Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

A car covers the first $2km$ of the total distance of $4km$ with a speed of $30km/h$ and the remaining distance of $2km$ with a speed of $20km/h$. The average speed of the car is:
A. $24km/h$
B. $25km/h$
C. $50km/h$
D. $26km/h$

seo-qna
Last updated date: 25th Jul 2024
Total views: 385.5k
Views today: 11.85k
Answer
VerifiedVerified
385.5k+ views
Hint:Distance covered by anybody is defined as the length of path covered by the person in moving from one point from another. And speed is defined as the rate of change of distance with respect to time.

Complete step by step answer:
Distance: Distance$\left( S \right)$ covered by anybody is defined as the length of path covered by the person in moving from one point to another. Its SI unit is metre$\left( m \right)$.
Speed: Speed$\left( v \right)$ is defined as the rate of change of distance with respect to time. Which means,
$v = \dfrac{{\Delta S}}{{\Delta t}}$

Instantaneous Velocity: Instantaneous velocity$\left( {{v_{inst}}} \right)$ is defined as the derivative of displacement over time at a particular moment in time. In a graph, it is defined as the slope of the graph at that particular time.
Average speed: Average speed$\left( {\overline v } \right)$ is defined as the ratio of total distance covered to total time taken to cover that distance.
So, this case is solved in two parts since there are two speeds given.

First part: Let the distance covered in the first part be ${S_1}$, which is covered in time ${t_1}$ with speed ${v_1}$. So,
${S_1} = 2km$
$\Rightarrow{v_1} = 30km/h$
$\because {v_1} = \dfrac{{{S_1}}}{{{t_1}}}$
$ \Rightarrow {t_1} = \dfrac{{{S_1}}}{{{v_1}}}$
$ \Rightarrow {t_1} = \dfrac{2}{{30}}h = \dfrac{1}{{15}}h$

Second part: Let the distance covered in the first part be ${S_2}$, which is covered in time ${t_2}$ with speed ${v_2}$. So,
${S_2} = 2km$
${v_2} = 20km/h$
$\because {v_2} = \dfrac{{{S_2}}}{{{t_2}}}$
$ \Rightarrow {t_2} = \dfrac{{{S_2}}}{{{v_2}}}$
$ \Rightarrow {t_2} = \dfrac{2}{{20}}h = \dfrac{1}{{10}}h$
So, average speed is,
$\overline v = \dfrac{{{S_1} + {S_2}}}{{{t_1} + {t_2}}}$
$ \Rightarrow \overline v = \dfrac{4}{{\dfrac{1}{{15}} + \dfrac{1}{{10}}}}$
$ \Rightarrow \overline v = \dfrac{4}{{\dfrac{{2 + 3}}{{30}}}}$
$ \Rightarrow \overline v = \dfrac{{4 \times 30}}{5}$
$\therefore \overline v = 24km/h$

Thus the correct answer is option A.

Note: For special cases like this there are simpler formulae to calculate average speed. For a case in which the journey is covered in equal time intervals with different speeds, the average speed can be given by, $\overline v = \dfrac{{{v_1} + {v_2}}}{2}$. And when the journey is covered in equal distance intervals with different speeds, the average speed can be given by, $\overline v = \dfrac{{{v_1}{v_2}}}{{{v_1} + {v_2}}}$.