# Physics 11 Speed and acceleraton

[SOLVED] Physics 11 Speed and acceleraton

## Homework Statement

The brakes on a car permit it to decelerate at the rate of -8.0m/s^2. How much distance is required to stop this car when it is travelling 60.0km/hr?

A=V/T
D=RT

## The Attempt at a Solution

I solved it this way
60km/h = 16.667m/s
A=V/T 16.667m/s / -.80m/s^2 = 20.834 seconds
D=RT 16.667m/s x 20.834seconds = 347.24m

This is wrong and I do not understand why. Am I doing it right or is the answer wrong?

Related Introductory Physics Homework Help News on Phys.org
The reason your answer is wrong is because you assumed the car maintained its 60.0 km/hr pace for the entire 20.834 seconds. It did not, because it was slowing down.

Since its deceleration is constant, you are justified in saying that the car traveled for 20.834 seconds at an average velocity of (16.667 + 0)/2 m/s, and you will come up with the right answer.

Another way to do this problem that avoids this potential pitfall is to just use the equation $v_f^2 = v_i^2 + 2ad$.

Adding to what Tedjn said, it was probably just a typo but make sure you use -0.8 ms^-2 for the acceleration or the answer is a tenth of what it's supposed to be. Your speed looks right.

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