Q: A body covers 100 cm in first 2 seconds and 128 cm in the next four seconds moving with constant acceleration. Find the velocity of the body at the end of 8 sec?

Sol. Distance in first two seconds is

$\large S_1 = u t_1 + \frac{1}{2}a t_1^2$

$\large 100 = 2 u + \frac{1}{2}a (4) $ …(i)

Distance in (2+4) sec from starting point is

$\large S_1 + S_2 = u ( t_1 + t_2 ) + \frac{1}{2}a (t_1 + t_2)^2$

$\large 228 = 6 u + \frac{1}{2}a (36) $ …(ii)

Solving (i) & (ii) we get ,

a = -6 cm/s^{2} and u = 56 cm/sec

v = u + a t

v = 56 – 6 x 8 = 8 cm/sec