Am I setting up this kinematic equation properly?

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The discussion revolves around calculating the acceleration, final speed, and stopping distance of a race car that starts from rest and travels a quarter mile in 12.1 seconds with constant acceleration. The calculated acceleration is approximately 5.5 m/s², and the final speed is 66.5 m/s. For the stopping distance, the user initially used the wrong reference point, leading to an impractical distance of 1507.8 meters, which was later corrected to 1105.5 meters after adjusting the starting point to zero. The conversation highlights that a braking acceleration of -2.00 m/s² is considered modest, with safer braking typically around 0.5g. The importance of terminology is also noted, emphasizing the correct use of "braking" instead of "breaking."
TaylorHoward21
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A race car starts from rest and goes a quarter mile (1/4 mi) in 12.1 seconds. Assume the acceleration of the car is constant. (a.) What is the acceleration of the car? (b.) What is the final speed of the car? (c.) If you hit the brakes, how far would it take to stop the car if the breaks cause an acceleration of -2.00 m/s2?

I believe I have parts a and b right, however I am not confident with my process of solving part C.


1/4 mi ≈ 402.3 m

a.) acceleration = 5.5m/s^2
b.) Final speed (at 12.1 seconds) = 66.5 m/s

c.) For part C, I reset the known values to:
a = -2.00m/s2
V0 = 66.5m/s
V = 0 (because solving for car stop which is zero velocity)
X0 = 402.3m

The equation I used: v2 - V20 = 2a(x - x0)
The answer I got was 1507.8m = x
For a car going 66.5m/s that is slowing down at -2.00 m/s2 does this answer seem practical? I am struggling to check my work here.

Many thanks.
 
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TaylorHoward21 said:
X0 = 402.3m
It asks how far the car goes after hitting the brakes. The distance already traveled is not relevant.
 
haruspex said:
It asks how far the car goes after hitting the brakes. The distance already traveled is not relevant.
So if I change X0 to be = 0 the new answer I get is 1105.5m. This still seems rather far for a car to stop from 66.5m/s to 0m/s. Is -2.00m/s^2 simply just a very slow breaking acceleration?
 
TaylorHoward21 said:
Is -2.00m/s^2 simply just a very slow breaking acceleration?
It is rather modest. Safe but serious braking is about 0.5g on a dry road. Emergency braking could go as high as 0.7-1g, depending on the vehicle.

And it is "braking", using "brakes", not "breaking".
 
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