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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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