Question

One car moving on a straight road covers one third of the distance with $20\,Kmh{r^{ - 1}}$ and the rest with $60\,Kmh{r^{ - 1}}$ The average speed is
(A) $40\,Kmh{r^{ - 1}}$
(B) $80\,Kmh{r^{ - 1}}$
(C) $46\dfrac{2}{3}\,Kmh{r^{ - 1}}$
(D) $36\,Kmh{r^{ - 1}}$

Answer 1

Hint: In order to solve this question, we will first calculate the time taken by the car to cover each interval distance with given speeds and then we will use the average speed formula to determine the average speed of the car.

Formula Used:
Average speed of a body is given by,
${v_{average}} = \dfrac{{{S_{total}}}}{{{t_{total}}}}$
where, S and t denote for total distance and total time taken by the body.

Complete step by step solution:
Let us assume that total distance covered by the car is $x\;Km$ then according to the question we have given that, car covers one third of the distance with $20\,Kmh{r^{ - 1}}$ so using speed distance formula we have,
${t_1} = \dfrac{{\dfrac{x}{3}}}{{20}} \\
\Rightarrow {t_1} = \dfrac{x}{{60}}hr \to (i) \\ $
Again, we have given that remaining distance which will be $(x - \dfrac{x}{3}) = \dfrac{{2x}}{3}Km$ is covered by the car with a speed of $60\,Kmh{r^{ - 1}}$ so, again using speed distance formula we have,
${t_2} = \dfrac{{\dfrac{{2x}}{3}}}{{60}} \\
\Rightarrow {t_2} = \dfrac{x}{{90}}hr \to (i) \\ $
Now, let total time taken by the car is $t\;hr$ and average speed is ${v_{average}}$ and total distance is $x\;Km$ so we have,
${v_{average}} = \dfrac{{{S_{total}}}}{{{t_{total}}}}$ here S is the total distance which we assumed is $x\;Km$ and total time will be sum of equations (i) and (ii) which is
${t_{total}} = {t_1} + {t_2} \\
\Rightarrow {t_{total}} = \dfrac{x}{{90}} + \dfrac{x}{{60}} \\
\Rightarrow {t_{total}} = \dfrac{{5x}}{{180}} \\ $
so putting these values in formula ${v_{average}} = \dfrac{{{S_{total}}}}{{{t_{total}}}}$ we get,
${v_{average}} = \dfrac{x}{{\dfrac{{5x}}{{180}}}} \\
\therefore {v_{average}} = 36Kmh{r^{ - 1}} $
Therefore, the average speed of the car is $36\,Kmh{r^{ - 1}}$.

Hence, the correct option is (D) i.e, $36\,Kmh{r^{ - 1}}$.

Note: It should be remembered that, while solving such numericals always convert the all physical quantities in same units and we can also convert given units in to S.I units of speed with the formula as $1Kmh{r^{ - 1}} = \dfrac{5}{{18}}m{\sec ^{ - 1}}$.