1. The State Police have set up an aerial speed trap along a highway where the speed limit is 55 mph. An officer in the plane measures 26 seconds as the time it takes a car to travel the distance between two marks on the highway that are 2500 feet apart. Was the driver exceeding the speed limit?

First, we need to find the car's average speed.

speed _average = d / t = 2500 ft / 26 s = 96 ft/s

But these units do not allow us to make the required comparison. We must use a conversion factor to change ft/s into mph.

96 ft/s (1 mph / 1.467 ft/s) = 66 mph

Now it is obvious that our overzealous driver was going a little too fast. At 11 mph over the speed limit he will probably get a ticket. Since the road could have been curved, this must be considered an average speed and not an average velocity because the direction of his motion might also be changing, or he might have slowed down or speeded up somewhat during this 26 second time interval.


2. A professional bowler throws his ball straight down a 60-foot alley and watches it strike the pins 4.2 seconds later. What is the average velocity of the bowling ball?

v_average = d / t = 60 ft / 4.2 s = 14 ft/s

The complete answer must include a direction. Here the best we can do is to state:

v_average = 14 ft/s "down the alley"

which is a valid direction because it tells us exactly which way the ball is traveling.

Because the textbook deals primarily with the SI, we could have converted 60 feet to 18.3 meters (Review Chapter 1 if you do not remember how to do this conversion). This would make the average velocity 4.4 m/s "down the alley." We left these examples in the British system to show you that the calculational methods shown here will work in all systems of units.

Acceleration is also a vector quantity, so it must have a direction specified for it, as shown in the next problem.


3. If a bicycle passes a man with a stopwatch at an initial velocity of 10 m/s and then moves downhill in a straight line toward the east, what will be the acceleration of the bicycle if its final velocity after 7.0 seconds is 45 m/s?

a = v / t = (v_F - v_o ) / t = (45 m/s - 10 m/s) / 7.0 s = 5.0 m/s² (east)

Again, to be complete we must give a direction. In this case either east or "down the hill" would do, but east seems more specific.

It is often useful to write down the quantities that are given in a problem before we begin the actual calculation. (See Appendix II in the textbook.) This gives us a chance to see just what it is that we already know and what we are trying to determine. Let's try it in the next problem.


4. A car that was originally at rest accelerates for 0.30 minutes at a rate of 2.5 m/s² straight south. What would be the velocity of the car at the end of this time period?

First we know v_o = zero, a = 2.5 m/s²(south), and t = 0.30 min = 18 s

Since v = v_F - v_o we can use the defining equation for acceleration and rewrite it as follows:

a = v/t = (v_F - v_o ) / t

so:v_F = v_o + a t = 0 + 2.5 m/s² (18 s) = 45 m/s (south)


5. What would be the final velocity of the car in question 4 if the car had not been at rest but was traveling at 20 m/s toward the south when the acceleration of 2.5 m/s² began?



Acceleration can speed things up or slow them down, depending on the direction in which the process is occurring. If the acceleration is in the same direction in which the object is already moving, the motion of the object will speed up (accelerate), but if the acceleration is in the opposite direction to that of the original motion, the object will slow down (decelerate).


6. If the car described above were traveling at 65 m/s and then experienced a deceleration of -1.8 m/s² for 6.0 seconds, what would be the final velocity of the car?



Notice that the car is going slower after the change in motion has occurred, but it is still moving in the same direction. The slowing down is caused by the negative acceleration (deceleration) of the car.

Free-fall motion occurs around us every day. Anytime an object is dropped, either accidentally or on purpose, it immediately begins to fall toward Earth (that is, downward).


7. A circus juggler throws a ball upward with an initial velocity of 22.0 m/s. How long does he have before it comes back down and he has to catch it? (Assume he catches it at the same height from which he threw it and that we can neglect air friction.)

At first this looks a little tricky. We don't seem to have enough information. But we know the acceleration (g) that is acting downward on the ball and, if we think for a moment, we will realize that the velocity of the ball becomes zero for an instant just at the top of its flight. If we let "up" be the positive direction, g must be negative because it is directed downward. So we know the following:

vo = 22.0 m/s,

vF = 0,

a = g = -9.80 m/s2

from

vF = vo + a t,

we can get

t = (vF - vo ) / a

since a = g, t = (v_F - v_o ) / g = ( 0 - 22.0 m/s) / -9.80 m/s² = 2.25 s

This is the time for the ball to go upward from his hand and reach the top of its flight path. The ball will require an equal amount of time to come back down, so the total time that he has to get ready and catch the ball is twice the 2.25 s calculated above, or 2 x (2.25 s) = 4.50 s.

Another type of motion must be considered before we move on to the study of force in Chapter 3. This is the type of motion that an object undergoing vertical free-fall would experience if it also had a horizontal component to its velocity. Such a motion is called projectile motion.


8. A student throws a stone horizontally with a velocity of 14.0 m/s from the top of a building that is 25.0 m high. Neglecting air resistance, how far has the stone fallen vertically and how far is it from the side of the building after 2.00 seconds of elapsed time?

First let's list what we know. Letting X represent horizontal components and Y refer to the vertical components of the motion:

vXo = 14.0 m/s		aX = 0
vYo = 0		aY = g = 9.80 m/s2
t = 2.00 s		height = 25.0 m

From this we can calculate the distance of vertical fall much like the example shown in Figure 2.14 in the textbook. Note that in this case we are using SI units so the acceleration due to gravity (g) is 9.80 m/s²

dY	= vYot + 1/2 g t2
	= 0 (2.00s) + 1/2 (9.80 m/s2) (2.00 s)2
	= 0 + 19.6 m = 19.6 m (downward)

Since the building is 25.0 m high, the stone will not have reached the ground yet. The stone is 19.6 m down from the top of the building or 25.0 m - 19.6 m = 5.4 m above ground level after 2.00 seconds of fall.

The horizontal distance that the stone is from the building after 2.00 seconds can be found by:

d_X = v_Xo t

V_Xo is considered to be constant during the entire fall because a_X is zero if there is no air resistance. Therefore:

d_X = 14.0 m/s (2.00 s) = 28.0 m (away from the building)

Working problems involving projectile motion is rather difficult. You must remember that the vertical (Y) and horizontal (X) portions of the motion are independent of each other except for the element of time. This problem shows that quite clearly. If the horizontal initial velocity had been larger, the stone would have been farther from the wall after 2.00 seconds.

Also remember that the acceleration of gravity (g) only affects the vertical part of the motion and always has a constant downward value (9.80 m/s² or 32 ft/s²) near Earth's surface.

Don't forget to work on the questions and problems in the textbook as well as the sample questions that follow in this study guide. The more practice you have using the concepts presented in this course, the better you will understand the material and the better grade you will earn in this course.

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