A particle is moving in a straight line with initial velocity `u` and uniform acceleration `f`. If the sum of the distances travelled in `t^(th) and (t + 1)^(th)` seconds is `100 cm`, then its velocity after `t` seconds, in `cm//s`, is.

Options

A. 80
B. 50
C. 20
D. 30

Answer

B

Explanation

The distance travel in `n^(th)` second is
`S_(n) = u + ½ (2n-1)a` …(1)
so distance travel in `t^(th)` & `(t+1)^(th)` second are
`S_(t)= u + ½ (2t-1)a `..(2)
`S_(t+1) = u + ½ (2t+ 1)a` …(3)
As per question,
`S_(t) + S_(t+1) = 100 = 2 (u + a t)` ...(4)
Now from first equation of motion the velocity of particle after time t, if it moves with an accleration a is `v= u + a t` ...(5) where u is initial velocity
So from eq(4) and (5), we get v = 50cm/sec.