Question

A point traversed half the distance with a velocity

v_{0} .

The remaining part of the distance was covered with velocity

v_{1}

for half the time, and with velocity

v_{2}

for the other half of the time. Find the average speed of the point average over the whole time of motion.

Accepted Answer

Consider $2 s$ is the total distance covered by the particle between its initial $v_{1}$ point and the final $(F)$ point.

First half distance $s$ is covered with velocity $v_{0}$ . So, time required to reach from point $A$ to point $B$ $t_{A B} = \frac{s}{v_{0}}$ .

Suppose the time required to reach the particle from point $B$ to point $F$ is $t_{B F}$ and particle covered this distance $s$ in two parts .
Now consider second part of journey $s = s_{1} + s_{2} . . . (1)$

First part of distance that is $s_{1}$ is covered with velocity $v_{1}$ in time $\frac{t_{B F}}{2}$

So, $s_{1} = v_{1} (\frac{t_{B F}}{2})$

Second part of distance that is $s_{2}$ is covered with velocity $v_{2}$ in time $\frac{t_{B F}}{2} .$

So, $s_{2} = v_{2} (\frac{t_{B F}}{2})$

Putting $s_{1}$ and $s_{2}$ in equation $(1)$ . We get

$s = v_{1} (\frac{t_{B F}}{2}) + v_{2} (\frac{t_{B F}}{2})$

$t_{B F} = \frac{2 s}{v_{1} + v_{2}}$

We know, $Average speed = \frac{Total distance}{Total time taken}$

Hence, average speed $= \frac{2 s}{t_{A B} + t_{B F}} = \frac{2 s}{\frac{s}{v_{0}} + \frac{2 s}{(v_{1} + v_{2})}}$

$Average speed = \frac{2 v_{0} (v_{1} + v_{2})}{(v_{1} + v_{2} + 2 v_{0})}$

B M Sharma Solutions for Chapter: Kinematics I, Exercise 10: ILLUSTRATION 4.10

B M Sharma Physics Solutions for Exercise - B M Sharma Solutions for Chapter: Kinematics I, Exercise 10: ILLUSTRATION 4.10

Questions from B M Sharma Solutions for Chapter: Kinematics I, Exercise 10: ILLUSTRATION 4.10 with Hints & Solutions

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