After brakes are applied how far does the car travel? - integrating twice

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SUMMARY

The discussion focuses on calculating the distance a car travels after brakes are applied, given the acceleration function a(t) = -6t - 4 ft/sec² and an initial speed of 32 ft/sec. The user correctly integrates the acceleration to find the velocity function v(t) = -3t² - 4t + 32 and determines that the car comes to a stop at t = 8/3 seconds. The total distance traveled before stopping is calculated to be approximately 52.1481 feet. The user questions the use of absolute value in their calculations, which is confirmed to be unnecessary.

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Homework Statement


t seconds after the brakes of a car are put on, the acceleration of the car is a(t) = -6t - 4 ft/sec^2. If the automobile was moving at the speed of 32 ft/sec when the brakes were put on, how far does it go before stopping?


Homework Equations


a(t) = -6t - 4 ft/sec^2


The Attempt at a Solution



integral of a(t) = -6t - 4 ft/sec^2 ----> |-3t^2 - 4t + c| = 32
I put the absolute value because it is the speed. is that right?

v(0) = c = 32

v(t) = -3t^2 -4t + 32

-3t^2 -4t + 32 = 0
t = 8/3

integral from 0 to 8/3 of -3t^2 -4t + 32

= 52.1481 ft

The main thing I am worried about is that it says speed instead of velocity, can anyone help me out here? Thanks! :)
 
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That should be correct what you did.

But you don't need the modulus sign really.
 

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