Question

A train starts from a station and stops at another in 4 minutes. The max speed attained is 72 kmph (20 m/s).The distance travelled with uniform velocity assuming acceleration nd retardation of the train to be numerically equal is?
1) 4.8 km
2) 4.2 km
3) 4.0 km
4) 3.6 km​

Answer 1

Answer:2.4 km

u=0

a=20m/s = 75km/ph

T=4min=240s

V=u+at

Substitute values below:-

20=0+a×240

a=1/12m/s2

Now

S=ut+1/2 a t square

S=0+1/2x 1/12x 240x 240

S=2400m

S=2.4 km

Explanation:

20m/s = 75km/ph

T=4min=240s

V=u+at

Substitute values below:-

20=0+a×240

a=1/12m/s2

Now

S=ut+1/2 a t square

S=0+1/2x 1/12x 240x 240

S=2400m

S=2.4 km

Explanation:

20m/s = 75km/ph

T=4min=240s

V=u+at

Substitute values below:-

20=0+a×240

a=1/12m/s2

Now

S=ut+1/2 a t square

S=0+1/2x 1/12x 240x 240

S=2400m

S=2.4 km

Explanation:

20m/s = 75km/ph

T=4min=240s

V=u+at

Substitute values below:-

20=0+a×240

a=1/12m/s2

Now

S=ut+1/2 a t square

S=0+1/2x 1/12x 240x 240

S=2400m

S=2.4 km

Explanation: