Question

If velocity of a particle is $v\left(t\right)=t^2-3t\text{m/s}$, the average speed and magnitude of average velocity of the particle in $6\text{sec}$ respectively are

• A. $6\text{m/s},3\text{m/s}$
• B. $4.5\text{m/s},4.5\text{m/s}$
• C. $4.5\text{m/s},3\text{m/s}$
• D. $3\text{m/s},3\text{m/s}$

Answer 1

The correct option is C $4.5\text{m/s},3\text{m/s}$

Given, $v\left(t\right)=t^2-3t$, we will check if the particle has changed its direction or not within the asked time interval, i.e. the time when velocity of the particle becomes zero.
$\Rightarrow t^2-3t=0\Rightarrow t=0,3\text{sec}$
Also, we know that
$x_f-x_i={\int }_0^{t}v\left(t\right)dt={\int }_0^t(t^2-3t)dt$
$\Rightarrow x_f-x_i=\frac{t^3}{3}-\frac{3t^2}{2}$
Let particle starts from origin i.e $x_i=0$
So, at $t=0,x_f=0$,
at $t=3,x_f=\frac{3^3}{3}-\frac{3\times 3^2}{2}=-\frac{9}{2}\text{m}$
at $t=6,x_f=\frac{6^3}{3}-\frac{3\times 6^2}{2}=18\text{m}$

So, the average speed of the particle is $\frac{|-\frac{9}{2}|+|-\frac{9}{2}|+18}{6}=\frac{27}{6}=\frac{9}{2}=4.5\text{m/s}$
and, average velocity is $\frac{18}{6}=3\text{m/s}$