Acceleration of particle moving rectilinearly is `a = 4-2x`(where x is position in metre and a in ms^(-2))`. It si at instantaneous rest at x = 0. At what position `x 9`in meter) will the particle again come to instantaneous rest?

Acceleration of particle moving rectilinearly is `a = 4-2x`(where x is

1.	Acceleration of particle moving rectilinearly is `a = 4-2x`(where x is position in metre and a in ms^(-2))`. It si at instantaneous rest at x = 0. At what position `x 9`in meter) will the particle again come to instantaneous rest?
Answer» Correct Answer - 4 `(vdv)/(dx) = 4-2x` `int_(0)^(v) vdv = int_(0)^(x) (4-2x)dx rArr v^2/2 = 4x - x^2` when `v = 0 , 4x - x^2 = 0` `x = 0, 4` `rArr Thus, at x = 4, the particle will again come to rest.

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Acceleration of particle moving rectilinearly is `a = 4-2x`(where x is position in metre and a in ms^(-2))`. It si at instantaneous rest at x = 0. At what position `x 9`in meter) will the particle again come to instantaneous rest?

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