A car starts from rest and moves with constant acceleration during the 5th second of its motion,it covers a distance of 36 meters. What is the acceleration

I think a 30 years old engineer can help you with a simple question like this...



d = 36 m


t = 5s


vi = 0 m/s



Find:
a =?
d = vi*t + 0.5*a*t2

36 m = (0 m/s)*(5s)+ 0.5*(a)*(5s)2

36 m = (12.5 s2)*a

a = (36 m)/(12.5 s2)

a = 36/12.5

a= 2.88

so the answer is 2.88

Quote:

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Originally Posted by John Bold

I think a 30 years old engineer can help you with a simple question like this....

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This answer is incorrect - what you've calculated is the acceleration for an object to travel 36 meters in 5 seconds, but that's not the question.

To the OP: the distance covered in 't' seconds from a standing start is d=1/2at^2. The distance covered in the 5th second is the diffence between the distance covered for t=5 seconds and for t=4 seconds, and that value is 36 meters. Plug and chug, and solve for 'a.'