Question

A body travelling along a straight line tranversed one third of the total distance with a velocity 4 ms-1. The remaining part of the distance was covered with a velocity 2 ms-1 for half the time and with velocity 6ms-1 for the other half of time. What is the mean velocity averaged over the whole time of motion?

Answer 1

First of all lets see the details of the question. Question says that a body travelling along a straight line traversed one third of the total distance with a velocity, so say,
$\frac{s}{3}\to v_0(4m/s)$
where s is total distance covered.
Now the question says that, the remaining part of the distance was covered with a velocity 2 ms^-1 for half the time and with velocity 6 ms^-1 for the other half of time, that is,
$$\begin{gathered} \frac{t_1}{2}\to v_1(2m/s) \\ and, \\ \frac{t_2}{2}\to v_2(6m/s) \end{gathered}$$

Now, the expression for the mean of average velocity is given as,

where t is total time taken and is given as,
$t=t_0+t_1+t_2$
So, the mean equation for av. velocity will become,
... (1)
Using equation for straight line motion and the data given in the question we can write,
$$\begin{gathered} \frac{s}{3}=v_0{t}_0 \\ and, \\ \frac{2}{3}s=v_1{t}_1+v_2{t}_2 \end{gathered}$$
As per the question, $t_1=t_2=\frac{1}{2}t$, so the above equation can be given as,
$\frac{2}{3}s=\frac{t}{2}\left(v_1+v_2\right)$
Similarly equation (1) can be rewritten as,

So, substituting for t and t₀ in the above equation we get,

On substituting the data in above equation we get,