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Physics 13 Online
OpenStudy (anonymous):

Medal+Fan An automobile driver puts on the brakes and decelerates from 30.0m/s to zero with an acceleration of -3m/s/s. What distance does the car travel?

OpenStudy (surry99):

look up kinematic equations of motion for constant acceleration

OpenStudy (anonymous):

That doesn't help surry.

OpenStudy (anonymous):

Use the formula s=ut - 0.5at^2 and v=u-at Where s is the distance traveled, a is acceleration, t is time , v is final speed, u is initial speed. So according to the values you have, we need to find time firstle and hence we use v=u-at, having 0=30-3t. t will be equal to 10 seconds. Resorting in the first equation, we will have: s= 30(10)-0.5[3(10^2)] giving you 150metres. Hope it did help.

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