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Chapter 2: Problem 50

How much time does it take for a car to accelerate from a standing start to \(22.2 \mathrm{~m} / \mathrm{s}\) if the acceleration is constant and the car covers \(243 \mathrm{~m}\) during the acceleration?

Short Answer

Expert verified
Answer: It takes approximately 21.9 seconds for the car to accelerate from a standing start to 22.2 m/s.

Step by step solution

01

Write down the given information.

We have the following information: - Initial velocity (u) = 0 m/s - Final velocity (v) = 22.2 m/s - Distance covered (s) = 243 m We need to find the time taken (t) and the constant acceleration (a).
02

Use the first kinematic equation to find acceleration.

We have the equation: \(v = u + at\) Since the initial velocity (u) is 0 m/s, this equation simplifies to: \(v = at\) We can solve for acceleration (a) by dividing both sides by the time (t): \(a = \frac{v}{t}\)
03

Substitute the first equation into the second equation.

Now, we can substitute the expression for acceleration from the first equation into the second equation: \(s = ut + \frac{1}{2}at^2\) Since the initial velocity (u) is 0 m/s, the equation becomes: \(s = \frac{1}{2}at^2\) Substituting the expression for acceleration found in Step 2: \(s = \frac{1}{2}(\frac{v}{t})t^2\)
04

Solve for the time.

We now have a single equation with one unknown, t: \(s = \frac{1}{2}(\frac{v}{t})t^2\) To solve for t, first multiply both sides by 2: \(2s = vt\) Now, divide both sides by the final velocity (v): \(t = \frac{2s}{v}\)
05

Plug in the given values and calculate the time.

Now, we can plug in the given values for the distance (s) and final velocity (v): \(t = \frac{2(243\,\mathrm{m})}{22.2\,\mathrm{m/s}}\) Calculating the time: \(t \approx 21.9\,\mathrm{s}\) So, it takes approximately 21.9 seconds for the car to accelerate from a standing start to 22.2 m/s, covering a distance of 243 meters during the acceleration.

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Most popular questions from this chapter

In 2005, Hurricane Rita hit several states in the southern United States. In the panic to escape her wrath, thousands of people tried to flee Houston, Texas by car. One car full of college students traveling to Tyler, Texas, 199 miles north of Houston, moved at an average speed of \(3.0 \mathrm{~m} / \mathrm{s}\) for one-fourth of the time, then at \(4.5 \mathrm{~m} / \mathrm{s}\) for another one-fourth of the time, and at \(6.0 \mathrm{~m} / \mathrm{s}\) for the remainder of the trip. a) How long did it take the students to reach their destination? b) Sketch a graph of position versus time for the trip.

The position versus time for an object is given as \(x=A t^{4}-B t^{3}+C\) a) What is the instantaneous velocity as a function of time? b) What is the instantaneous acceleration as a function of time?

Bill Jones has a bad night in his bowling league. When he gets home, he drops his bowling ball in disgust out the window of his apartment, from a height of \(63.17 \mathrm{~m}\) above the ground. John Smith sees the bowling ball pass by his window when it is \(40.95 \mathrm{~m}\) above the ground. How much time passes from the time when John Smith sees the bowling ball pass his window to when it hits the ground?

You toss a small ball vertically up in the air. How are the velocity and acceleration vectors of the ball oriented with respect to one another during the ball's flight up and down?

A particle starts from rest at \(x=0\) and moves for \(20 \mathrm{~s}\) with an acceleration of \(+2.0 \mathrm{~cm} / \mathrm{s}^{2}\). For the next \(40 \mathrm{~s}\), the acceleration of the particle is \(-4.0 \mathrm{~cm} / \mathrm{s}^{2} .\) What is the position of the particle at the end of this motion?
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