Starting from rest, acceleration of a particl | vedclass

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Question

(a)$ \Rightarrow dv = 2(t - 1)\;dt$
$ \Rightarrow 
u = \int_0^5 2 \;(t - 1)dt = 2\left[ {\frac{{{t^2}}}{2} - t} ight]_0^5$$ = 2\left[ {\frac{{25}}

Vedclass · Accepted Answer

$15$

Vedclass · Answer

(a)$ \Rightarrow dv = 2(t - 1)\;dt$
$ \Rightarrow 
u = \int_0^5 2 \;(t - 1)dt = 2\left[ {\frac{{{t^2}}}{2} - t} ight]_0^5$$ = 2\left[ {\frac{{25}}{2} - 5} ight]$ $ = 15 \,m/s$

Starting from rest, acceleration of a particle is $a = 2(t - 1).$ The velocity of the particle at $t = 5\,s$ is.........$m/sec$

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