Question

You are travelling at city speed in your car (35 mi/h). Approaching a stop sign, you slwo down at a uniform rate and come to a stop in 3.9 seconds. From the time you step on the brakes, to when you come to a complete stop, what distance does the car travel?

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Answer 1

Here ,

initial speed , u = 35 mi/hr = 35 * 1609/3600 m/s

u = 15.6 m/s

time taken , t = 3.9 s

let the acceleration is a

Using first equation of motion

v = u + at

0 = 15.6 + 3.9 * a

a = -4 m/s^2

Now, for the distance travelled

distance travelled = u *t + 0.50 *a * t^2

distance travelled = 15.6 * 3.9 - 0.50 * 4 * 3.9^2

distance travelled = 30.4 m

the distance travelled is 30.4 m