Question

If a particle has velocity $v\left(t\right)=18-6t\text{m/s}$, the magnitude of average velocity in $10\text{sec}$ is

• A. $-10\text{m/s}$
• B. $12\text{m/s}$
• C. $10\text{m/s}$
• D. $-8\text{m/s}$

Answer 1

The correct option is B $12\text{m/s}$

We know that the average speed is given by $\overline{v}=\frac{\text{Total displacement}}{\text{Total time taken}}$
Also, total displacement is given by
$x_f-x_i={\int }_0^t(18-6t)dt=[18t-3t^2{]}_0^t=18t-3t^2\text{m}$
Let particle be at origin at $t=0\Rightarrow x_i=0$
and, at $t=0,x_f=0$
also, at $t=10,x_f=18\times 10-3\times {10}^2=-120\text{m}$
and total time taken is $10\text{sec}$
Thus, $\overline{v}=-\frac{120}{10}=-12\text{m/s}$
Hence, the magnitude of the average velocity is $12\text{m/s}$