Question

If $v=(4t^2+5t+1)m/s$ is the expression according to which the velocity of a particle varies, then the expression for instantaneous displacement at any time $t$ will be (Assume $s=0$ at $t=0$)

• A. $\frac{4}{3}{t}^3+\frac{5}{2}{t}^2+t$
• B. $4t^3+t+1$
• C. $\frac{4}{3}{t}^3+\frac{5}{2}{t}^2+t+1$
• D. $3t^3+5t^2+t$

Answer 1

The correct option is A $\frac{4}{3}{t}^3+\frac{5}{2}{t}^2+t$

Velocity of particle can be written as
$v=(4t^2+5t+1)=\frac{ds}{dt}$ (where $s$ is displacement)
$\int ds=\int (4t^2+5t+1)dt$
$s=\frac{4}{3}{t}^3+\frac{5}{2}{t}^2+t+c$

At $t=0,s=0$ (given)
$0=0^3+0+0+c$
$c=0$

$\therefore s=\frac{4}{3}{t}^3+\frac{5}{2}{t}^2+t$