Question

A particle moves with uniform acceleration along a straight line `AB`. Its velocities at `A and B` are `2 m//s and 14 m//s` respectively. `M` is the mid-point of `AB`. The particle takes `t_1` seconds to go from `A` to `M` and `t_2` seconds to go from `M` to `B`. Then `t_2 : t_1` is.
A. `1 : 1`
B. `2 : 1`
C. `1 : 2`
D. `3 : 1`

Answer 1

Correct Answer - C
`v^2 = 2^2 + 2 as`
`rArr v^2 - 4 = as`
`14^2 - v^2 = (2as)/(2)`…(i)
`rArr 196 - v^2 = as`….(ii)
From (i) and (ii)
`v^2 - 4 = 196 - v^2 rArr v = 10 m//s`
Now `t_1 = (v - 2)/(a) = (10 - 2)/(a) = (8)/(a)`
`t_2 = (14 - v)/(a) = (14 - 10)/(a) = (4)/(a) rArr (t_2)/(t_1) = (4//a)/(8//a) = (1)/(2)`.