please assume the free-fall acceleration \(g=9.80 \mathrm{m} / \mathrm{s}^{2}\) unless a more precise value is given in the problem statement. Ignore air resistance. Superman is standing \(120 \mathrm{m}\) horizontally away from Lois Lane. A villain throws a rock vertically downward with a speed of $2.8 \mathrm{m} / \mathrm{s}\( from \)14.0 \mathrm{m}$ directly above Lois. (a) If Superman is to intervene and catch the rock just before it hits Lois, what should be his minimum constant acceleration? (b) How fast will Superman be traveling when he reaches Lois?

Short Answer

Expert verified
Answer: The minimum constant acceleration Superman needs is approximately 23.56 m/s², and his speed when he reaches Lois is approximately 41.96 m/s.

Step by step solution

01

Calculate the time it takes for the rock to fall

First, we need to find out how long it takes for the rock to fall from 14.0 m above Lois Lane. We can use this equation of motion: $$ y_{final} = y_{initial} + v_{initial} * t - 0.5 * g * t^{2} $$ Using the given information, we have: $$ 0 \mathrm{m} = 14.0 \mathrm{m} + 2.8 \mathrm{m / s} * t - 0.5 * 9.80 \mathrm{m / s^{2}} * t^{2} $$ Solving for t, we get: $$ t \approx 1.781 \mathrm{s} $$ It takes approximately 1.781 seconds for the rock to reach Lois Lane.
02

Calculate the minimum constant acceleration for Superman

Now, we need to find the minimum constant acceleration Superman needs to catch the rock before it reaches Lois. Using the equation of motion: $$ x_{final} = x_{initial} + v_{initial} * t + 0.5 * a * t^{2} $$ According to the problem statement, Superman is standing 120 m horizontally away from Lois. His initial position is 0 m, and his final position should be 120 m. Superman's initial velocity is 0 m/s since he starts from rest. We want to find a, the minimum constant acceleration, and we know the time t should be less than or equal to the time it takes for the rock to fall (1.781 s). Plugging in the values, we get: $$ 120 \mathrm{m} = 0 \mathrm{m} + 0 \mathrm{m / s} * t + 0.5 * a * (1.781 \mathrm{s})^{2} $$ Solving for a, we get: $$ a \approx 23.56 \mathrm{m / s^{2}} $$ Superman's minimum constant acceleration should be approximately 23.56 m/s².
03

Calculate Superman's speed when reaching Lois

Finally, we need to find Superman's speed when he reaches Lois. We can use the equation of motion: $$ v_{final} = v_{initial} + a * t $$ Superman's initial velocity is 0 m/s, and we found his constant acceleration to be approximately 23.56 m/s². Using the time it takes for the rock to fall (1.781 s), we can find his final speed: $$ v_{final} = 0 \mathrm{m / s} + 23.56 \mathrm{m / s^{2}} * 1.781 \mathrm{s} $$ Solving for \(v_{final}\), we get: $$ v_{final} \approx 41.96 \mathrm{m / s} $$ Superman's speed when he reaches Lois is approximately 41.96 m/s.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Find the point of no return for an airport runway of \(1.50 \mathrm{mi}\) in length if a jet plane can accelerate at \(10.0 \mathrm{ft} / \mathrm{s}^{2}\) and decelerate at \(7.00 \mathrm{ft} / \mathrm{s}^{2} .\) The point of no return occurs when the pilot can no longer abort the takeoff without running out of runway. What length of time is available from the start of the motion in which to decide on a course of action?
A runner, jogging along a straight line path, starts at a position $60 \mathrm{m}$ east of a milestone marker and heads west. After a short time interval he is \(20 \mathrm{m}\) west of the mile marker. Choose east to be the positive \(x\) -direction. (a) What is the runner's displacement from his starting point? (b) What is his displacement from the milestone? (c) The runner then turns around and heads east. If at a later time the runner is \(140 \mathrm{m}\) east of the milestone, what is his displacement from the starting point at this time? (d) What is the total distance traveled from the starting point if the runner stops at the final position listed in part (c)?
In the problems, please assume the free-fall acceleration $g=9.80 \mathrm{m} / \mathrm{s}^{2}$ unless a more precise value is given in the problem statement. Ignore air resistance. A car traveling at \(29 \mathrm{m} / \mathrm{s}(65 \mathrm{mi} / \mathrm{h})\) runs into a bridge abutment after the driver falls asleep at the wheel. (a) If the driver is wearing a seat belt and comes to rest within a 1.0 -m distance, what is his acceleration (assumed constant)? (b) A passenger who isn't wearing a seat belt is thrown into the windshield and comes to a stop in a distance of \(10.0 \mathrm{cm} .\) What is the acceleration of the passenger?
please assume the free-fall acceleration \(g=9.80 \mathrm{m} / \mathrm{s}^{2}\) unless a more precise value is given in the problem statement. Ignore air resistance. A brick is thrown vertically upward with an initial speed of $3.00 \mathrm{m} / \mathrm{s}\( from the roof of a building. If the building is \)78.4 \mathrm{m}$ tall, how much time passes before the brick lands on the ground?
A rocket engine can accelerate a rocket launched from rest vertically up with an acceleration of \(20.0 \mathrm{m} / \mathrm{s}^{2}\) However, after 50.0 s of flight the engine fails. (a) What is the rocket's altitude when the engine fails? (b) When does it reach its maximum height? (c) What is the maximum height reached? [Hint: A graphical solution may be easiest.] \((\mathrm{d})\) What is the velocity of the rocket just before it hits the ground?
See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free