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

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The discussion focuses on how initial velocity influences a car's stopping distance. A car traveling at 7 m/s slides 1.5 meters before stopping, while at 14 m/s, it would slide approximately 6.01 meters under the same braking force. The key equation used is v^2 = Vo^2 + 2(a)(S-So), which relates initial velocity, final velocity, acceleration, and distance. The acceleration calculated is -16.3 m/s^2, confirming that doubling the velocity results in a stopping distance that is four times greater. This illustrates the quadratic relationship between initial velocity and stopping distance.
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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...
 
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Hint: Which of the equations of motion relates initial velocity, final velocity, acceleration, and distance moved?
 
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
 
public_enemy720 said:
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.
 
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