Question

A train is running at a constant speed of 90 km/h on a straight track. A person standing at the top of a boggey moves in the direction of of motion of the train such that he covers 1m on the train each second . The speed of the person with respect of ground is


explain in detail plzzzzzz

Answer 1

Speed of the train = 90 km /s =90*(5/18) m/s
=25 m/s
Speed of the person relative to train = 1m/s

Now speed of the person relative to ground
= (25+1)m/s =26m/s
I hope it's help

Answer 2

This is the case of the motion of body on body. When a train is running with speed 

V=90 km/hr×(5 / 18 m/s)

=25 m/s


and

the speed of the boy running at top of the train's car is 

v=1m/s,

then the velocity of the boy with respect to ground will be,
 v (BG) =V+v

v (BG) =25 m/s+1m/s

vBG=26 m/s