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.

The car travel travels 211.22 extra feet before coming to stop

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