heathorcutt
Physics Student
Posts: 3 
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#13
Oct 28, 2014 20:57:25 GMT -6
Post by heathorcutt on Oct 28, 2014 20:57:25 GMT -6
To find acceleration, would you rearrange the second orange equation as [(2(deltaX)/(t^2)]-Vot=a?
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#13
Oct 28, 2014 21:07:22 GMT -6
Post by Mr. Askey on Oct 28, 2014 21:07:22 GMT -6
answering . . .
∆x = vo t + 1/2 a t^2
to solve for a: { 2 [ ∆x - vo t ] } / t^2
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heathorcutt
Physics Student
Posts: 3
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#13
Oct 28, 2014 21:12:32 GMT -6
Post by heathorcutt on Oct 28, 2014 21:12:32 GMT -6
thanks!
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