Question

A man covers a distance of 40 km.he covers first 10km at 10km/h, second 10km at 20km/h, third 10km at 30km/h and last 10km at 40km/h. What is the average speed of man?

Answer 1

Required average speed = 19.2 km/h

Answer 2

$\Large\underline{\underline{\sf \blue{Given}:}}$

• Distance covered by man = 40km

• He covers

First distance $\sf{(s_1)}$ =10km

Speed of first Distance $\sf{(v_1)}$=10km/h

Second distance $\sf{(s_2)}$=10km

Speed of second Distance $\sf{(s_2)}$=20km/h

Third distance $\sf{(s_3)}$=10km

Speed of third distance $\sf{(v_3)}$=30km/h

Last Distance $\sf{(s_4)}$=10km

Speed of last Distance $\sf{(v_4)}$=40km/h

$\Large\underline{\underline{\sf \blue{To\:Find}:}}$

• Average Speed of man = ?

$\Large\underline{\underline{\sf \blue{Solution}:}}$

We know :

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

Total Time :-

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

$\implies{\sf t_1=\dfrac{v_1}{s_1} }$

$\implies{\sf t_1=\dfrac{10}{10}\:\:\implies \: 1hr }$

$\implies{\sf t_2=\dfrac{s_1}{v_2}}$

$\implies{\sf t_2=\dfrac{10}{20}\:\implies\:\dfrac{1}{2}hr}$

$\implies{\sf t_3=\dfrac{s_3}{v_3}}$

$\implies{\sf t_3=\dfrac{10}{30}\:\:\implies\dfrac{1}{3}hrs}$

$\implies{\sf t_4=\dfrac{s_4}{v_4} }$

$\implies{\sf t_4=\dfrac{10}{40}\:\:\implies\:\dfrac{1}{4}hr }$

${\boxed{\sf Total\:time(t)=t_1+t_2+t_3+t_4 }}$

$\implies{\sf t=1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}}$

$\implies{\sf t=\dfrac{25}{12}hr }$

Total Distance

s = 40km

Average Speed :

$\implies{\sf V_{av}=\dfrac{40}{\dfrac{25}{12}}}$

$\implies{\sf V_{av}=\dfrac{480}{25} }$

$\implies{\sf V_{av}=19.2\:km/h }$

$\Large\underline{\underline{\sf \blue{Answer}:}}$

•°• Average speed of man is 19.2km/h