are interested only in the height of the projectile then we can consider only the motion of the projectile along the vertical direction (axis y). Let us introduce the launch speed of the projectile as . Then the y component of the initial velocity of the projectile is
................................................(1)
The motion along axis y is a free fall motion and it is described by the following equations (only two equations are independent):
We know the initial height (y-coordinate) of the projectile: .
Then we introduce the final point – the point at which the projectile has the maximum height. We know this height: .
We also know that the y-component of final velocity (at the maximum height) is zero. We substitute these values in the third equation and obtain
From this equation we can find :
Then from equation (1) we can find the launch speed of the projectile:
Problem 80. 
You travel on the highway at a rate of 60 mph for 1 hour and at 50 mph for 2 hours and 40 mph for 3 hours. What is the total distance you have traveled? What is your average speed during the trip?
Solution: 
We divide the whole trip into three intervals:
(1) travel at a speed of 60 mph for 1 hour.
(2) travel at a speed of 50 mph for 2 hours,
(3) travel at a speed of 40 mph for 3 hours.
For each interval we know speed and time, then we can easily find the traveled distance (for each interval). The distance is equal to the product of time and speed (motion with constant velocity):
Then
(1) speed is 60 mph, time is 1 hour. Then
(2) speed is 50 mph, time is 2 hour. Then
(3) speed is 40 mph, time is 3 hour. Then
Then the total traveled distance is
The average speed is defined as the ratio of total traveled distance and the total traveled time. The total traveled time is
Now we can find the average speed:
Problem 81. 
An object travels a distance of 5 km towards east, then 4 km towards north and finally 10 km towards east
1) what is total traveled distance?
2) what is resultant displacement?
Solution: 
We show each displacement (travelled paths) in the figure.
(1) The total traveled distance is the sum of all traveled distances. It does not matter what the relative directions of displacements are, we just need to add the magnitude of displacements:
Problem 82. 
Can two displacements (vectors) of different magnitudes be combined to give a zero displacement (resultant)?
Solution: 
The sum of two vectors (displacements) is zero only if
1. vectors have opposite directions;
2. vectors have the same magnitude.
Indeed, the sum of two vectors is shown in the figure below for a fixed orientation of the first vector and different orientations of the second vector. In the picture we assume that the magnitude of the second vector is less than the magnitude of the first vector.







We can see that the sum of two vectors has the smallest magnitude when the vectors have opposite directions. In this case the sum is zero only if the vectors have the same magnitude.

Problem 83. 
What is the speed (m/sec) needed for a stunt driver to launch from a 20 degree ramp to land 15 m away?
What is his maximum height?
Solution: 
Initial speed.
This is the projectile motion. There are two sets of equations, which describe the motion of the projectile (stunt driver):
Set 1: motion along horizontal axis (axis x – see figure). This is the motion with constant velocity. There is only one equation, which describe this motion:
..............................(1)
Here and .
Set 2: motion along vertical axis (axisy – see figure). This is the motion with constant acceleration – free fall motion. There are three equations, which describe this motion. Only two equations are independent, but it is convenient to write all three equations:
................(2)
...........................(3)
...........................(4)







We know that the y-coordinate of the final point (point B) is 0 and the x-coordinate of the final point is 15 m. We substitute these values in equations (1) and (2) and obtain
Then from the first equation we can find :
Substitute this expression into the second equation:
From this equation we can easily find the initial velocity:
Maximum height.
The condition that the projectile is at the point with the maximum height is that the y-component of its velocity at this point is zero. It is easier to find the maximum height from equation (4). Indeed, we substitute in this equation and obtain:
Then
We know , then we can find the maximum height:
Problem 84. 
A body moves 4 km towards East from a fixed point A and reaches point B. Then it covers 5 km towards North and arrives at point C. Find the distance and directions of the net displacement.
Solution: 
We show two displacements (travelled paths) in the figure.


 



The net (resultant) displacement is the vector sum of two displacements: the first displacement () and the second displacement ():
The easiest way to find the net displacement is to introduce coordinate system (axes x and y) and then find the x and y components of the net vector-displacement. The x and y components of the displacement () and displacement () are the following:
Then the x and y components of the net displacement is
Then the magnitude of the net displacement is
The direction of the net displacement is characterized by angle (shown in the figure), which can be found from the known x and y components of the net displacement:
Problem 85. 
A baseball player hits a homerun, and the ball lands in the left field seats, which is 120 m away from the point at which the ball was hit. The ball lands with a velocity of 20 m/s at an angle of 30 degrees below horizontal. Ignoring air resistance
(A) find the initial velocity and the angle above horizontal with which the ball leaves the bat;
(B) find the height of the ball relatively to the ground.
Solution: 


(A) Initial velocity.
Without air resistance this is simple projectile motion. In the present problem we do not know initial velocity: we do not know the magnitude of the velocity (speed) and we do not know its direction.
There are two sets of equations, which describe the motion of the projectile (ball).
Set 1: motion along horizontal axis (axis x – see figure). This is the motion with constant velocity. There is only one equation, which describe this motion:
................................................(1)
Here .
Since the motion along the axis x is the motion with constant velocity then the x-component of the velocity is constant. We know the velocity at the final point. Then we can find the x-component of the velocity at the final point:
This x-component of the velocity is equal to the x-component of the initial velocity:
...........(2)
We also know the x-coordinate of the final point (point B): it is 120 m. We substitute this value in equation (1) and obtain
From this equation we can find the time of travel from point A to point B:
Now we need to analyze the second set of equations.
Set 2: motion along vertical axis (axis y – see figure). This is the motion with constant acceleration – free fall motion. There are three equations, which describe this motion. Only two equations are independent, but it is convenient to write all three equations:
.............(3)
................................................(4)
Since the initial y-coordinate is zero, then
.............................................(5)
 

We know the y component of the final velocity
This is the y-component of the velocity at the moment of time . We substitute these values in equation (4) and obtain
From this equation we can find the y-component of the initial velocity:
Finally we know the x- and y-components of the initial velocity:
From these expressions we can find the magnitude of the initial velocity and the direction (angle) of the initial velocity:
Now we know the initial velocity.
(B) Final height.
We need to find the final height of the ball (the final y-coordinate). To find the final height we can use equation (3). We just need to substitute the y-component of the initial velocity and the traveled time in this equation:


Problem 86. 
A body covers 1/4 journey with a speed of 40 km/h, 1/2 of it with 50 km/h and remaining with the speed of 60 km/h. Calculate average speed for entire journey.
Solution: 
In this problem we need to use the definition of the average speed. The average speed is equal to the ratio of the total travelled distance and the total traveled time:
We introduce the total traveled distance as and then calculate the total traveled time.
We know that 1/4 of the journey a body moves with a speed of 40 km/h. It means that the body moves a distance of with the speed of 40 km/h. Then we can find the time of this motion:
Then the body moves 1/2 of the journey with a speed of 50 km/h. It means that the body moves a distance of with the speed of 50 km/h. We can find the time of this motion:

Then the body moves the rest of the journey (which is 1/4 of the journey) with a speed of 60 km/h. It means that the body moves a distance of with the speed of 60 km/h. We can find the time of this motion:  

Then the total traveled time is  


Then the average speed is