Question

If velocity of a particle is $v\left(t\right)=t^2-5t\text{m/s}$, the average speed and magnitude of average velocity of the particle in $10\text{sec}$ respectively are

• A. $\frac{25}{2}\text{m/s},\frac{25}{3}\text{m/s}$
• B. $\frac{25}{3}\text{m/s},\frac{25}{2}\text{m/s}$
• C. $\frac{125}{12}\text{m/s},\frac{25}{12}\text{m/s}$
• D. $\frac{25}{12}\text{m/s},\frac{125}{12}\text{m/s}$

Answer 1

The correct option is A $\frac{25}{2}\text{m/s},\frac{25}{3}\text{m/s}$

Given, $v\left(t\right)=t^2-5t$, 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.
$\Rightarrow t^2-5t=0\Rightarrow t=0,5\text{sec}$
Also, we know that
$x_f-x_i={\int }_0^{t}v\left(t\right)dt={\int }_0^t(t^2-5t)dt$
$\Rightarrow x_f-x_i=\frac{t^3}{3}-\frac{5t^2}{2}$
Let particle starts from origin i.e $x_i=0$
So, at $t=0,x_f=0$,
at $t=5,x_f=\frac{5^3}{3}-\frac{5\times 5^2}{2}=-\frac{125}{6}\text{m}$
at $t=10,x_f=\frac{{10}^3}{3}-\frac{5\times {10}^2}{2}=\frac{500}{6}\text{m}$
The motion of the particle can be represented as shown

So, the average speed of the particle is $\frac{|-\frac{125}{6}|+|-\frac{125}{6}|+\frac{500}{6}}{10}=\frac{25}{2}=12.5\text{m/s}$
and, average velocity is $\frac{\frac{500}{6}}{10}=\frac{25}{3}\text{m/s}$