Question

A practical travel 1st half distance with uniform speed of 40m/s and next half distance with uniform speed of 80m/s calculate average speed.​

Answer 1

Let total distance = x m

1st half distance = x/2 m

Speed for 1st half distance = 40 m/s

2nd half distance = x - x/2 = x/2 m

Speed for next half (2nd) distance = 80 m/s

Average Speed : It is defined as total distance travelled per total time

$\to \sf A_{Speed}=\dfrac{Total\ Distance}{Total\ Time}$

$\to \sf A_{Speed}=\dfrac{x}{t_{1st\ 1/2}+t_{2nd\ 1/2}}$

• Time = Distance / Speed

$\to \sf A_{Speed}=\dfrac{x}{\frac{x}{2\times 40}+\frac{x}{2\times 80}}\\\\\to \sf A_{Speed}=\dfrac{x}{x\bigg(\frac{1}{80}+\frac{1}{160}\bigg)}\\\\\to \sf A_{Speed}=\dfrac{160}{3}\\\\\leadsto \sf \pink{A_{Speed}=53.34\ m/s}\ \; \bigstar$

Answer 2

$\underline{\underline{ \sf{\color{lime}{GIVEN:-}} }}$

A particle travels 1st half distance with uniform speed of 40 m/s and next half distance with uniform speed of 80 m/s.

• v₁ = 40 m/s
• v₂ = 80 m/s

$\underline{\underline{ \sf{\color{lime}{TO\ FIND:-}} }}$

The average speed.​

$\underline{\underline{ \sf{\color{lime}{SOLUTION:-}} }}$

Let the particle travel a total distance of x m.

CASE - 1

• Distance, s = x/2
• Speed, v₁ = 40 m/s
• Time, t₁ = s/v₁

$\sf{\longrightarrow t_1=\dfrac{x}{2} \div 40}$

$\sf{\longrightarrow t_1=\dfrac{x}{2} \times \dfrac{1}{40}}$

$\sf{\longrightarrow t_1=\dfrac{x}{80}}$

___________________

CASE - 2

• Distance, s = x/2
• Speed, v₂ = 80 m/s
• Time, t₁ = s/v₂

$\sf{\longrightarrow t_2=\dfrac{x}{2} \div 80}$

$\sf{\longrightarrow t_2=\dfrac{x}{2} \times \dfrac{1}{80}}$

$\sf{\longrightarrow t_2=\dfrac{x}{160}}$

___________________

$\sf{Total\ time=t_1+t_2}$

$\sf{\longrightarrow Total\ time=\dfrac{1x}{80}+\dfrac{1x}{160}}$

$\sf{\longrightarrow Total\ time=\dfrac{3x}{160} \ s}$

___________________

◘ Average speed = Total distance / Total time

→ Average speed = x ÷ 3x/160

→ Average speed = x × 160/3x

→ Average speed = 160/3 m/s

Therefore, the average speed of the particle is 160/3 m/s.