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A particle moves along a straight line. The relation between the position of a particle and time is x^2 = (8t^2 + 8t +4 )m^2 Find acceleration of a particle in (m/s^2) at x = 2m

aruna kumari Samarla 22nd Feb, 2020

Answer (1)

Devraj Student Expert 22nd Feb, 2020

Hii

First all taking all the terms to one side we get

x= square root of (8t^2+8t+4)

On diffrentiating one time

dx/dt= (8t+4)(8t^2+8t+4)

Double diff.

d^2x/dt^2 =8/sqrt(8t^2+8t+4) - (8t+4)*(8t^2+8t+4)^-3/2.

Now, put the values t=2

You will get

8*0.138-2.667*10^-3*20

Its 1.05066

Hope it helps

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