If 250 feet are required to stop a car traveling 60 miles per hour,how many feet are required to stop a car traveling 96 miles per hour?

Assuming a constant effective braking force (hence acceleration), the braking time is in direct variation with the initial speed. Suppose that T seconds are required in the 60 mph case. Then 96T/60 is required in the second case.

96 mph breaking distance
= (1/2)a(96T/60)²
= (1/2)aT²(96/60)²
= 250(96/60)²
= 640 feet

Another way to look at it, braking distance times the braking force is the work done by the brakes, which is equal to the initial kinetic energy of the car. The kinetic energy is in direct variation with the square of the velocity. Again assuming constant force, the braking distance must also vary directly with the square of the velocity.

Yes, I got that same answer,
1.set up a proportion: 250/60 = x/96
2.cross multiply: 250(96)=60x
3.answer: x=400