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A train is approaching a platform with a speed of 10 km per hour when it is at a distance of 2 km from the platform brakes are applied so that a uniform retardation is produced at the same time a bird sitting on a pole near the platform price with a speed of 50 km per hour towards the train and touches the front of the train and comes back to the pole if the if the motion is repeated until the train comes back to rest the total distance travelled by the body is A 10 km B 60 km C 20 km D 15 km

A train is approaching a platform with a speed of 10 km per hour when it is at a distance of 2 km from the platform brakes are applied so that a uniform retardation is produced at the same time a bird sitting on a pole near the platform price with a speed of 50 km per hour towards the train and touches the front of the train and comes back to the pole if the if the motion is repeated until the train comes back to rest the total distance travelled by the body is
A 10 km
B 60 km
C 20 km
D 15 km

Grade:11

2 Answers

Arun
25750 Points
5 years ago
V^2 = u^2 – 2as
 
using this v =0 and u=10km/hr s=2
we get -100 = 2a(2)
a=-25
time to stop the train 
v=u +at 
0=10 - 25t
t=2/5hr 
so bird will travel for d distance
d=vt
d=50(2/5)
d=20km
Saurabh Koranglekar
askIITians Faculty 10335 Points
5 years ago

V2=u2+2as
0=100+2×a×2
Hence a=-25km/hr2

v=u+at
t= -10/(-25)
= 2/5

bird traveled for (2/5 hrs )@ 50km/hr

dist coveredis 0.4*50 = 20 km
ans c


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