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Chapter 2: Problem 38

A car accelerates uniformly from rest to a speed of \(40.0 \mathrm{mi} / \mathrm{h}\) in \(12.0 \mathrm{~s}\). Find (a) the distance the car travels during this time and (b) the constant acceleration of the car.

Short Answer

Expert verified
The car travels approximately 107.28 meters, and the constant acceleration of the car is approximately 1.49 m/s^2.

Step by step solution

01

- Convert speed to m/s

The speed is given in mi/hr but our calculations demand that it be in m/s. The conversion factor is 1 mi/hr = 0.44704 m/s. Hence, \( 40 \mathrm{mi/hr} \) converts to \( 40 \tx{0.44704 m/s} = 17.88 \mathrm{m/s} \).
02

- Find the distance using the equation of motion

For uniform acceleration from rest, the distance traveled is given by the equation \( x = ut + \frac{1}{2} a t^{2} \), where \( x \) is the distance, \( u \) is the initial speed, \( a \) is the acceleration, and \( t \) is the time. Here, the car starts from rest, so the initial speed \( u = 0 \). Therefore, the equation simplifies to \( x = \frac{1}{2} a t^{2} \). We know that \( t = 12 \mathrm{s} \) but the value of \( a \) is not known yet. So, we'll need to find \( a \) before we can calculate \( x \).
03

- Find the acceleration

Acceleration is given by the equation \( a = \frac{(v - u)}{t} \), where \( v \) is final velocity, \( u \) is initial velocity, and \( t \) is time. Here, \( v = 17.88 \mathrm{m/s} \), \( u = 0 \), and \( t = 12.0 \) seconds. Substituting these values in the equation, we have \( a = \frac{(17.88 - 0)}{12} = 1.49 \mathrm{m/s^2} \).
04

- Calculate the distance

Now that we have obtained the acceleration \( a = 1.49 \mathrm{m/s^2} \), we can substitute this value in the simplified equation from step 2 to find the distance: \( x = \frac{1}{2} a t^{2} = 0.5 \tx{1.49 m/s^2} \tx{(12 s)^2} = 107.28 \mathrm{m} \).

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