During a test, the car, moves in a straight line such that its velocity is defined by v=0.3 (9t2 + 2t) where 't' is in seconds. Determine the position and acceleration when t=3 sec. take at t=0, x=0.

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written 4.5 years ago by

teamques10 ★ 70k

modified 3.7 years ago by

DevarenjiniP • 3.2k

engineering physics

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written 4.5 years ago by

teamques10 ★ 70k

Given the car moves in a straight line such that its velocity is defined by $v=0.3 (9t^2 + 2t)$ where 't' is in seconds

To find

The position and acceleration when t = 3 sec. taking at$t=0s, x=0m$
We know that$v = {dx \over dt} $and$a= {dv \over dt}$
So$x …

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