Question

3. A particle is moving in a straight line with
initial velocity u and uniform acceleration a.
If the sum of the distances travelled in t
and (t + 1)th seconds is 100 cm, then its
velocity after t seconds in cm s-1 is
(A) 20
(B) 30
(C) 50
(D) 80​

Answer 1

Answer:

50cm/s

Explanation:

distance traveled in nth sec

Sn=u+1/2(2n-1)a

hence distance traveled in t and (t+1)th sec

St=u+1/2(2t-1)a

St+1=u+1/2(2(t+1)-1)a

=u+1/2(2t+2-1)a

=u+1/2(2t+1)a

St+St+1=u+1/2(2t-1)a + u+1/2(2t+1)a

St+St+1=u+ta-a/2 + u+ta+a/2

St+St+1=u+u+ta+ta+a/2-a/2

St+St+1=2u+2ta

St+St+1=2(u+ta)=100 --->(1) (given)

v=u+at ------>(2)

hence in (1)

St+St+1=2(v)=100

v=50cm/s