Question

If velocity of a particle is $v\left(t\right)=(4t+6)\text{m/s}$, the magnitude of average velocity of the particle in $10\text{sec}$ is

• A. $20\text{m/s}$
• B. $30\text{m/s}$
• C. $22\text{m/s}$
• D. $26\text{m/s}$

Answer 1

The correct option is D $26\text{m/s}$

Given, $v\left(t\right)=(4t+6)$
As we know that,
Displacement $s={\int }_0^{10}(4t+6)dt$
$\Rightarrow s={[2t^2+6t]}_0^{10}$
$\Rightarrow s=(2\times {10}^2+6\times 10)-0=260\text{m}$
time taken by the particle is $t=10\text{sec}$
Hence, average velocity of the particle is given by
$\frac{260}{10}=26\text{m/s}$