fan (rf22492) 1 - 1D Motion carroll (1065) This print-out should have 27 questions.

ns. Multiple-choice questions may continue on the next column or page nd all choices before answering. 001 (part 1 of 2) 10.0 points Consider a moving object whose position x is plotted as a function of the time t. The object moved in dierent ways during the time intervals denoted I, II and III on the gure. 6 4 2 I 2 II 4 III 6 t x 2. During interval III only 3. During interval II only 4. During interval I only 5. During none of the three intervals

Explanation: For each of the three intervals I, II or III, the x(t) curve is linear, so its slope (the velocity v) is constant. Between the intervals, the velocity changed in an abrupt manner, but it did remain constant during each interval. 003 (part 1 of 2) 10.0 points Consider the following graph Displacement vs Time 10 8 displacement (m) 6 4 2 0 2 4 6

During these three intervals, when was the objects speed highest? Do not confuse the speed with the velocity. 1. During interval II 2. Same speed during each of the three intervals. 3. During interval III correct 4. During interval I 5. Same speed during intervals II and III Explanation: The velocity v is the slope of the x(t) curve; the magnitude v = |v| of this slope is the speed. The curve is steepest (in absolute magnitude) during the interval III and that is when the object had the highest speed. 002 (part 2 of 2) 10.0 points During which interval(s) did the objects velocity remain constant? 1. During each of the three intervals correct

8 10 12 14 16 18 20 time (s) What is the displacement at 10 s? 1. 4 m correct 2. 1 m 3. 1 m 4. 3 m 5. 0 m 6. 3 m 7. None of these

8 10

0 2

4 6

fan (rf22492) 1 - 1D Motion carroll (1065) 8. 2 m 9. 2 m 10. Unable to determine Explanation: Read the displacement from the graph. 004 (part 2 of 2) 10.0 points What is the velocity at 10 s? 1. 2 m/s 2. None of these 3. 0 m/s correct 4. 1 m/s 5. Unable to determine 6. 4 m/s 7. 1 m/s 8. 2 m/s 9. 3 m/s 10. 3 m/s Explanation: At 10 s, the slope is 0. 005 10.0 points One swimmer in a relay race has a 0.47 s lead and is swimming at a constant speed of 3.96 m/s. The swimmer has 58.2 m to swim before reaching the end of the pool. A second swimmer moves in the same direction as the leader. What constant speed must the second swimmer have in order to catch up to the leader at the end of the pool? Correct answer: 4.09082 m/s. Explanation: Correct answer: 14.3333 km. Let : t = 0.47 s , Explanation: v2 = 58.2 m x 1 = t 1 t 14.697 s 0.47 s x1 = 58.2 m , and v1 = 3.96 m/s .

The lead swimmer will reach the end of the pool in t 1 = x 1 58.2 m = = 14.697 s . v1 3.96 m/s

The second swimmers distance is given by x2 = v2 t + x1 and the velocity is v2 t + x1 x 2 = t 1 t 1 v2 t1 = v2 t + x1 v2 (t1 t) = x1 v2 =

= 4.09082 m/s .

006 (part 1 of 2) 10.0 points A runner is jogging at a steady 7.2 km/hr. When the runner is 8.6 km from the nish line, a bird begins ying from the runner to the nish line at 36 km/hr (5 times as fast as the runner). When the bird reaches the nish line, it turns around and ies back to the runner.
vb vr finish line L

How far does the bird travel? Even though the bird is a dodo, assume that it occupies only one point in space (a zero length bird) and that it can turn without loss of speed.

fan (rf22492) 1 - 1D Motion carroll (1065)

Let :

vr = 7.2 km/hr , L = 8.6 km , and vb = 5 vr .

bird will y 5 times as far as the runner until the next encounter. This pattern repeats over the entire original distance, so db = 5 L = 5 (8.6 km) = 43 km . 008 10.0 points A car makes a 219 km trip at an average speed of 35.4 km/h. A second car starting 1 h later arrives at their mutual destination at the same time. What was the average speed of the second car for the period that it was in motion? Correct answer: 42.2255 km/h. Explanation: Let : d = 219 km , v = 35.4 km/h , t = 1 h . d . t

finish line dr1 db1 L1

The runner travels a distance x until the encounter with the bird. In that time, the bird has traveled a distance L + (L x) = 2 L x . The bird travel 5 times as fast as the runner during this time frame, so db = 5 dr 2L x = 5x 2L = 6x 1 x= L 3 and the bird ies a distance of 1 5 5 db = 2 L L = L = (8.6 km) 3 3 3 = 14.3333 km . 007 (part 2 of 2) 10.0 points After this rst encounter, the bird then turns around and ies from the runner back to the nish line, turns around again and ies back to the runner. The bird repeats the back and forth trips until the runner reaches the nish line. How far does the bird travel from the beginning (including the distance traveled to the rst encounter)? Correct answer: 43 km. Explanation: The distance remaining for the runner after 2 the rst encounter is L1 = L , and again the 3

and

Average velocity is v =

t = t 1 t 2 =

d d v1 v2

d d = t v2 v1 v2 1 v1 = d d v t 1 v1 d v1 v2 = d v1 t (219 km)(35.4 km/h) = 219 km (35.4 km/h)(1 h) = 42.2255 km/h .

009 10.0 points A car travels along a straight stretch of road. It proceeds for 12.9 mi at 56 mi/h, then 27.4 mi at 47 mi/h, and nally 32.1 mi at 32.4 mi/h. What is the cars average velocity during the entire trip?

fan (rf22492) 1 - 1D Motion carroll (1065) Correct answer: 40.1313 mi/h. Explanation: Let : dA vA dB vB dC vC = 12.9 mi , = 56 mi/h , = 27.4 mi , = 47 mi/h , = 32.1 mi , and = 32.4 mi/h . v= 4 m (2 m) = 3 m/ s . 2s0s

The instantaneous velocity is the slope of the tangent line at that point.

011 (part 2 of 4) 10.0 points Find the instantaneous velocity at 2.5 s. Correct answer: 2 m/s. Explanation: 2 m (4 m) = 2 m/s . 3s2 s

The total time the car spent on the road is dA dB dC t = + + vA vB vC 12.9 mi 27.4 mi 32.1 mi = + + 56 mi/h 47 mi/h 32.4 mi/h = 1.80408 h , so the average velocity is v= dA + dB + dC d = t t 12.9 mi + 27.4 mi + 32.1 mi = 1.80408 h = 40.1313 mi/h .

v=

012 (part 3 of 4) 10.0 points Find the instantaneous velocity at 4.5 s. Correct answer: 0 m/s. Explanation: 2 m (2 m) = 0 m/ s . 6s3s

010 (part 1 of 4) 10.0 points The position versus time for a certain object moving along the x-axis is shown. The objects initial position is 2 m. 6 position (m) 4 2 0 2 4 0 1 2 3 4 5 time (s) 6 7 8 9

v=

013 (part 4 of 4) 10.0 points Find the instantaneous velocity at 8 s. Correct answer: 1 m/s. Explanation: 0 m (2 m) = 1 m/ s . 9s7s

v=

Find the instantaneous velocity at 1 s. Correct answer: 3 m/s. Explanation: 014 10.0 points Consider the acceleration of an object starting from rest.

fan (rf22492) 1 - 1D Motion carroll (1065) 5 4 Acceleration (m/s2 ) 3 2 1 0 1 2 3 4 5

Let :

vi = 0 vf = 76.8 mi/h , t = 7.62 s .

and

vf = vi + a t vf vf vi = a= t t 76.8 mi/h = 4.5056 m/s2 . = 7.62 s 0 1 016 (part 2 of 2) 10.0 points Find the distance that the car travels during this time. Correct answer: 130.808 m. Explanation: x = vi t + = 1 2 1 2 at = at 2 2

2 3 4 5 Time (s) Other than at t = 0, when is the velocity of the object zero? 1. 5.0 s 2. At no other time on this graph correct 3. 3.5 s 4. 4.0 s 5. During the interval from 1.0 s to 3.0 s Explanation:
t

1 (4.5056 m/s2 ) (7.62 s) 2 = 130.808 m .

vt =
0

a dt is the area between the accel-

eration curve and the t axis during the time period from 0 to t. If the area is above the horizontal axis, it is positive; otherwise, it is negative. In order for the velocity to be zero at any given time t, there would have to be equal amounts of positive and negative area between 0 and t. According to the graph, this condition is never satised. 015 (part 1 of 2) 10.0 points A car accelerates uniformly from rest to a speed of 76.8 mi/h in 7.62 s. Find the constant acceleration of the car. Correct answer: 4.5056 m/s2 . Explanation:

017 10.0 points An electron, starting from rest and moving with a constant acceleration, travels 3.6 cm in 10 ms. What is the magnitude of this acceleration? Correct answer: 0.72 km/s2 . Explanation: Let : d = 3.6 cm = 3.6 105 m t = 10 ms = 0.01 s . and

Starting from rest with constant acceleration, the distance is 1 d = a t2 2 2 (3.6 105 m) 1 km 2d a= 2 = t (0.01 s)2 1000 m = 0.72 km/s2 .

fan (rf22492) 1 - 1D Motion carroll (1065) The distance is

018 (part 1 of 2) 10.0 points A plane cruising at 233 m/s accelerates at 1 15 m/s2 for 9.1 s. x = x0 + v0 t + a t2 2 What is its nal velocity? 2 (x x0 v0 t) a= t2 Correct answer: 369.5 m/s. 2 [7.78 cm 4.01 cm (16.5 cm/s) (2.62 s)] Explanation: = (2.62 s)2 Let : v = 233 m/s , a = 15 m/s2 , t = 9. 1 s . and = 16.0305 cm/s2 .

v = v0 + a t = 233 m/s + (15 m/s2 ) (9.1 s) = 369.5 m/s . 019 (part 2 of 2) 10.0 points How far will it have traveled in that time? Correct answer: 2741.38 m. Explanation: sf = so + v0 t + = vt+ 1 2 at 2 1 2 at 2 1 (15 m/s2 ) (9.1 s)2 2

021 (part 1 of 4) 10.0 points Two students are on a balcony 18.6 m above the street. One student throws a ball vertically downward at 15.3 m/s. At the same instant, the other student throws a ball vertically upward at the same speed. The second ball just misses the balcony on the way down. What is the magnitude of the velocity of the rst ball as it strikes the ground? Correct answer: 24.4749 m/s. Explanation: Assume that the position of the balcony is yi = 0 m; then the position of the ground is yf = 18.6 m: Let : yf 1 = yf 2 = 18.6 m , vi1 = 15.3 m/s , vi2 = +15.3 m/s and a = 9.81 m/s2 .

= (233 m/s) (9.1 s) + = 2741.38 m .

020 10.0 points A body moving with uniform acceleration has a velocity of 16.5 cm/s when its x coordinate is 4.01 cm. If its x coordinate 2.62 s later is 7.78 cm, what is the x-component of its acceleration? Correct answer: 16.0305 cm/s2 . Explanation: Let : v0 = 16.5 cm/s , x = 7.78 cm , x0 = 4.01 cm , and t = 2.62 s .

For the rst ball,

= (15.3 m/s)2 + 2 9.81 m/s2 (18.6 m) = 599.022 m2 /s2 vf,1 = 599.022 m2 /s2 = 24.4749 m/s .

fan (rf22492) 1 - 1D Motion carroll (1065) Since its direction is downward, the velocity is 24.4749 m/s and its speed is v = 24.4749 m/s . 022 (part 2 of 4) 10.0 points What is the magnitude of the velocity of the second ball as it strikes the ground? Correct answer: 24.4749 m/s. Explanation: For the second ball,
2 2 vf, 2 = vi2 + 2 a y

The dierence in the times is t = t 2 t 1 = 4.05453 s (0.935263 s) = 24.4749 m/s .

024 (part 4 of 4) 10.0 points How far apart are the balls 0.525 s after they are thrown? Correct answer: 16.065 m. Explanation: For each of the balls, Let : For the rst ball, y1 = vi,1 t + 1 a (t)2 2 = (15.3 m/s) (0.525 s) 1 9.81 m/s2 (0.525 s)2 + 2 = 9.38444 m , t = 0.525 s .

= (15.3 m/s)2 + 2 9.81 m/s2 (18.6 m) = 599.022 m2 /s2 vf,2 = 599.022 m2 /s2 = 24.4749 m/s . Since its direction is downward, the velocity is 24.4749 m/s and its speed is v = 24.4749 m/s . 023 (part 3 of 4) 10.0 points What is the dierence in the time the balls spend in the air? Correct answer: 3.11927 s. Explanation: For the rst ball, vf 1 vi1 v1 = t 1 t 1 vf 1 vi1 t 1 = a (24.4749 m/s) (15.3 m/s) = 9.81 m/s2 = 0.935263 s . a= For the second ball, vf 2 vi2 v2 = t 2 t 2 vf 2 vi2 t 2 = a (24.4749 m/s) (15.3 m/s) = (9.81 m/s2 ) = 4.05453 s . a=

which is 9.38444 m below the balcony. For the second ball, 1 y2 = vi,2 t + a(t)2 2 = (15.3 m/s) (0.525 s) 1 + 9.81 m/s2 (0.525 s)2 2 = 6.68056 m . The distance between the balls is y2 y1 = 6.68056 m (9.38444 m) = 16.065 m .

025 10.0 points At a certain instant of time, a toy car is traveling at a constant speed of v0 to the right, while a distance D ahead is another toy car starting from rest and traveling to the left with constant acceleration a. The cars are

fan (rf22492) 1 - 1D Motion carroll (1065) moving along the same line, directly toward each other. At what time do they collide, in terms of D, v0 and a? 1. v0 a

026 10.0 points The acceleration due to gravity on planet X is one fth that on the surface of the earth. If it takes 4.3 s for an object to fall a certain distance from rest on earth, how long would it take to fall the same distance on planet X? Correct answer: 9.61509 s. Explanation:

2. The cars will never collide. 2D 3. a 4. 5. 6. 7. 8. 1 2 2 D v0 + a a v0 a

Let :

te = 4 . 3 s .

2 D v0 correct + a a
2

Because the acceleration due to gravity is uniform near the surface of both planets, the distance an object falls in a time t is given by xf all = gplanet t2 . 2

2D v0 a a v0 a
2

v0 a

2 D v0 a a

Since the distance xf all is the same for both planets, then g e t2 e 2 t2 e = g x t2 x 2 1 g e t2 x 5 = 2

v0 2 2 D v0 + + a a a Explanation: The rst toy car starts at a position x = 0 and moves to the right at a constant speed v0 , so its position is given by x1 = v0 t. Meanwhile, the second toy car starts at a distance D ahead of the rst and accelerates to the left with an acceleration a, so its position is given 1 by x2 = D a t2 . 2 The two cars collide when x1 = x2 v0 t = D 1 2 at 2 0 = a t2 + 2 v0 t 2 D t= = 2 v0 v0 2a
2 +2Da v0

1 2 t 5 x tx = 5 te = 5 (4.3 s) = 9.61509 s . 027 10.0 points Judy, by chance, spots a potted plant falling outside the window at 1.76 m/s . Further down the same building, Judys boyfriend clocks the pot at 68 m/s . How far apart are the friends? The acceleration of gravity is 9.8 m/s2 . Correct answer: 235.76 m. Explanation: All the motion is down, so we can consider down to be positive. Let : v0 = 1.76 m/s , vf = 68 m/s , and g = 9.8 m/s2 .

2 + 8Da 4 v0

a v0 v0 2 2 D = + + a a a t > 0 when the two cars collide, so the negative root must be rejected.

fan (rf22492) 1 - 1D Motion carroll (1065)

2 2 vf = v0 + 2gy

y=

2g (68 m/s)2 (1.76 m/s)2 = 2 (9.8 m/s2 ) = 235.76 m .

2 v2 vf 0