Why are you using conservation of energy? It is going up in smoke here. Use F=m a.
average speed during stop = (7+0)/2
= 3.5 m/s
so
time to stop = 1.5 / 3.5 = .4286 seconds
v = Vi + a t
0 = 7 +a(.4286)
a = -16.3333 m/s^2
same force in second case, same mass so same acceleration
average speed now = 7
so d = 7 t
7 t = 14 t - (1/2)(-16.33)(t^2)
8.1666 t^2 -7 t = 0
t = 0 or t = 7/8.1666
d = 7 t = 6 meters
A car is traveling at 7.0 m/s when the driver applies the brakes. The car moves 1.5 m before it comes to a complete stop. If the car had been moving at 14 m/s, how far would it have continued to move after the brakes were applied? Assume the braking force in both cases is constant and the same.
The answer is 6m BUT I keep getting 4m.
This is my work:
Wnc= (Delta)KE + (Delta)PE of both cases.
[f(k)d(2)]/[f(k)d(1)] = [(1/2)m(v2^2(final)-v2^2(initial))/(1/2)m(v1^2(final)-v1^2(initial))]
d(2)/d(1)= [(v2^2(final)-v2^2(initial)]/[(v1^2(final)-v1^2(initial)]
d(2)/(1)= [(0)^2-(14)^2]/[(0^2)-(7)^2]
1 answer