The speed of a car traveling in a straight line is reduced from 45 to 30 mph in a distance of 264 feet. find the distance in which the car can be brought to rest from 30 mph, assuming the same constant acceleration.

Explanation:

We have equation of motion v² = u² + 2as

Initial velocity, u = 45 mph = 20 m/s

Acceleration, a = ?

Final velocity, v = 30 mph = 13.33 m/s

Displacement,s = 264 feet = 80.47 m

Substituting  

v² = u² + 2as

13.33² = 20² + 2 x a x 80.47

a = -1.38 m/s²

Acceleration is -1.38 m/s²

Now we have

We have equation of motion v² = u² + 2as

Initial velocity, u = 13.33 m/s

Acceleration, a = -1.38 m/s²

Final velocity, v = 0 m/s

Displacement,s = ?

Substituting  

v² = u² + 2as

0² = 13.33² + 2 x -1.38 x s

s = 64.38 m = 211.22 feet

The car travel travels 211.22 extra feet before coming to stop

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