Question

A body moves for 5 hours at the speed of 30km/hr and 29km/hr next 5hours then its average speed will be

Answer 1

GIVEN :–

• A body moves for 5 hours at the speed of 30km/hr and 29km/hr next 5hours.

TO FIND :–

• Average speed = ?

SOLUTION :–

• We know that –

$\\ \longrightarrow \: \large { \boxed{ \sf Speed = \dfrac{Distance}{Time} }}$

▪︎ Distance when body moves with 30km/hr for 5 hr. is –

$\\ \implies \: \sf 30 = \dfrac{D_1}{5}$

$\\ \implies \: \sf D_1 = 30 \times 5$

$\\ \implies \: \sf D_1 = 150 \: km$

▪︎ Distance when body moves with 29km/hr for 5 hr. is –

$\\ \implies \: \sf 29 = \dfrac{D_2}{5}$

$\\ \implies \: \sf D_2 = 29 \times 5$

$\\ \implies \: \sf D_2 = 145 \: km$

• Total Distance = 150 + 145 = 295 km

• Total time = 5 + 5 = 10 hr.

▪︎ As we know –

$\\ \longrightarrow \: \large { \boxed{ \sf Average \: \: Speed = \dfrac{Total \: \: Distance}{Total \: \: Time} }}$

$\\ \implies \: \sf Average \: \: Speed = \dfrac{295}{10}$

$\\ \implies \large{ \boxed{ \sf Average \: \: Speed = 29.5 \: \dfrac{km}{hr} }}$

Answer 2

$\red{\color{white}{\fcolorbox{cyan}{black}{Answer:-}}}$

${\boxed{\sf{Distance = Speed \times Time}}}$

• Time taken (t₁) = 5 h
• Speed (v₁) = 30 km/h
• Distance (d₁) = 30 × 5 = 150 km

• Time taken (t₂) = 5 h
• Speed (v₂) = 29 km/h
• Distance (d₂) = 29 × 5 = 145 km

${\boxed{\sf{Average~ Speed = \frac{TotalDistance}{Total~ ~Time~Taken}}}}$

$\sf~ Average \: Speed = \frac{d1+d2}{t1+t2} \\\\ \sf~Average \: Speed = \frac{150+145}{5+5} \\\\ \sf~Average \: Speed = \frac{295}{10}$

${\red{\sf{Average \: Speed = 29.5 ~km/h}}}$