Question

a body goes from A to B with a velocity of 40 m/s and comes back from B to A with the velocity of 60 m/s .what is the (i) average velocity during the whole journey (ii) average speed during the whole journey

Answer 1

24km/h is the answer and

Here, Displacement is zero so Average Velocity= Total          displacement/Total Time=0

(here displacement is 0) and 0/x=0

Now, AV Speed = 1st speed × 2nd speed/1st speed + 2nd speed

60×40/60+40

2400/100=24km/h

Answer 2

Given:

• Speed from A to B = 40 m/s
• Speed from B to A = 60 m/s

To find:

• Average speed and average velocity

Calculation:

Average velocity is the total displacement divided by the total time taken to cover the displacement.

$avg. \: v = \dfrac{displacement}{time}$

• Here, the starting and stopping point of the object is same after the whole journey. Hence the displacement is zero.

$\implies \: avg. \: v = \dfrac{0}{t}$

$\implies \: avg. \: v = 0$

Now, average speed will be :

$avg. \: s = \dfrac{distance}{time}$

$\implies \: avg. \: s = \dfrac{d + d}{ \dfrac{d}{40} + \dfrac{d}{60} }$

$\implies \: avg. \: s = \dfrac{2d}{ \dfrac{d}{40} + \dfrac{d}{60} }$

$\implies \: avg. \: s = \dfrac{2}{ \dfrac{1}{40} + \dfrac{1}{60} }$

$\implies \: avg. \: s = \dfrac{2}{ \dfrac{5}{120} }$

$\implies \: avg. \: s = 48 \: m {s}^{ - 1}$

So, average velocity is zero & average speed is 48 m/s .