Question

1. a car is being driven at a rate of 40 mph when the brakes are applied. the car decelerates at a constant rate of 10ft/sec². how long before the car stops? how far did it travel in the time it takes to stop?

Explanation:

Step1: Convert the initial velocity to ft/sec

We know that \(1\) mile \( = 5280\) feet and \(1\) hour \(=3600\) seconds.
The initial velocity \(v_0\):
\(v_0 = 40\space mph=\frac{40\times5280}{3600}\space ft/sec=\frac{176}{3}\space ft/sec\)

Step2: Find the time when the car stops

The acceleration \(a=- 10\space ft/sec^{2}\) (negative because it is decelerating), and the final velocity \(v = 0\).
Using the formula \(v=v_0+at\)
\(0=\frac{176}{3}-10t\)
\(10t=\frac{176}{3}\)
\(t=\frac{176}{30}=\frac{88}{15}\approx5.87\space sec\)

Step3: Find the distance traveled

Using the formula \(s = v_0t+\frac{1}{2}at^{2}\)
Substitute \(v_0=\frac{176}{3}\), \(a = - 10\) and \(t=\frac{88}{15}\)

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Answer:

The time before the car stops is \(\frac{88}{15}\space sec\approx5.87\space sec\) and the distance traveled is \(\frac{7744}{45}\space ft\approx172.09\space ft\)