A train travels with a speed of 60 km h^-1 from station A to station B and then comes back with a speed 80 km h^-1 from station B to station A.

Find —

(i) the average speed, and

(ii) the average velocity of train.

As we know,

Avg Speed = $\dfrac{\text{Total Distance}}{\text{Total time}}$

Given,

From, A to B

speed_AB = 60 km h^-1

distance travelled = d_AB = d

time taken = $\dfrac{\text{d}}{60}$

From, B to A

speed_BA = 80 km h^-1

distance travelled = d_BA = d

time taken = $\dfrac{\text{d}}{80}$

Total distance travelled = d + d = 2d    [Eqn. 1]

Total time = $\dfrac{\text{d}}{60}$ + $\dfrac{\text{d}}{80}$    [Eqn. 2]

Substituting the values from the equations 1 and 2 in the formula we get,

$\text {Avg Speed} = \dfrac{\text {2d}}{\dfrac{\text {d}}{60} + {\dfrac{\text {d}}{80}}} \\[0.5em] \text {Avg Speed} = \dfrac{\text {2d}}{\dfrac {\text {4d + 3d}}{240}} \\[0.5em] \text {Avg Speed} = \dfrac{\text {2d}}{\dfrac {\text {7d}}{240}} \\[0.5em] \text {Avg Speed} = \dfrac{\text {2}}{\dfrac {\text {7}}{240}} \\[0.5em] \text {Avg Speed} = \dfrac {2 \times 240}{7} \\[0.5em] \text {Avg Speed} = \dfrac {480}{7} \\[0.5em] \Rightarrow \text {Avg Speed} = 68.57 \text{ km h}^{-1} \\[0.5em]$

Hence, Avg Speed = 68.57 km h^-1

(ii) We know,

Average velocity = $\dfrac{\text {displacement}}{\text {total time}}$

As the train starts from station A and comes back to same station, hence, displacement is zero.

Therefore, the average velocity is also zero.