# A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 38 ft/s2...A car is traveling at 50 mi/h when the brakes are fully applied, producing a...

A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 38 ft/s2...

A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 38 ft/s2. What is the distance covered before the car comes to a stop? (Round your answer to one decimal place.)

_____________ft.

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For the unit to be consistent, convert 38 ft/s^2 to mi/h^2. Use the conversion factor,

1 mile = 5280 feet and 1 hour = 3600 seconds.

So,

`38 `  `(ft)/s^2`  `xx`  `(1 mi)/(5280ft)`  `xx` `( 3600s)^2/(1h)^2` `=` `93272.27` `(mi)/h^2`

Then, use the formula of constant acceleration to determine the distance travelled (s) by the car before it came to stop. The formula is:

`v_2^2=v_1^2 + 2as`

In the problem, the initial velocity of car is 50mi/h `(v_1=50)` . When the car comes to stop its final velocity then is zero `(v_2=0)`. And since the given is deceleration, not acceleration, then the value of a is negative. So `a = -93272.27` .

Substituting these values to the formula yields:

`0^2=50^2+2(-93272.27)s`

`0=2500-186544.54s`

Then, isolate the s.

`185644.54s= 2500`

`s = 2500/(185644.54)`

`s=0.0134`

Hence, the car travelled a distance of 0.0134 miles before it came to stop.

If we are going to express the distance travelled in feet, use the conversion factor 1 mile =5280 ft.

So

`0.0134 mi ` `xx` `(5280 ft)/(1 mi)` `=` `70.8 ft`

In feet, the car travelled a distance of 70.8 ft before it came to stop.