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A force $F_{1}$ accelerates a particle from rest to a velocity v. Another force $F_{2}$ decelerates the same particle from v to rest, then

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$F_{2} m a y b e s m a l l e r t h a n, g r e a t e r t h a n o r e q u a l t o F_{1}$

$F_{2} i s g r e a t e r t h a n F_{1}$

$F_{1} i s a l w a y s e q u a l t o F_{2}$

$F_{2} c a n n o t b e e q u a l t o F_{1}$

answer is C.

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Detailed Solution

During acceleration,

$v = a_{1} t_{1} = \frac{F_{1}}{m} t_{1} F_{1} = \frac{m v}{t_{1}}$

During retardation,

$0 = v - a_{2} t_{2} = v - \frac{F_{2}}{m} t_{2} F_{2} = \frac{m v}{t_{2}}$

$If$

$t_{1} = t_{2} then F_{1} = F_{2} t_{1} < t_{2} then F_{1} > F_{2} and t_{1} > t_{2} then F_{1} < F_{2}$

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