Question

a body goes from A to B with a speed of 60 km per hour then returns the starting point at a speed of 40 km per hour find i- the average speed and ii-the average velocity of the body​

Answer 1

GiveN :

• A body goes from A to B with speed of 60 km/h
• And returns from B to A with speed of 40 km/h

To FinD :

• Average speed
• Average Velocity

SolutioN :

$\underbrace{\sf{Average \: Speed}}$

let, the distance from A to B be x m, then we know that :

$\implies \sf{Speed \: = \: \dfrac{Distance}{Time}} \\ \\ \implies \sf{Speed \: = \: \dfrac{2x}{\dfrac{x}{40} \: + \: \dfrac{x}{60}}} \\ \\ \implies \sf{Speed \: = \: \dfrac{2x}{\dfrac{3x \: + \: 2x}{120}}} \\ \\ \implies \sf{Speed \: = \: \dfrac{2x}{\dfrac{5x}{120}}} \\ \\ \implies \sf{Speed \: = \: \dfrac{2x \: \times \: 120}{5x}} \\ \\ \implies \sf{Speed \: = \: \dfrac{2x \: \times \: 24}{x}} \\ \\ \implies \sf{Speed \: = \: 2 \: \times \: 24} \\ \\ \implies \sf{Speed \: = \: 48 \: kmh^{-1}}$

___________________________

$\underbrace{\sf{Average \: Velocity }}$

Average velocity is 0 km/h, because the displacement is zero. As the body starts from A to B and again come to A from B.

$\implies \sf{Velocity \: = \: \dfrac{Displacement}{Time}} \\ \\ \implies \sf{Velocity \: = \: \dfrac{0}{Time}} \\ \\ \implies \sf{Velocity \: = \: 0 \: kmh^{-1}}$

Answer 2

Given :-

A body goes from point A to B at a speed of 60km/h

And again B to A at a speed of 4km/h

To Find :-

Average speed and average velocity

$\bigstar{\boxed{\sf{Average\ speed =\dfrac{Total \ Distance }{Total \ time }}}}$

• Let the distance be n
• Speed = 60km/h and 40km/h

• Now time

$\implies\sf Time _1(A \ to B)= \dfrac{Distance }{speed}\\ \\ \implies \sf T_1= \dfrac{n}{60}\\ \\ \implies \sf Time_2(B\ to \ A)= \dfrac{n}{40}$

Now the average speed

$\implies\sf Averege \ speed =\dfrac{Total \ Distance }{Total \ time }\\ \\ \implies \sf Average\ speed = \dfrac{n+n}{\dfrac{n}{60}+\dfrac{n}{40}}\\ \\ \implies \sf Average\ speed= \dfrac{2n}{\dfrac{2n+3n}{120}}\\ \\ \implies \sf Average\ speed = \dfrac{2n}{\dfrac{5n}{120}}\\ \\ \implies\sf Average \ speed = \dfrac{2\cancel{n}\times 120}{5\cancel{n}} \\ \\ \implies\sf Average\ speed =\cancel{\dfrac{240}{5}}\\ \\ \implies\sf Average \ speed = 48km/h$

$\underline{\textsf{\textbf{\dag \ Average \ speed = 48km/h}}}$

• Now we have to find the average velocity

$\bigstar{\boxed{\sf{Average\ velocity =\dfrac{Displacement }{Time }}}}$

• Displacement becomes zero because he come back to his initial point ! He start from point A and turn back to point A

$\underline{\textsf{\textbf{\dag \ Average \ velocity = 0km/h}}}$