hw - motion in 1d 2 answers · title: hw - motion in 1d 2 answers author: katharine macdonald...
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Honors Physics Name _____________________________ Motion in One Dimension HW #2 Date _______________________________ Complete the following problems on a separate sheet of paper. Significant figures are to be used.
1. A pronghorn antelope has been observed to run with a top speed of 97 km/h. Suppose an antelope runs
1.5 km with an average speed of 85 km/h, and then runs 0.80 km with an average speed of 67 km/h.
a. How long will it take the antelope to run the entire 2.5 km?
Two different time frames.
First part of run Second part of run
d = 1.5 km
s = 85 km/h
𝑠 =𝑑𝑡; 𝑡 =
𝑑𝑠=
1.5 𝑘𝑚85 𝑘𝑚/ℎ
= 0.017647 ℎ𝑜𝑢𝑟𝑠
d = 0.80 km
s = 67 km/h
𝑠 =𝑑𝑡; 𝑡 =
𝑑𝑠=0.80 𝑘𝑚67 𝑘𝑚/ℎ
= 0.011940 ℎ𝑜𝑢𝑟𝑠
𝑡!"!#$ = 𝑡! + 𝑡! = 0.017647 ℎ𝑜𝑢𝑟𝑠 + 0.011940 ℎ𝑜𝑢𝑟𝑠 = 0.029587 ℎ𝑜𝑢𝑟𝑠 = 0.030 ℎ𝑜𝑢𝑟𝑠
b. What is the antelope’s average speed during this time?
𝑠 =𝑑𝑡=1.5 𝑘𝑚 + 0.80 𝑘𝑚0.030 ℎ𝑜𝑢𝑟𝑠
= 0.77𝑘𝑚ℎ𝑟
2. An ostrich can run at speeds up to 72 km/h. How long will it take an ostrich to run 1.5 km at this top
speed?
𝑠 =𝑑𝑡, 𝑡 =
𝑑𝑠=
1.5 𝑘𝑚72 𝑘𝑚/ℎ
= 0.020 ℎ𝑜𝑢𝑟𝑠 = 1.2 𝑚𝑖𝑛𝑢𝑡𝑒𝑠
3. Draw a motion diagram showing a jogger moving to the right at a constant speed. Provide the signs for
speed, velocity, and acceleration.
4. Draw a motion diagram showing a jogger moving to the left and slowing down at a constant rate. Provide
the signs for speed, velocity, and acceleration.
d t
t v
d t
t v
2
5. The table below shows the velocity of a student walking down the hallway between classes.
a) What is happening to the student’s speed during t = 60.0 s and t = 61.0 s?
Speed is decreasing. Accelerating in the negative direction.
b) What is his acceleration between t = 10.0 s and t = 20.0 s?
𝑎 = ∆𝑣∆𝑡
=1.5𝑚𝑠 − 1.5
𝑚𝑠
20.0 𝑠 − 10.0 𝑠= 0
𝑚𝑠!
c) What is his acceleration between t = 60.0 s and t = 61.0 s?
𝑎 = ∆𝑣∆𝑡
=0𝑚𝑠 − 3.0
𝑚𝑠
61.0 𝑠 − 60.0 𝑠= −3.0
𝑚𝑠!
d) Assuming constant acceleration, how far did he walk during the first 5.0 s?
∆𝑥 =12𝑣! + 𝑣! ∆𝑡 =
120.75
𝑚𝑠+ 0 5.0 𝑠 = 1.9 𝑚
Time (s) Velocity (m/s)
0.0 0.0
10.0 1.5
20.0 1.5
30.0 1.5
31.0 0.0
40.0 0.0
50.0 3.0
60.0 3.0
61.0 0.0
6. Use the velocity-time graph below to calculate the velocity of the object whose motion is plotted on the
graph.
a) What is the acceleration between the points on the graph labeled A and B?
𝑎 = ∆𝑣∆𝑡
=300.0𝑚𝑠 − 0
𝑚𝑠
20.0 𝑠 − 0.0 𝑠= 15
𝑚𝑠!
b) What is the acceleration between the points on the graph labeled B and C?
𝑎 = ∆𝑣∆𝑡
=300.0𝑚𝑠 − 300.0
𝑚𝑠
30.0 𝑠 − 20.0 𝑠= 0
𝑚𝑠!
c) What is the acceleration between the points on the graph labeled D and E?
𝑎 = ∆𝑣∆𝑡
=500.0𝑚𝑠 − 0
𝑚𝑠
80.0 𝑠 − 40.0 𝑠= 12.5
𝑚𝑠!
3
d) What is the total distance that the object travels between points B and C?
∆𝑥 =12𝑣! + 𝑣! ∆𝑡 =
12300.0
𝑚𝑠+ 300.0
𝑚𝑠
30.0 𝑠 − 20.0 𝑠 = 3.00 𝑥 10! 𝑚
7. A car is traveling at 20.0 m/s when the driver sees a ball roll into the street. From the time the driver
applies the brakes, it takes 2.0 s for the car to come to a stop.
a) What is the average acceleration of the car during that period?
𝑎 = ∆𝑣∆𝑡
=0𝑚𝑠 − 20
𝑚𝑠
2.0 𝑠 − 0.0 𝑠= −1.0 𝑥 10!
𝑚𝑠!
b) How far does the car travel while the brakes are being applied?
∆𝑥 =12𝑎 ∆𝑡! + 𝑣!∆𝑡 =
12(−1.0 𝑥 10!
𝑚𝑠!)(2.0𝑠)! + 20.0
𝑚𝑠
2.0𝑠 = 2.0 𝑥 10!𝑚
8. A sudden gust of wind increases the velocity of a sailboat relative to the water surface from
3.0 m/s to 5.5 m/s over a period of 30.0 s.
a) What is the average acceleration of the sailboat?
𝑎 = ∆𝑣∆𝑡
=5.5𝑚𝑠 − 3.0
𝑚𝑠
30.0 𝑠 − 0.0 𝑠= 0.083
𝑚𝑠!
b) How far does the sailboat travel during the period of acceleration?
∆𝑥 =12𝑎 ∆𝑡! + 𝑣!∆𝑡 =
12(0.083
𝑚𝑠!)(30.0𝑠)! + 3.0
𝑚𝑠
30.0𝑠 = 130 𝑚
9. During a serve, a tennis ball leaves a racket at 180 km/h after being accelerated for 0.80 s.
a) What is the average acceleration on the ball during the serve in m/s2?
180 𝑘𝑚ℎ𝑟
𝑥 1000 𝑚1 𝑘𝑚
𝑥 1 ℎ𝑟3600 𝑠
= 5.0 𝑥 10!𝑚𝑠
4
𝑎 = ∆𝑣∆𝑡
=5.0 𝑥 10!𝑚𝑠 − 0
𝑚𝑠
0.80 𝑠 − 0.0 𝑠= 62
𝑚𝑠!
b) How far does the ball move during the period of acceleration?
∆𝑥 =12𝑎 ∆𝑡! + 𝑣!∆𝑡 =
12(62
𝑚𝑠!)(0.80𝑠)! + 0
𝑚𝑠
0.80𝑠 = 2.0 𝑥 10! 𝑚
10. The world’s fastest warship belongs to the United States Navy. This vessel, which floats on a cushion of
air, can move as fast as 1.7 x 102 km/h. Suppose that during a training exercise the ship accelerates
+2.67 m/s2, so that after15.0 s its displacement is +6.00 x 102m. Calculate the ship’s initial velocity just
before the acceleration. Assume that the ship moves in a straight line.
𝑎 = ∆𝑣∆𝑡; ∆𝑣 = 𝑎 ∆𝑡; 𝑣! − 𝑣! = 𝑎∆𝑡; 𝑣! = 𝑣! − 𝑎∆𝑡 = 47.2
𝑚𝑠
− 2.67𝑚𝑠! 15.0𝑠 = 7.2
𝑚𝑠
11. A German stuntman named Martin Blume performed a student called “the wall of death.” To perform it,
Blume rode his motorcycle for seven straight hours on the wall of a large vertical cylinder. His average
speed was 45.0 km/h. Suppose that in a time interval of 30.0s Blume increases his speed steadily from
30.0 km/h to 42.0 km/h while circling inside the cylindrical wall. How far does Blume travel in that time
interval?
30.0𝑠 ∗1 ℎ𝑟3600 𝑠
= 0.0083333 ℎ𝑟
∆𝑥 =12𝑣! + 𝑣! ∆𝑡 =
1242.0
𝑘𝑚ℎ𝑟
+ 30.0𝑘𝑚ℎ𝑟
0.0083333 𝑠 = 0.300 𝑚
12. An automobile that set the world record for acceleration increased speed from rest to 96 km/h in 3.07 s.
How far had the car traveled by the time the final speed was achieved?
∆𝑥 =12𝑣! + 𝑣! ∆𝑡 =
1296𝑘𝑚ℎ+ 0
3.07𝑠3600𝑠
= 0.041 𝑚
13. In 1993, bicyclist Rebecca Twigg of the United States traveled 3.00 km in 217.347s. Suppose Twigg
travels the entire distance at her average velocity and that she then accelerates at -1.72 m/s2 to come to
a complete stop after crossing the finish line. How long does it take Twigg to come to a stop?
𝑣 =∆𝑥∆𝑡
=3.00 𝑥 10! 𝑚217.347 𝑠
= 13.8028𝑚𝑠
𝑎 = ∆𝑣∆𝑡; ∆𝑡 =
∆𝑣𝑎= 0𝑚𝑠 − 13.8028
𝑚𝑠
−1.72 𝑚𝑠!= 8.02 𝑠𝑒𝑐𝑜𝑛𝑑𝑠
14. The first supersonic flight was performed by then Captain Charles Yeager in 1947. He flew at a speed of
3.00 x 102 m/s at an altitude of more than 12 km, where the speed of sound in air is slightly less than
3.00 x 102 m/s. Suppose Captain Yeager accelerated 7.20 m/s2 in 25.0 seconds to reach a final speed of
3.00 x102 m/s. What was his initial speed?
𝑣! = 𝑎∆𝑡 + 𝑣!; 𝑣! = 𝑣! − 𝑎∆𝑡 = 3.00 𝑥10!𝑚𝑠
− 7.20𝑚𝑠!
25.0 𝑠 = 1.20 𝑥 10!𝑚𝑠