Average velocity - Definition, Formulae, Practice problems, FAQs

Let's say Virat Kohli plays ODI and scores 300 runs in 3 innings. He scores 100 runs in each innings. If one were to ask you what is his average runs, then you would say it is 100. But what if he scored 0, 180, 120 runs in the first, second and third innings respectively ? Then in that case calculating the average runs would become complicated. The same logic applies to calculating the velocity. What if a car traveled at one third of the displacement during a journey at a velocity of 10 km/hr, and the next two thirds at a velocity of 20 km/hr? Or what if it travels with 10 km/hr in the first 1 hour, and 20 km/hr in the next two hours? In such cases, the calculation of average velocity becomes complicated. In this article, we will explore average velocity in detail.

Table of contents

Velocity
Average velocity
Average velocity for different cases
Practice problems
FAQs

Velocity

Generally speaking,

Velocity (v) is the ratio of the displacement to the time taken.

$v = \frac{D i s p l a c e m e n t}{T i m e t a k e n}$

The unit of velocity is m/s. It is a vector quantity.

Average velocity

Average velocity vavg is defined as the ratio of the total displacement to the total time taken.

$v_{a v g} = \frac{T o t a l d i s p l a c e m e n t}{T o t a l t i m e t a k e n}$

Average velocity for different cases

If the body travels different displacements s1, s2, s3 with velocities v1, v2 and v3. The total displacement traveled would be=s1+s2+s3+...

Total time taken $= \frac{s_{1}}{v_{1}} + \frac{s_{2}}{v_{2}} + \frac{s_{3}}{v_{3}}$

Average velocity $v_{a v g} = \frac{s_{1} + s_{2} + s_{3}}{(\frac{s_{1}}{v_{1}} + \frac{s_{2}}{v_{2}} + . .)}$

Assuming s1=s2=s, i.e the body travels equal displacements with different velocities,

$v_{a v g} = \frac{2 s}{s (\frac{1}{v_{1}} + \frac{1}{v_{2}})}$

Simplifying we get, $v_{a v g} = \frac{2 v_{1} v_{2}}{v_{1} + v_{2}}$

If a body travels with velocities v1, v2, v3…. during time intervals t1, t2, t3... etc, the total displacement traveled in that case would be $s = {v_1 t_1 + v_2 t_2 + v_3 t_3 + . . .}_$

Total time taken, t1+t2+t3+...

$v_{a v g} = \frac{v_{1} t_{1} + v_{2} t_{2} + v_{3} t_{3} + . . .}{t_{1} + t_{2} + t_{3} + . . .}$

3. If t1=t2=t3=tn=t , then

$v_{a v g} = \ f r a c {v_1 t_+ v_2 t_+ v_3 t_+ . . .} {t_+ t_+ t_+ . . .} = \ f r a c {{(v}_1 + v_2 + v_3 + . . . v_n) t_} {n t}$

$= \frac{{(v}_{1} + v_{2} + v_{3} + . . . v_{n})}{n}$

Practice problems

Q. Jagadeesh, on driving his way to school, calculates the velocity for the trip to be 20 km/hr. After reaching the school he found the school was closed. So he immediately started returning home. While on his return trip, due to less traffic, he calculates the velocity to be 40 km/hr. Calculate Jagadeesh's average velocity for the entire journey.

26.67 kmhr-1 (b) 33.34 kmhr-1 (c) 50 kmhr-1 (d)30 kmhr-1

A. a

Let t1 and t2 be the time taken for Jagadeesh to go to school and then back home. If s is the displacement, then

$t_{1} = \frac{s}{20}$ hour and $t_{2} = \frac{s}{40}$ hour

Total time taken $t_{1} + t_{2} = \frac{s}{20} + \frac{s}{40} = \frac{3 s}{40}$

Total displacement=2 s

∴ Average Velocity $= \frac{2 s}{3 s} \times 40 = \frac{80}{3} = 26.67 k m {h r}^{- 1}$

Q. A particle covers half the displacement with a velocity v0. The remaining part of the displacement was covered with a velocity v1 for half the time and with velocity v2 for the remaining half of the time. Calculate the average velocity of the particle.

A. According to the question,

$\frac{s}{2} = v_{0} t_{1}, t_{1} = \frac{s}{2 v_{0}}$

If t is the time taken by the particle to travel remaining displacement s/2, then

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

Average velocity =st1+t=s(s2v0)+sv1+v2=2v0(v1+v2)v1+v2+2v0

Q. A car travels first half of the displacement between two places with a velocity of 40 km/hr and the second half at 60 km/hr. Calculate the average velocity for the entire journey.

26.67 kmhr-1 (b) 30 kmhr-1 (c) 48 kmhr-1 (d) 40 kmhr-1

A. Given,

v1=40 km/hr and v2=60 km/hr.

The average velocity when the car covers equal displacements would be,

$v_{a v} = \frac{s}{\frac{s / 2}{v_{1}} + \frac{s / 2}{v_{2}}}$

$v_{a v} = \frac{2 v_{1} v_{2}}{v_{1} + v_{2}} = \frac{2 (40) (60)}{40 + 60} = 48 k m / h r$

Q. A girl is running around a circular track with a uniform speed of 10 ms-1. What is the average velocity for movement of the girl from A to B ( in ms-1)?

(a) 10 (b)40 (c) 10 (d) None of these

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A. d

The displacement made by the girl is AB

$| \vec{A B} | = d = 2 r = 2 \times 20 = 40 m$

$v_{a v} = \frac{d i s p l a c e m e n t}{t i m e} = \frac{2 R}{π R / v} = \frac{2 v}{π} = \frac{2 \times 10}{π} = \frac{20}{π} m / s$

FAQs

Q. What is the difference between average speed and average velocity ?
A. Average speed is the ratio of the total distance to the total time taken. On the other hand, average velocity is the ratio of the total displacement to the total time taken. Average speed is a scalar, and average velocity is a vector.

Q. Can the average velocity be zero?
A. Yes, when the total displacement of the body is zero, then its average velocity is zero. For instance, when a body completes a circular path, then its total displacement is zero, meaning its average velocity should be zero.

Q. Why do we need average velocity?
A. The displacement of the body may be different at different intervals of time during the journey. Hence to describe the entire motion properly, we need to take the total displacement into account. Hence, average velocity comes into picture in such a case.

Q. A body while in motion has its initial and final velocity positive. Can its average velocity be negative?
A. Yes. If the object has moved in the negative direction, then its velocity would have been negative. If the resultant displacement is negative, then its average velocity will be negative despite the fact that its initial and final velocity are positive.

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