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public_enemy720
Nov9-06, 06:21 PM
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
Nov9-06, 07:22 PM
Hint: Which of the equations of motion relates initial velocity, final velocity, acceleration, and distance moved?
public_enemy720
Nov9-06, 08:16 PM
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
Nov9-06, 08:22 PM
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.