Question

A train travels from one station to another
at a speed of 40 kmph and returns to the
first station at a speed of 60 km ph. Then
average speed of the train is....... kmph.​

Answer 1

Answer:

48 km/h

Explanation:

Net displacement of the train in coming back to the first station is zero i.e. S=0 .

Average velocity V

avg

=

t

S

=0

Average speed of the train when in moving equal distances is given by ∣V∣

avg

=

v

1

+v

2

2v

1

v

2

⟹ ∣V∣

avg

=

40+60

2×40×60

=48 km/h

Answer 2

Average speed =

Total Distance ÷ Total Time

Let 'D 'be the distance from one station to another

Let t1 & t2 be the time taken to reach and return respectively

first case

speed 40 kmph

40= D÷ t1

second case

speed 60 kmph

60= D/t2

you have D= 40*t1

also D = 60*t2

if you equate it you have t2 = (2/3) t1

u have Total Distance = D+D = 2D

Total Time= t1 + t2 = t1 + (2/3)t1 = (5/3)t1

substituting

Average speed = 2*(40 t1) ÷ (5/3) t1

Average speed = 48 kmph.....