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
Answers
Answered by
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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:
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