Discussion
A particle of mass m is initially at rest. A time varying force \(F = kt\) acts on it. Its velocity after time \(t\) is
Answer: A
The velocity is \(\dfrac{kt^{2}}{2m}\).
The acceleration is \(a = \frac{F}{m} = \frac{kt}{m}\), which changes with time, so integrate it:
\[ v = \int_{0}^{t} \frac{kt}{m}\,dt = \frac{kt^{2}}{2m} \]
Equivalently, the impulse \(\int F\,dt = \frac{kt^{2}}{2}\) equals the momentum gained. \(\frac{kt^{2}}{m}\) forgets the \(\frac{1}{2}\) from integration (it treats the force as constant at its final value), and the options with \(k^{2}\) are dimensionally wrong.
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