Question

multiple choice
identify the choice that best completes the statement or answers the question.

1. a railroad train travels forward along a straight track at 80.0 m/s for 1 000 m and then travels at 50.0 m/s for the next 1 000 m. what is the average velocity?

a. 61.5 m/s b. 63.7 m/s c. 65.0 m/s d. 70.0 m/s

1. a 50 - g ball traveling at 25.0 m/s is bounced off a brick wall and rebounds at 22.0 m/s. a high - speed camera records this event. if the ball is in contact with the wall for 3.50 ms, what is the average acceleration of the ball during this time interval?

a. 20 m/s² b. 13 400 m/s² c. 6 720 m/s² d. 857 m/s²

1. a cart is given an initial velocity of 5.0 m/s and experiences a constant acceleration of 2.0 m/s². what is the magnitude of the carts displacement during the first 6.0 s of its motion?

a. 66 m b. 80 m c. 10 m d. 55 m

1. a rock, released at rest from the top of a tower, hits the ground after 1.5 s. what is the speed of the rock as it hits the ground? (g = 9.8 m/s² and air resistance is negligible)

a. 31 m/s b. 20 m/s c. 39 m/s d. 15 m/s

1. as an object falls freely in a vacuum, its

a. velocity b. acceleration c. both a and b d. none of the increases. increases.
above.

1. refer to the following position - time graph to answer the next 4 questions

at what time is the object farthest from its starting point
a. 3s b. 4s c. 2s d. 6s

1. at what time is the velocity of the object zero?

a. 2s b. 0s c. 3s d. 5s

1. at what time is the object moving with the greatest velocity?

a. 0.5s b. 3.5s c. 1.5s d. 2.5s

1. when is the object moving in a negative direction?

a. 0s - 2s b. 0s - 3s c. 2s - 3s d. 3s - 6s

Explanation:

1.

Step1: 計算總位移

總位移 \(s = s_1 + s_2=1000 + 1000 = 2000\ m\)

Step2: 計算總時間

\(t_1=\frac{s_1}{v_1}=\frac{1000}{80}=12.5\ s\)，\(t_2=\frac{s_2}{v_2}=\frac{1000}{50}=20\ s\)，總時間 \(t=t_1 + t_2=12.5+20 = 32.5\ s\)

Step3: 計算平均速度

平均速度 \(v=\frac{s}{t}=\frac{2000}{32.5}\approx61.5\ m/s\)

Step1: 確定速度變化量

\(\Delta v=v - u=- 22-25=-47\ m/s\)（設初速度方向為正），時間 \(t = 3.5\times10^{-3}\ s\)

Step2: 計算加速度

加速度 \(a=\frac{\Delta v}{t}=\frac{-47}{3.5\times10^{-3}}\approx - 13400\ m/s^{2}\)，加速度大小為 \(13400\ m/s^{2}\)

Step1: 代入位移公式

根據位移公式 \(s=ut+\frac{1}{2}at^{2}\)，\(u = 5\ m/s\)，\(a = 2\ m/s^{2}\)，\(t = 6\ s\)

Step2: 計算位移

\(s=5\times6+\frac{1}{2}\times2\times6^{2}=30 + 36=66\ m\)

Answer:

a. \(61.5\ m/s\)

2.