Close to the moon the acceleration due to gravity is 1.6 m s-2. The first time that a meteor is detected it is falling directly toward the moon's surface at 93 m s-1. The next time it is detected it is falling at 151 m s-1. How much time had passed between the two times that the meteor was detected ? Give your answer to one decimal place.

Question

Close to the moon the acceleration due to gravity is 1.6 m s-2. The first time that a meteor is detected it is falling directly toward the moon's surface at 93 m s-1. The next time it is detected it is falling at 151 m s-1. How much time had passed between the two times that the meteor was detected ? Give your answer to one decimal place.

BlackTom AI · Accepted Answer

Using v = u + at for constant acceleration, we can relate the two measured speeds and the elapsed time.

1) Relevant knowledge points and concepts
- For motion with constant acceleration a, the velocity v after time t from initial velocity u is v = u + at.
- Here acceleration due to gravity near the Moon is constant at a = 1.6 m s^-2, so the formula applies directly.
- Units must be consistent (m s^-1 for velocity, s for time, m s^-2 for acceleration).

2) Step-by-step reasoning
- Let the first detected speed be u = 93 m s^-1 and the later speed be v = 151 m s^-1.
- The elapsed time Δt between the two detections satisfies v = u + a Δt.
- Solve for Δt: Δt = (v - u) / a.

3) Calculations with intermediate steps
- Compute the difference in speed: v - u = 151 − 93 = 58 m s^-1.
- Divide by acceleration: Δt = 58 / 1.6 s.
- 58 ÷ 1.6 = 36.25 s.

4) Why the correct answer is correct
- Because acceleration is constant (given) the linear relation between change in velocity and time holds exactly. The computed time 36.25 s is the exact elapsed time; rounding to one decimal place gives 36.3 s.

5) Why other approaches would be incorrect
- Using s = ut + 0.5at^2 or average speed without knowing displacement is unnecessary and could lead to errors.
- Using Earth's gravity (9.8 m s^-2) instead of the Moon's 1.6 m s^-2 would give a much smaller time and be wrong.
- Forgetting to subtract the velocities in the correct order or mishandling units would also produce incorrect values.

Conclusion: The elapsed time between detections is 36.3 s, which matches the provided answer.

类似问题

The velocity of a car is given by [math: dxdt=3t2−6t+2]. The value of the displacement, x metres, in the first 5 seconds is:

A ball is thrown downwards at 8.3 m s-1 from the top of a building. After 1.2 seconds accelerating at 9.81 m s-2, what is its speed? Give your answer to two decimal places.

A rocket reaches a maximum height of 866 m in the 14.2 s after it is launched. What acceleration was the rocket engine able to achieve? Give your answer to one decimal place.

A car moved 830 m in 39.6 s. Using the formula [math: speed=distance movedtime taken] calculate the speed of the car to the appropriate number of significant figures.

A car moved 654 m in 24.1 s. Using the formula speed=distance movedtime taken calculate the speed of the car to the appropriate number of significant figures.

Find the displacement ,[math: x], of an object which has an initial velocity ,[math: u], of 7.2 ms-1 and moves with a uniform acceleration ,[math: a], of 7.2 ms-2 to reach a final velocity ,[math: v], of 34.2 ms-1.

更多留学生实用工具

智能学习助手

风格化写作助手

论文查重助手

文献引用助手

课堂转译助手

课堂笔记助手

Quiz搜索助手

学校历年真题

智能刷题助手

智能匹配练习

希望你的学习变得更简单