Question

If velocity of a particle is $v\left(t\right)=(3t^2-2t+4)\text{m/s}$, the magnitude of average velocity of the particle in $10\text{sec}$ is

• A. $94\text{m/s}$
• B. $88\text{m/s}$
• C. $104\text{m/s}$
• D. $90\text{m/s}$

Answer 1

The correct option is A $94\text{m/s}$

Given, $v\left(t\right)=(3t^2-2t+4)$
As we know that,
Displacement $s={\int }_0^{10}(3t^2-2t+4)dt$
$\Rightarrow s={[t^3-t^2+4t]}_0^{10}$
$\Rightarrow s=({10}^3-{10}^2+4\times 10)-0=940\text{m}$
time taken by the particle is $t=10\text{sec}$
Hence, average velocity of the particle is given by
$\frac{940}{10}=94\text{m/s}$