An object moves with velocity of 20km/hr in the first 1/3rd of total distance and moves with 60km/hr in the second 1/3rd of total distance. What is average velocity up to 2/3 rd of total distance?

ArunrajuNaspuri
Assume total length=`L kilometers`
Average velocity=Total distance up to`2/3`part /Time up to `2/3`part
`V_(avg)=(d_1+d_2)/(t_1+t_2)` `" "color(blue)((1))`

where,
`v_1`=velocity up to `1/3`rd part =20 km/hr,

`t_1`=time taken for car up to `(1/3)^(rd)` part,

`d_1`=distance up to `(1/3)^(rd)` part =`L/3`kilometers

`v_2`=velocity from `(1/3)^(rd)`part to `(2/3)^(rd)`part=60km/hr,

`t_2`=time taken from `(1/3)^(rd)`part to `(2/3)^(rd)`part,

`d_2`=distance from `(1/3)^(rd)`part to `(2/3)^(rd)`part= `L/3`km.

For first `1/3`rd part:
given that,

`v_1=20(km)/(hr)`

`d_1/t_1=20`km/hr [since from velocity definition]

`L/3/t_1=20`km/hr [since`d_1=L/3`]

`L/(t_1)=60`km/hr `" "color(blue)((2))`

from `(1/3)^(rd)`part to `(2/3)^(rd)`part:

given that,

`v_2=60`km/hr,

`d_2/t_2`=60km/hr,

`L/3/t_2`=60km/hr [since `d_2=L/3km`]

`L/(3t_2)`=60km/hr `" "color(blue)((3))`

Divide equation(2) with equation(1),

We get `t_1=3t_2` `" "color(blue)((4))`

From equation(1),

`V_(avg)=(d_1+d_2)/(t_1+t_2)`

`V_(avg)=((L/3)+(L/3))/(3(t_2)+t_2)`[since from equation(4)]

`V_(avg)=1/2*(L/(3t_2))`

`V_(avg)=1/2*(60(km)/(hr))`[since from equation (3)]

`V_(avg)=30(km)/(hr)`

A car cover 1/3 distance with speed 20km/hr and 2/3 with 60km/hr. average speed is `" "color(blue)30` km/hr.