2 Description of Motion

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2 Describing Motion

You may prefer to plot v vs. t to get the slope. The result 1 Average and Instantaneous speed will look better than getting a from successive interval data as 2 Velocity above. 3 Acceleration The vector nature of displacement can be demonstrated 4 Graphing Motion by putting a golf ball into a cup. Students can easily see that 5 Uniform Acceleration one straight shot is equivalent to two or three bad putts Everyday Phenomenon: Transitions in Traffic Flow before the ball enters the cup. Having established how Everyday Phenomenon: The 100-Meter Dash displacements combine, it is now easy to present the directional concepts of velocity and acceleration. Care spent in developing the concepts of kinematics in Students may find the units of acceleration confusing this chapter will be rewarded in future chapters on dynamics. when presented as m/s2. You can start out with an example I personally think it important, though, to relate things to the that gives mixed time units such as those in which car upcoming idea of energy transfer. This helps connect performance is expressed. A car which can start from rest concepts students generally otherwise see as completely and reach a velocity of 90 km/hr in 10 seconds, has an discrete and unconnected. average acceleration of 9 km/hr-s which means that on the average the velocity increases by 9 km/hr each second. This can then be converted to the more useful equivalent Suggestions for presentation form of 9000 m/hr-s = 2.5 m/s2. The concept of speed (both average speed and then instantaneous speed) can be easily demonstrated. A two- Answers to Questions meter air track with photocell timers is ideal but one can get by with a croquet ball rolling across a board or a small metal Q1 a. Speed is distance divided by time, so it will be car on a track, a meter stick and a timer. The key point measured in marsbars/zots. students must note is that a speed determination involves b. Velocity has the same units as speed, so it will two measurements, distance and time. Show first for a body also be measured in marsbars/zots. moving on a horizontal surface that the average speed is the c. Acceleration is change in speed divided by 2 same over various portions of its path. Though this is time, so the units will be marsbars/zots . apparent without taking actual measurements, it will more Q2 a. The unit system of inches and days would give likely be striking if measurements are taken. velocity units of inches per day and To study acceleration, set the track or board at a small acceleration of inches per day squared. angle. About a 1/2 cm drop in 2.0 meters works well. With b. This would be a terrible choice of units! The the body starting from rest, measure time successively to distance part would be huge while the time part travel predetermined distances. Rather than using equal would be miniscule! spatial distances, it is more instructive to choose distances Q3 Since fingernails grow slowly, a unit such as that will involve equal time intervals such as 10, 40, 90 and mm/month may be appropriate. 160 cm. It is good to try this in advance since some tracks Q4 a. The winner of a race must have the greater may not be quite straight and will give you problems. It is a average speed, so the plodding tortoise is the good idea to take more than one measurement for the time leader here. on each of the distances so that students can see what the b. Since the hare has the higher average speed uncertainties are. Since the instantaneous velocity is equal to taken over short intervals, he is likely to have the average velocity at the mid-point in the time interval, the the greater instantaneous speed. acceleration in each interval is found from the changes in Q5 In England "doing 70" likely means driving at 90 average velocity and corresponding changes in the time- km/hour which would be a reasonable highway midpoints (actually the change in time-midpoint is the same speed. In the US it means 70 mi/hr. Stating a as the changes in the time interval itself. Students will number without the proper units leaves us uncertain probably find this reasonable without going into the rationale as to what it means. behind it). Typical values with such an air track timing by Q6 A speedometer measures instantaneous speed; the hand are as follows: speed that you are driving at a particular instant of time. You can note how it responds immediately as 2 you speed up (accelerate) or slow down (brake). s (cm) t (sec) vav (cm/s) a (cm/s ) Q7 In low density traffic, the speed is more likely to be 0 0 constant, therefore the average and the (1.4)3.6 instantaneous speed will be close for relatively long 10 2.8 2.4 periods. In high density traffic, the speed is likely to (4.2) 10.3 be constantly changing so that only for short periods 40 5.7 2.1 the instantaneous speed will equal the average. (7.2) 16.7 Q8 The radar gun measures instantaneous speed - the 90 8.7 2.7 speed at the instant the radar beam hits the car and (10.1) 24.5 is reflected from it. An airplane spotter measures 160 11.6

3 Full file at http://TestbankCollege.eu/Solution-Manual-Physics-of-Everyday-Phenomena-6th-Edition- Griffith average speed; timing a car between two points 4 and 6 sec. Between 2 and 4 sec. The slope is which are a known distance apart. zero so the velocity in that interval does not change. Q9 Yes. When the ball is reflected from the wall (bounces back), the direction of its velocity is different than when it approaches the wall. Q22 Between 2 and 4 sec the car travels the greatest Q10 a. Yes. As it moves in a circle the velocity of the distance. Distance traveled can be determined from ball at each instant is tangent to the circular path. a velocity-time graph and is represented by the area Since this direction changes, the velocity changes under the curve, and between 2 and 4 sec. The even though the speed remains the same. area is the largest. The car travels the next greatest b. No. Since the velocity is changing, the distance between 4 and 6 sec. acceleration is not zero. Q23 a. Yes, during the first part of the motion where the Q11 a. No. The velocity changes because the speed instantaneous speed is greatest. The average and direction of the ball are changing. speed for the entire trip must be less than during b. No. The speed is constantly changing. At the this interval since for the rest of the trip the speeds turn-around points the speed is zero. are less. Q12 No. When a ball is dropped it moves in a constant b. Yes. The velocity changes direction. Even if the direction (downwards), but the magnitude of its magnitude of the velocity (speed) is the same, the velocity (speed) increases as it accelerates due to different directions make the velocities different. gravity. Actually, the change is negative so the car Q13 Yes. Acceleration is the change in velocity per unit decelerates. time. Here the change in velocity is negative (the Q24 No. This relationship holds only when the velocity decreases), so the acceleration is also acceleration is constant. negative and opposite to the direction of velocity. Q25 Velocity and distance increase with time when a car This is often called a deceleration. accelerates uniformly from rest as long as the Q14 No. In order to find the acceleration, you need to acceleration is positive. As long as it accelerates know the change in velocity that occurs during a uniformly, the acceleration is constant. time interval. Knowing the velocity at just one Q26 No. The acceleration is increasing. A constant instant tells you nothing about the velocity at a later acceleration is represented by a straight line. Here instant of time. the curve shown has an increasing, or positive, Q15 No. If the car is going to start moving, its slope. acceleration must be non-zero. Otherwise the Q27 Yes. For uniform acceleration the acceleration is velocity would not change and it would remain at constant. Since acceleration does not change, the rest, or stopped. average acceleration equals this constant Q16 No. As the car rounds the curve, the direction of its acceleration. velocity changes. Since there is a change in Q28 The distance covered during the first 5 sec is velocity (direction even if not magnitude), it must greater than the distance covered during the second 2 have an acceleration. 5 sec. Thus since distance = vot + ½at , even Q17 The turtle. As long as the racing car travels with though acceleration and time are the same for both constant velocity (even as large a velocity as 100 intervals, the initial velocity at which the car starts MPH), its acceleration is zero. If the turtle starts to the second interval is less than at the beginning of move at all, its velocity will change from zero to the first, so it will cover a shorter distance. something else, and thus it does have an Q29 acceleration. a. v Q18 a. Yes. Constant velocity is represented by the horizontal line which indicates the velocity does not change. b. Acceleration is greatest between 2 and 4 sec where the slope of the graph is steepest. Q19 a. Yes. The velocity is represented by the slope of a line on a distance-time graph. You can also see t that sometime after pt. B the line has a negative b. a slope indicating a negative velocity. b. Greater. The instantaneous velocities can be compared by looking at their slopes. The steeper slope indicates the greater instantaneous velocity. t Q20 Yes. The velocity is constant during all three different time intervals, that is in each interval where there is a straight line. Note that while the velocities are constant in these intervals, they are not the same in each. Q21 a. No. The car has a positive velocity during the Q30 The second runner. If both runners cover the same entire time shown. distance in the same time interval, then their b. At pt. A. The acceleration is greatest since the average velocity has to be the same and the area slope between 0 and 2 sec. is greater than between under the curves on a velocity-time graph are the same. If the first runner reaches maximum speed

4 Full file at http://TestbankCollege.eu/Solution-Manual-Physics-of-Everyday-Phenomena-6th-Edition- Griffith quicker, the only way the areas can be equal is if b. 3 m the second runner reaches a higher maximum E17 9.09 s speed which he then maintains over a shorter portion of the interval. E18 a. Speed: 3 m/s, 6 m/s, 9 m/s, 12 m/s, 15 m/s b. Distance: 1.5 m, 6 m, 13.5 m, 24 m, 37.5 m Q31 20 v

) 15 s

second runner / m (

first runner d e e

0 0 2 4 6 Tim e (s) t

Q32 40

V ) 30 m (

c 20 n a t s i 10 D

Time 0 0 2 4 6 Tim e (s)

Answers to Synthesis Problems A SP1 a. 21s b. A v. 5 Time S 4 p e

e Answers to Exercises d 16 E1 57.5 MPH t (s) E2 3.6 km/hr c. E3 0.4 cm/day E4 135 mi v E5 200 s 5 E6 4.84 hr E7 4.320 km E8 a. 0.022 km/s b. 79.2 km/hr E9 93.3 km/hr t (s) E10 3.5 m/s2 E11 21 m/s E12 -3 m/s2 -4 E13 a. 17 m/s b. 29 m E14 a. 4.4 m/s d. b. 6.4 m a E15 a. 21 m/s b. 76.5 m E16 a. 3 m/s

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SP2 a. 1.0 m/s2 b. 2.0 m/s2 c. 1.5 m/s2 d. No. This acceleration should be calculated as a . . time average, aav = (a1 t1 + a2 t2)/(t1 + t2) not aav = (a1 + a2)/2. Note the acceleration here can also be calculated as aav = (v2 - v1)/(t2 - t1).

SP3 a.

30 25 )

s 20 / m ( 15 d e

e 10 p S 5 0 0 10 20 30 40 b. Tim e (s)

2 2 s / m (

0 n o i

t 0 10 20 30 40

a -2 r e l e

c -4 c A -6 Tim e (s)

c. Yes. The car never has a negative velocity (that is, it moves backwards), so its distance must increase.

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SP4 a. 5 s since t = (v2 - v1) / a b. 95 m c.

) 80 m (

e 60 c n

a 40 t s i 20 D 0 0 2 4 6 Tim e (s)

Note the parabolic shape of this curve is not obvious over the given time span, but it is indeed a parabola!

SP5 a. Time (s) Distance (m) Car A Car B

1 2.25 10 2 9.0 20 3 0.25 30 4 36 40 5 56.2 50

b. Car A passes car B at approximately 4.5 sec. c. To find a better time you could graph the distance versus time for each car and see where the two curves cross. To find the exact time when the distance is the same for both 2 cars, note that then dA = dB = ½ aAt = vBt. Thus tmeet = 0 s and (2 vB/aA ) = 5.0 s.