Question

A car travels a distance 50 km with a velocity 25 km per hour and then 60 km with a velocity 20 km per hour in the same direction. Calculate the average velocity of car.
(A) $11km{h^{ - 1}}$
(B) $22.5km{h^{ - 1}}$
(C) $22km{h^{ - 1}}$
(D) $45km{h^{ - 1}}$

Answer 1

Hint :
In order to solve this problem first we have to calculate the time taken by car to travel 50 km distance with 25 km per hour.
After then calculate the time taken by car to travel 60 km distance with 20 km per hour.
At last put these values in following expression, we get average velocity of car i.e.,
Average velocity $ = $ Total displacement $/$ Total time

Complete step by step solution :
Given that the car travels first 50 km with velocity 25 km per hour. So, the time taken by car is
Time $ = $ Displacement $/$ Velocity
${t_1} = \dfrac{{{d_1}}}{{{v_1}}}$
$ \Rightarrow {t_1} = \dfrac{{50}}{{25}} = 2hr$ …..(1)

Now the car travels the next 60 km with velocity 20 km per hour. So, time taken by car is
${t_2} = \dfrac{{{d_2}}}{{{v_2}}}$
$\Rightarrow {t_2} = \dfrac{{60}}{{20}} = 3hr$ …..(2)

We know that average velocity is given as
Average velocity $ = $ Total displacement $/$ Total time
Average velocity $ = \dfrac{{{d_1} + {d_2}}}{{{t_1} + {t_2}}}$
$ = \dfrac{{50 + 60}}{{2 + 3}}$
$\Rightarrow {v_{avg}} = \dfrac{{110}}{5}$
$\Rightarrow {v_{avg}} = $ 22 km per hour

Hence, the average velocity of a car is 22 km per hour.

So, option C is the correct answer.

Note :
- Many times, students may get confused between average velocity and instantaneous velocity.
- Average velocity $ = $ Total displacement travelled by body $/$ Total time taken by body
${\overrightarrow v _{avg}} = \dfrac{{\Delta \overrightarrow r }}{{\Delta t}}$
When $\Delta t \to 0$ then average velocity is converted into instantaneous velocity.
So, ${\overrightarrow v _{inst}} = \dfrac{{d\overrightarrow r }}{{dt}}$
Rate of change in displacement with respect to time.