How does initial velocity affect a car's stopping distance?

[public_enemy720]

A car is traveling at 7 m/s when the wheels are locked up. The car slides 1.5 meters before coming to rest. If the car had been moving at 14 m/s, how far would the car slide, assuming the same breaking force is used?

Not sure where to go with this problem...

 

[OlderDan]

Hint: Which of the equations of motion relates initial velocity, final velocity, acceleration, and distance moved?

 

[public_enemy720]

I used the equation v^2=Vo^2+2(a)(S-So) and solved for acceleration, then plugged in new value for initial velocity and used the found acceleration. Is this correct?

accel. = -16.3m/s^2
distance traveled @ 14 m/s = 6.01 m

 

[OlderDan]

I used the equation v^2=Vo^2+2(a)(S-So) and solved for acceleration, then plugged in new value for initial velocity and used the found acceleration. Is this correct?

accel. = -16.3m/s^2
distance traveled @ 14 m/s = 6.01 m

Close enough, and that's exactly the correct approach. Since the velocity doubled, the stopping distance would be 2^2 = 4 times as great for the same acceleration.