Question

A particle moving in a straight line covers half the distance with a speed of $3\text{m/s}$. The other half of the distance is covered in two equal time intervals with speed of $4.5\text{m/s}$ and $7.5\text{m/s}$ respectively. The average speed of the particle during this motion is

• A. $4.0\text{m/s}$
• B. $5.0\text{m/s}$
• C. $5.5\text{m/s}$
• D. $4.8\text{m/s}$

Answer 1

The correct option is A $4.0\text{m/s}$

Let total distance $=x\text{m}$
and half of distance is travelled in $t_1\text{sec}$.
$\Rightarrow v_1\times t_1=\frac{x}{2},v_1=3\text{m/s}$
$\Rightarrow 3\times t_1=\frac{x}{2}$
$\therefore t_1=\frac{x}{6}\text{s}$
Other half of the distance is covered in two equal time intervals i.e $t_2$ and $t_3\text{sec}$.
where $t_2=t_3$

$\Rightarrow \text{Distance travelled in}{t}_2\text{sec}+\text{Distance travelled in}{t}_3\text{sec}=\frac{x}{2}$
$\therefore v_2\times t_2+v_3+t_3=\frac{x}{2}$
where,
$\left[\begin{array}{c}{v}_2=4.5\text{m/s}\\ v_3=7.5\text{m/s}\end{array}\right]$
$\Rightarrow 4.5\times t_2+7.5\times t_2=\frac{x}{2}$
$\Rightarrow t_2=t_3=\frac{x}{24}\text{s}$
$\text{Average Speed}=\frac{\text{Total Distance travelled}}{\text{Total time taken}}$
$v_{\text{avg}}=\frac{x}{\frac{x}{6}+\frac{x}{24}+\frac{x}{24}}$
$\therefore v_{\text{avg}}=\frac{x}{\left(\frac{x}{4}\right)}=4\text{m/s}$