The car starts from rest and moves in a straight line such that for a short time its velocity is defined by $v = ( 9 t^2+2t) m/s.$ Where t is in seconds.

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written 8.4 years ago by

teamques10 ★ 68k

• modified 4.5 years ago

Determine its position and acceleration when t = 3 sec.

engineering mechanics

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written 8.4 years ago by

teamques10 ★ 68k

$v=9t^2+2t -------(i)$ For displacement at t=3s, $ \dfrac { dx}{dt}=9t^2+2t \ x=∫_0^3 (9t^2+2t)dt \space \space [\text {Assuming initial position at } x=0] \ =[3t^3+t^2]_0^3 \ \therefore x=90m \Rightarrow Ans $ For acceleration at t=3s Differentiate (i) w.r.t ‘t’ $\therefore a=(18t+2)_{t=3} \ \therefore a=56m/s^2 Ans$

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