# What is the distance covered before the car comes to a stop? A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 42 ft/s2. What is the distance...

What is the distance covered before the car comes to a stop?

A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 42 ft/s2. What is the distance covered before the car comes to a stop? (Round your answer to one decimal place.)

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You need to remember what is the formula of velocity in uniformly accelerated motion such that:

`v^2 final = v^2 initial + 2*a*d`

The problem provies the initial velocity v initial = 50 mi/h and the final velocity needs to be zero since the car comes to a stop such that:

`0 = 50^2 + 2a*d`

You need to convert the miles in feet since the velocity is given in miles per hour and decceleration is gven in feet/`s^2` .

`1 mi = 5280 feet =gt 50 (mi)/(hour) = 5280*50(feet)/(60 sec) = 4400 (feet)/(sec)` `0 = 4400^2 - 2*42*d =gt d = 4400^2/84`

`d = 230476.1 feet`

**Hence, evaluating the distance covered before the car comes to a stop yields `d = 230476.1` feet.**