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Carnegie Mellon University

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Problem 5 Easy Difficulty

A car is traveling east at 25.0 $\mathrm{m} / \mathrm{s}$ when it turns due north and accelerates to $35.0 \mathrm{m} / \mathrm{s},$ all during a time of 6.00 s. Calculate the magnitude of the car's average acceleration.

Answer

$\left| \vec { a } _ { \mathrm { av } } \right| = 7.17 \mathrm { ms } ^ { - 2 }$

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Video Transcript

so the car's average acceleration. We can say that the average acceleration in the ex direction would be equaling the change in the velocity in the ex direction divided by Delta T. This would be equaling zero minus 25.0 meters per second because of course, the final velocity in the ex direction is zero. This would be divided by 6.0 seconds. So this is equaling negative 4.17 meters per second squared. This would be the average acceleration in the ex direction. The average acceleration in the UAE direction would then be equaling Elsa Visa. Why divided by Delta t. So this would be equaling here. It would be 35.0 meters per second minus zero. Considering that the initial velocity in the UAE direction of zero meters per second divided by the same time interval of 6.0 seconds and this is giving us 5.83 meters per second squared, this would be our answer for the average acceleration in the UAE direction. That is the end of the solution. Thank you for watching

Carnegie Mellon University
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