Question

A particle moving in a straight line covers half the distance with a speed of $3m{s}^{-1}$. The other half of the distance is covered in two equal time intervals with the speed of $4.5m{s}^{-1}$ and $7.5m{s}^{-1}$, respectively. The average speed of particles during this motion is

• A. $4m{s}^{-1}$
• B. $5m{s}^{-1}$
• C. $5.5m{s}^{-1}$
• D. $4.8m{s}^{-1}$

Answer 1

The correct option is D

$4.8m{s}^{-1}$

Step 1: Given data and assumptions.

Let the total distance covered by the particle be $S$.

The speed of the particle at the first half distance $\frac{S}{2}$ is $3m{s}^{-1}$.

The speed of the particle at the next half distance $\frac{S}{2}$ with two equal interval times is $4.5m{s}^{-1}$ and $7.5m{s}^{-1}$.

Step 2: Finding the average speed of the particle.

We know that,

$V_{avg}=\frac{Totaldis\tan ce}{Totaltimetaken}$

Where $V_{avg}$ is the average speed.

Since, from the figure total time,

$t=t_1+\left(\frac{t_2}{2}+\frac{t_2}{2}\right)$

$\Rightarrow$$t=t_1+t_2$

$\Rightarrow$$t=\frac{{\displaystyle \raisebox{1ex}{$S$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{3}+\frac{{\displaystyle \raisebox{1ex}{$S$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{4.5+7.5}$ $\left[\because time=\frac{dis\tan ce}{speed}\right]$

$\Rightarrow$$t=\frac{5S}{24}$

Now,

$V_{avg}=\frac{S}{t}$

$\Rightarrow$ $V_{avg}=\frac{S}{{\displaystyle \raisebox{1ex}{$5S$}\!\left/ \!\raisebox{-1ex}{$24$}\right.}}$ $\left[\because t=\frac{5S}{24}\right]$

$\Rightarrow$ $V_{avg}=4.8m{s}^{-1}$

Hence, option$\left(D\right)$ is correct.