Physics 11 Speed and acceleraton

AI Thread Summary
The discussion centers on calculating the stopping distance of a car decelerating at -8.0 m/s² from a speed of 60.0 km/h. The initial attempt incorrectly assumed the car maintained its speed throughout the deceleration period, leading to an overestimation of the stopping distance at 347.24 m. The correct approach involves using the average velocity during deceleration, which is half of the initial speed, to accurately calculate the distance. An alternative method using the equation v_f² = v_i² + 2ad is also suggested for clarity. The correct stopping distance is confirmed to be 173.6 m.
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[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 traveling 60.0km/hr?

The answer given is 173.6m

Homework Equations



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?
 
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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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