Question

A car travels half of the distance with constant velocity 40km/h and another half with a constant velocity of 60km/h along a straight line. Find the average velocity of the car???

Answer 1

let x be the distance travelled by the car..x/2 be half the distance now time t taken by the car with a velocity 40km/hr is x/80hr (time=distance/speed) time t taken by the car with a velocity 60km/hr is x/120 hr. 
now avg velocity= x/2+x/2
                           x/80+x/120

                          = x*9600
                                200x

                           = 48km/hr 

Answer 2

"The average velocity of the car is $48 \mathrm{kmhr}^{-1}$

Solution:

We know that,

Velocity $=\frac{\text {Displacement}}{\text {Time}}$

Let d be the distance travelled by the car.

Let $t_{1} and t_{2}$ indicate the time period for which the velocity is $40 {kmhr}^{-1} and 60 {kmhr}^{-1}$respectively.

$t_{ 1 }\quad =\quad \frac { \left( \frac { d }{ 2 } \right)}{ 40 } =\frac { d }{ 80 }$

$t_{ 2 }\quad =\quad \frac { \left( \frac { d }{ 2 } \right)}{ 60 } =\frac { d }{ 120 }$

Average velocity $=\frac{d}{t_{1}+t_{2}}$

Average velocity$=\frac{d}{\left[\left(\frac{d}{80}\right)+\left(\frac{d}{120}\right)\right]}$

$=\frac{240}{[3+2]}$

$=\frac{240}{5}$

Average velocity =$48 {kmhr}^{-1}$"