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The acceleration of a particle moving in...

The acceleration of a particle moving in straight line is defined by the relation `a= -4x(-1+(1)/(4)x^(2))`, where a is acceleration in `m//s^(2)` and `x` is position in meter. If velocity `v = 17 m//s` when `x = 0`, then the velocity of the particle when `x = 4` meter is :

A

`12 m//s`

B

`15 m//s`

C

`20 m//s`

D

`25 m//s`

Text Solution

Verified by Experts

The correct Answer is:
B
The acceleration ……………
`a = -4x(-1+ 0.25 x^(2))`
`int_(17)^(v) vdv = int_(0)^(4)(4x-x^(3))dx`
`(1)/(2)(v^(2)-17^(2))= (4)/(2)(x^(2))_(0)^(4) - (1)/(4)(x^(4))_(0)^(4)`
`v^(2)-17^(2)=4(4^(2))-(1)/(2)(4^(4))= 64 - 128`
`V^(2)= 298-64 = 225 , V = 15 m//sec`
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