Question

A particle moves in a straight line with an acceleration 'a' $m/s^2$ given as function of time, $a=\frac{-1}{t^2}$. At time t = 1 s the particle has velocity of 3 m/s, then find the velocity at t = 4 s.

• A. 2.25 m/s
• B. 5 m/s
• C. 4.5 m/s
• D. 7 m/s

Answer 1

The correct option is A 2.25 m/s

$a=\frac{dV}{dt}=\frac{-1}{t^2}$

$\Rightarrow dV=\frac{-1}{t^2}dt$
or integrating both sides w.r.t time with limits V = 3 m/s at t = 1 sec and V at t = 4 sec

$\underset{3}{\overset{V}{\int }}dV=\underset{1}{\overset{4}{\int }}\frac{-1}{t^2}dt$

$\Rightarrow [V{]}_3^V=-{\left[\frac{t^{-2+1}}{-2+1}\right]}_1^4$

$orV-3={\left[\frac{1}{t}\right]}_1^4$

$orV-3=\frac{1}{4}-1=\frac{-3}{4}$

$\therefore V=3-\frac{3}{4}=2.25m/s$

Why this question? Tip: Acceleration is varrying with time here, hence apply $a=\frac{dV}{dt}$ and use method of integration to obtain velocity at time 't'.