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Question

Physics Question on physical world

Starting from origin, a body moves along xx -axis. Its velocity at any time is given by v=4t32tm/sv = 4t^3 - 2t \,m/s Acceleration of the particle when it is 2m2 \,m away from the origin is

A

28ms228 \,ms^{-2}

B

12ms212 \,ms^{-2}

C

22ms222 \,ms^{-2}

D

14ms214 \,ms^{-2}

Answer

22ms222 \,ms^{-2}

Explanation

Solution

The velocity at any time
v=4t22tv = 4t^2 - 2t

v=dxdt=4t22tv = \frac{dx}{dt} = 4t^2 - 2t
x=4t442t22\Rightarrow x = \frac{4t^4}{4} - \frac{2t^2}{2}
t4t2=2\Rightarrow t^4 - t^2 = 2
t4t22=0[x=2]\Rightarrow t^4 -t^2 -2 = 0 \,\,\,[\because x =2]
t2=1±1+82=1±32\Rightarrow t^2 = \frac{ 1 \pm \sqrt{1+8}}{2} = \frac{ 1\pm 3}{2}
t=2st = \sqrt{2} s
Now, a=dvdt=12t22a = \frac{dv}{dt} = 12t^2 - 2
At t=2s t = 2s, accelaration of the particle
aort=2x=2=12×22a |_{or\, t =\sqrt{2}}^ {x= 2} = 12 \times 2 - 2
=22m/s2= 22\, m/s^2