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A 15 Kg mass is accelerated from rest with a force of 100 N. As it moves faster, friction and air resistance create an oppositely directed retarding force given by $F_{R} = A + BV$ where A=25 N and $B = 0.5 \frac{N}{m / s}$ . At what velocity does the acceleration equal to one half of the initial acceleration ?

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50ms^-1

100ms^-1

75ms^-1

25ms^-1

answer is B.

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Detailed Solution

$\begin{array}{l} a_{i} = \frac{F}{m} = \frac{100}{15} \\ {ma}_{f} = F - (A + Bv) = \frac{{ma}_{i}}{2} \\ 100 - 25 - 0.5 v = \frac{100}{2} \\ 25 - 0.5 v = 0 \\ v = \frac{25}{0.5} = \frac{250}{5} = 50 {ms}^{- 1} \end{array}$

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