Question

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.

Accepted Answer

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.

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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...

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