Question

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 -

1. The average speed
2. The average velocity of the train.

Answer 1

(i) Given,

From, A to B

speed_AB = 60 km h^-1

distance travelled = d_AB = d

time taken = `"d"/60`

From, B to A

speed_BA = 80 km h^-1

distance travelled = d_BA = d

time taken = `"d"/80`

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

Total time = `"d"/60 + "d"/80`  ...[Eqn. 2]

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

Average speed = `"Total Distance"/"Total time"`

= `"2d"/("d"/60 + "d"/80)`

= `"2d"/(("4d" + "3d")/240)`

= `("2d"/"7d")/240`

= `(2/7)/240`

= `(2 xx 240)/7`

= `480/7`

= 68.57 km h^-1

Hence, Average Speed = 68.57 km h^-1

(ii) Average velocity = `"Displacement"/"total time taken"`

Because the train starts and ends at the same station, the displacement is zero. Thus the average velocity is zero.