Question

The acceleration of a particle, starting from rest, varying with time according to the relation $a=kt+c$. Then the velocity $v$ of the particle after a time $t$ will be:

• A. $2k{t}^2+ct$
• B. $\frac{1}{2}k{t}^2+ct$
• C. $\frac{1}{2}(k{t}^{-2}+ct)$
• D. $k{t}^2+ct$

Answer 1

The correct option is B $\frac{1}{2}k{t}^2+ct$

We know that $a=\frac{dv}{dt}$ so change in velocity in time interval $dt$ will be $dv=adt$

integrating it for ${\int }_{v=0}^{V}dv={\int }_{t=0}^t(kt+c)dt=\frac{1}{2}(k{t}^2+ct)$

or $V=\frac{1}{2}k{t}^2+ct$ i.e. velocity at the end of time $t$