The magnitude of force acting on a particle moving along x-axis varies with time (t) as shown in figure. If at t=o the velocity of [particle is $v_{0}$ , then its velocity at $t = T_{0}$ will be

v_{0} + \frac{π F_{0} T_{0}}{4 m}

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v_{0} + \frac{π F_{0}}{2 m}

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v_{0} + \frac{π T_{0}^{2}}{4 m}

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v_{0} + \frac{π F_{0} T_{0}}{m}

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Solution

The correct option is A $v_{0} + \frac{π F_{0} T_{0}}{4 m}$
REF. Image.
Area under F - T graph = Impulse
Area $= \frac{(π) (F_{0}) (T_{0} / 2)}{2}$
$= \frac{π F_{0} T_{0}}{4}$
Impulse = change in momentum
$\frac{π F_{0} T_{0}}{4} = m (V_{f} - V_{i})$
$V_{f} = V_{i} + \frac{π F_{0} T_{0}}{4 m}$
$= V_{0} + \frac{π F_{0} T_{0}}{4 m}$
Option (A)

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