Question

For particle moving along $x-axis$, velocity is given as a function of time as $v=2t^2+\sin t$. At $t=0$, particle is at origin, if the position as a function of time is $\frac{2t^3}{m}-\cos t$, then find $m$.

• A. $3$
• B. $2$
• C. $1$
• D. $4$

Answer 1

The correct option is A $3$

$\begin{array}{c}\frac{ds}{dt}=2t^2+\sin t\left[\because v=\frac{ds}{dt}\right]\\ \Rightarrow {\int }_0^{s}ds=\int \left(2t^2=\sin t\right)dt\\ position\left(s\right)=\frac{2t^2}{3}-\cos t.....\left(1\right)\\ givenpositionofparticles\left(\frac{2t^2}{m}-\cos t\right)....\left(2\right)\\ compareequation\left(1\right)and\left(2\right)\\ \Rightarrow \frac{2t^3}{3}-\cos t=\frac{2t^3}{m}-\cos t\\ m=3\end{array}$