Question

A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 26 ft/s2. what is the distance covered before the car comes to a stop? (round your answer to one decimal place.)

Answer 1

The distance traveled during a period of constant acceleration is given by

`d=((v_(2))^(2)-(v_(1))^(2))/(2a)`

Your initial speed is 50 mi/h = 73 1/3 ft/s, so the stopping distance can be computed as

`d=(0-(73(1)/(3))^(2))/(2\cdot(-26))=(5777(7)/(9))/(52)\approx 103.4`

The stopping distance is 103.4 ft.