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

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answer is 4.

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Detailed Solution

vdvdx=4−2x ∫0v vdv=∫0x (4−2x)dx⇒v22=4x−x2when v = 0, 4x - x2 = 0 x = 0, 4⇒ Thus, at x = 4, the particle will again come to rest.

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