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Q. Starting from origin, a body moves along $ x $ -axis. Its velocity at any time is given by $ v = 4t^3 - 2t \,m/s $
Acceleration of the particle when it is $ 2 \,m $ away from the origin is

AMUAMU 2016Motion in a Straight Line

Solution:

The velocity at any time
$v = 4t^2 - 2t $
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$ v = \frac{dx}{dt} = 4t^2 - 2t$
$\Rightarrow x = \frac{4t^4}{4} - \frac{2t^2}{2}$
$\Rightarrow t^4 - t^2 = 2$
$\Rightarrow t^4 -t^2 -2 = 0 \,\,\,[\because x =2]$
$\Rightarrow t^2 = \frac{ 1 \pm \sqrt{1+8}}{2} = \frac{ 1\pm 3}{2} $
$ t = \sqrt{2} s$
Now, $a = \frac{dv}{dt} = 12t^2 - 2$
At $ t = 2s$, accelaration of the particle
$a |_{or\, t =\sqrt{2}}^ {x= 2} = 12 \times 2 - 2 $
$ = 22\, m/s^2$