IIT JEE , AIEEE Forum - Physics , Mechanics

ashish_banga

let acceleration in first 100 s = a

then distance covered in first 100 s = 1/2 a X 100 X100
= 5000a m

velocity at the end of 100s = a X 100
= 100a m/ s

distance covered by it when it is moving with constant speed
= 100 a X 5 X 60 m
= 30000 m

let deceleration = A
then 0 = v - A X 150
v = A X 150
100 a = A X 150
A = 2/3 a
distance traveled during this time interval = 1/2 A X 150 X150
= 7500a

total distance traveled = 5000a + 30000a + 7500a
= 42500 a
= 4250 m
this gives a = 1/10 m/ second square

rest u can find

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