multiple choice\nidentify the choice that best completes the statement or answers the question.\n1. a…

multiple choice\nidentify the choice that best completes the statement or answers the question.\n1. 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?\n a. 61.5 m/s b. 63.7 m/s c. 65.0 m/s d. 70.0 m/s\n2. 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?\n a. 20 m/s² b. 13 400 m/s² c. 6 720 m/s² d. 857 m/s²\n3. 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?\n a. 66 m b. 80 m c. 10 m d. 55 m\n4. 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)\n a. 31 m/s b. 20 m/s c. 39 m/s d. 15 m/s\n5. as an object falls freely in a vacuum, its\n a. velocity b. acceleration c. both a and b d. none of the increases. increases.\nabove.\n6. refer to the following position - time graph to answer the next 4 questions\nat what time is the object farthest from its starting point\n a. 3s b. 4s c. 2s d. 6s\n7. at what time is the velocity of the object zero?\n a. 2s b. 0s c. 3s d. 5s\n8. at what time is the object moving with the greatest velocity?\n a. 0.5s b. 3.5s c. 1.5s d. 2.5s\n9. when is the object moving in a negative direction?\n a. 0s - 2s b. 0s - 3s c. 2s - 3s d. 3s - 6s

multiple choice\nidentify the choice that best completes the statement or answers the question.\n1. 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?\n a. 61.5 m/s b. 63.7 m/s c. 65.0 m/s d. 70.0 m/s\n2. 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?\n a. 20 m/s² b. 13 400 m/s² c. 6 720 m/s² d. 857 m/s²\n3. 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?\n a. 66 m b. 80 m c. 10 m d. 55 m\n4. 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)\n a. 31 m/s b. 20 m/s c. 39 m/s d. 15 m/s\n5. as an object falls freely in a vacuum, its\n a. velocity b. acceleration c. both a and b d. none of the increases. increases.\nabove.\n6. refer to the following position - time graph to answer the next 4 questions\nat what time is the object farthest from its starting point\n a. 3s b. 4s c. 2s d. 6s\n7. at what time is the velocity of the object zero?\n a. 2s b. 0s c. 3s d. 5s\n8. at what time is the object moving with the greatest velocity?\n a. 0.5s b. 3.5s c. 1.5s d. 2.5s\n9. when is the object moving in a negative direction?\n a. 0s - 2s b. 0s - 3s c. 2s - 3s d. 3s - 6s

Answer

1.

Explanation:

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)

Answer:

a. (61.5\ m/s)

2.

Explanation:

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})

Answer:

b. (13400\ m/s^{2})

3.

Explanation:

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. (66\ m)

4.

Explanation:

Step1: 利用自由落體速度公式

自由落體速度公式 (v=gt)，(g = 9.8\ m/s^{2})，(t = 1.5\ s)

Step2: 計算速度

(v=9.8\times1.5 = 14.7\approx15\ m/s)

Answer:

d. (15\ m/s)

5.

Brief Explanations:

自由落體（真空）中，加速度 (g) 為定值（重力加速度），速度 (v = gt) 隨時間增加。

Answer:

c. both A and B

6.

Brief Explanations:

位置 - 時間圖中，縱軸表示位置，(t = 3s) 時位置離起始點（(x = 0)）最遠。

Answer:

a. (3s)

7.

Brief Explanations:

位置 - 時間圖斜率表示速度，(t = 3s) 時斜率為 (0)（圖像頂點處），速度為 (0)。

Answer:

c. (3s)

8.

Brief Explanations:

位置 - 時間圖斜率絕對值表示速度大小，(t = 1.5s) 附近斜率絕對值最大。

Answer:

c. (1.5s)

9.

Brief Explanations:

位置 - 時間圖斜率為負表示負方向運動，(t = 3s - 6s) 斜率為負。

Answer:

d. (3s - 6s)