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Physics 23 Online
OpenStudy (anonymous):

A car starts from rest and accelerates at a constant rate. After the car has gone 50 m it has a speed of 21 m/s. What is the acceleration of the car? A. 4.4 m/s2 B. 2.4 m/s2 C. 0.23 m/s2 D. 5.5 m/s2

OpenStudy (anonymous):

If you award medals, you WILL get much more attention! :)

OpenStudy (anonymous):

I do give medals to the ones that help me. @HomeschoolGrad

OpenStudy (schrodingers_cat):

Remember for a particle under constant acceleration Vf =Vi +at and Xf -Xi = 1/2(Vi +Vf)t where 1/2(Vi +Vf) is the average acceleration Substituting the first equation into the second for t yields Vf^2 -Vi^2 = 2a(Xf -Xi) Plugging in the numbers you get (21^2 -0)/100 =( 4.4 m/s^2) Hope this helps :)

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