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A particle is moving in a straight line and passes through a point O with a velocity of $6 m s^{- 1}$ . The particle moves with a constant retardation of $2 m s^{- 2}$ for $4$ s and there after moves with constant velocity. How long after leaving O does the particle return to O ?

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Solution

$V_{f} = 6 - 2 \times 4$ $(V = u - a t)$
$V_{f} = - 2 m / s$
This means it returns after $4$ sec
$t_{1} = 4$ sec
Now, distance travelled in $4$ sec
$d = u t + \frac{1}{2} a t^{2}$
$d = 6 \times 4 - \frac{1}{2} (2) (4)^{2}$
$= 24 - 16$
$d = 8 m$
Now after returning body moves with constant velocity
$V = \frac{d}{t_{2}}$
$t_{2} = \frac{d}{V} = \frac{8}{2} = 4$ sec
$t_{2} = 4$ sec
Total time, $T = t_{1} + t_{2} = 8$ sec

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