BITSAT 2011 Paper

The distance travelled in n‌th ‌ second is
Sn=u+‌

1

2

(2n−1)a
So distance travelled in tth&(t+1)‌th ‌ second are
St=u+‌

1

2

(2t−1)a
St+1=u+‌

1

2

(2t+1)a
As per question,
St+St+1=100=2(u+at)‌‌‌‌‌⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(1)
Now from first equation of motion the velocity of particle after time t , if it moves with an acceleration a is
v=u+at‌‌‌‌‌⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(2)
where u is initial velocity So from equation (1) and (2) , we get v=50cm∕s