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Chapter 2: Problem 47

On a dry road a car with good tires may be able to brake with a deceleration of \(11.0 \mathrm{mi} / \mathrm{h} / \mathrm{s}\left(=4.92 \mathrm{~m} / \mathrm{s}^{2}\right) .(a)\) How long does it take such a car, initially traveling at \(55 \mathrm{mi} / \mathrm{h}(=24.6\) \(\mathrm{m} / \mathrm{s}\) ), to come to rest? \((b)\) How far does it travel in this time?

Short Answer

Expert verified
The time the car takes to stop is approximately \(5\) seconds and the distance travelled in this time is about \(61.5\) meters.

Step by step solution

01

Identify given information

The initial speed of the car is given as \(24.6 m/s\). The deceleration or negative acceleration is given as \(-4.92 m/s^{2}\) (it is negative since it's slowing down the car). The final speed of the car would be \(0 m/s\) as it comes to rest.
02

Calculate time taken to stop

We use the first equation of motion, which is \(v = u + at\), where \(v\) is the final speed, \(u\) is the initial speed, \(a\) is the acceleration and \(t\) is the time taken. Rearranging for \(t\), we get \(t = (v - u) / a\). Substituting the given values, we calculate \(t = (0 - 24.6) / -4.92\).
03

Compute the distance traveled

For this part, we use the third equation of motion, which is \(v^{2} = u^{2} + 2as\), where \(s\) is the distance traveled. Rearranging for \(s\), we get \(s = (v^{2} - u^{2}) / (2a)\). Substituting the values, we compute \(s = [(0)^{2} - (24.6)^{2}] / [2*(-4.92)]\).

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