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]
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
Hmm, it seems like there might be a misunderstanding in your calculations. Let me help you calculate it correctly.
The initial velocity is 7 m/s, and it stops after moving 1.5 m. So, we can calculate the deceleration using the equation v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance.
For the first case:
v^2 = 0^2 + 2a(1.5)
0 = 2a(1.5)
a = 0 m/s^2
Since the acceleration is 0, the car is not slowing down. Therefore, it would continue to move forever (or until it hits something).
Now, let's calculate it for the second case:
v^2 = 0^2 + 2a(s)
(14)^2 = 2a(s)
196 = 2a(s)
Since the braking force is constant in both cases, we can assume the acceleration is the same. Therefore, we can set up a ratio:
d(2)/d(1) = s(2)/s(1)
d(2)/1.5 = s(2)/s(1)
d(2) = (s(2)/s(1)) * 1.5
d(2) = (196/49) * 1.5
d(2) = 6 m
So, the correct answer is 6 meters. The car would have continued to move 6 meters after the brakes were applied.
Your mistake is in the calculation of the numerator:
[(0)^2-(14)^2] is not equal to -14^2, it should be 14^2 - 0^2.
So the correct calculation should be:
d(2)/d(1) = (14^2 - 0^2) / (7^2 - 0^2)
= 196 / 49
= 4
Therefore, the car would have continued to move for 4 meters after the brakes were applied, not 6 meters.
To calculate how far the car would have continued to move after the brakes were applied when it was moving at 14 m/s, you can use the formula for distance traveled under constant acceleration:
d = (v^2 - u^2) / (2a)
Where:
d = distance traveled
v = final velocity
u = initial velocity
a = acceleration
In this case, the car is coming to a complete stop, so the final velocity, v, is 0 m/s. The initial velocity, u, is 14 m/s.
So, plugging in the given values:
d = (0^2 - 14^2) / (2 * a)
Now, we know that the braking force is constant and the same in both cases, so we can assume that the acceleration, a, is the same. Since the car comes to a stop in 1.5 m when it is moving at 7.0 m/s, we can find the acceleration using the same formula:
1.5 = (0^2 - 7.0^2) / (2 * a)
Simplifying this equation:
1.5 = (-49) / (2 * a)
1.5 = -24.5 / a
Cross-multiplying:
a = -24.5 / 1.5
a = -16.33 m/s^2
Now we can go back to the original equation for the distance traveled at 14 m/s:
d = (0^2 - 14^2) / (2 * -16.33)
d = (-196) / (-32.66)
d = 6 m
Therefore, the car would have traveled a distance of 6 m after the brakes were applied when it was moving at 14 m/s.
It seems there was an error in your calculation at this step:
d(2)/(1) = [(0)^2-(14)^2]/[(0^2)-(7)^2]
Instead, it should be:
d(2)/(1) = [(0^2)-(14)^2]/[(0^2)-(7)^2]