Question

A motorcyclist drives from A to B with a uniform speed of $30km{h}^{-1}$ and returns with a speed of $20km{h}^{-1}$. Find the average speed.
$A. 24km{h}^{-1}$
$B. 25km{h}^{-1}$
$C. 12km{h}^{-1}$
$D. 37km{h}^{-1}$

Answer 1

Hint: The total distance driven by motorcyclist is twice the distance between A and B. To solve this problem, use the formula for average speed in terms of total distance covered and total time taken. Now, find the total time taken by adding the time taken to drive from A to B and time taken to drive from B to A. Calculate the values of time taken to drive from A to B and from A to B. Now substitute these values in the formula for average speed and find its value.

Formula used:
${S}_{avg}= \dfrac {D }{T}$
$S= \dfrac{d}{t}$

Complete answer:
Given: Speed of the motorcyclist while driving from A to B, ${S}_{AB}= 30km{h}^{-1}$
           Speed of the motorcyclist while driving from B to A, ${S}_{BA}= 20km{h}^{-1}$
Let the distance between point A and B be x.
$\therefore d=x$
We know, the formula for average speed is given by,
${S}_{avg}= \dfrac {D }{T}$ …(1)
Where, ${S}_{avg}$ is the average speed of the motorcyclist
             D is the total distance covered
             T is the total time taken
The motorcyclist dives from A to B and returns to A. So, the total distance will be given by,
$D= x+x$
$\Rightarrow D= 2x$ …(2)
Time taken by the motorcyclist to drive from A to B and B to A will be given by,
$T= {t}_{AB}+{t}_{BA}$ …(3)
Where, ${t}_{A}$ is the time taken to drive from A to B
            ${t}_{BA}$ is the time taken to drive from B to A
The formula for speed is given by,
$S= \dfrac{d}{t}$
Rearranging the above equation we get,
$t= \dfrac {d}{S}$ …(4)
Using equation. (4), the time taken to drive from A to B will be,
${t}_{AB}= \dfrac {d}{{S}_{AB}}$
Substituting values in above equation we get,
${t}_{AB}= \dfrac {x}{30}$ …(5)
Similarly, the time taken to drive from B to A will be,
${t}_{BA}= \dfrac {d}{{S}_{BA}}$
Substituting values in above equation we get,
${t}_{BA}= \dfrac {x}{20}$ …(6)
Now, substituting the equation. (5) and (6) in equation. (3) we get,
$T= \dfrac {x}{30}+\dfrac {x}{20}$
$\Rightarrow T= \dfrac {50x}{600}$
$\Rightarrow T= \dfrac {x}{12}$ …(7)
Substituting equation. (2) and (7) we get,
${S}_{avg}= \dfrac {2x}{\dfrac {x}{12}}$
$\Rightarrow {S}_{avg}= 2 \times 12$
 $\Rightarrow {S}_{avg}= 12km{h}^{-1}$
Thus, the average speed is $12km{h}^{-1}$.

So, the correct answer is option C i.e. $12km{h}^{-1}$.

Note:
Students generally get confused between average speed and average velocity. So, they must have clear knowledge of these two terms. There are few similarities and few differences between average speed and average velocity. Both these terms are an average of some length by time taken. S.I. units and other standard units of measurement are the same for both. While the difference between them is that the average speed is a scalar quantity and thus, does not depend on the direction of the object. Whereas average velocity is a vector quantity thus, it depends on the direction of the object.