The discussion revolves around the relationship between initial velocity and stopping distance for a car under constant braking force, specifically comparing a scenario where the car travels at 7 m/s versus 14 m/s.
Some participants have provided hints and guidance regarding relevant equations. There is an ongoing exploration of the implications of doubling the initial velocity on stopping distance, with some participants confirming the approach taken by others.
Participants are working under the assumption of a constant braking force and are discussing the implications of this assumption on the calculations and results.
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.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