A body travels in a straight line from point A to point B with an initial velocity zero and uniform acceleration, covering 1m during the first second and 39m during the last second. The distance between A and B in metre is

Options

A. 50
B. 100
C. 390
D. 400

Answer

D

Explanation

u= 0

Distance travelled in first 1s,
`s_(1) = 1m = (1)/(2) "at"^(2)`
`rArr` a = acceleration `= 2ms^(-2)`
Velocity after 1s, `u + "at"`,
`=0 + 2 xx 1 = 2 ms^(-1)`
At the end of last second, let velocity is `v ms^(-1)` Then, from
`s = u t+ (1)/(2)"at"^(2)`
we have, `39 = v xx 1 + (1)/(2) (2) xx 1^(2) rArr v = 38 ms^(-1)`
So, in unknown time duration of t seconds, velocity changes from `2 ms^(-1)` to `38 ms^(-1)` with an acceleration of `2 ms^(-2)`
As `a = (v_(f) - v_(i))/(t) rArr = (v_(f) - v_(i))/(a)`
or `t = (38 -2)/(2) = (36)/(2) = 18s`
Distance covered in these 18s is
`s = u t + (1)/(2) "at"^(2)`
`=2 xx 18 + (1)/(2) (2) xx 18^(2)`
`= 36 + 324 = 360 m`
So, distance between points A and B,
`=1 + 39 + 360 = 400m`