A particle moving with a non-zero velocity and constant acceleration on straight line travels 15 m in first 3 s and 33 m in next 3 s in the same direction, find
(i) initial velocity of particle
(ii) acceleration of particle
Let initial velocity u and acceleration a.
Using s=ut+12at2
⇒15=3u+12×a×32 ⇒6u+9a=30 .....1
Distance travelled in 6s, =15+33=48m
⇒48=6u+12×a×62 ⇒6u+18a=48 .....2
Subtracting equation 1 from 2, we get ⇒9a=18 ⇒a=2m/s2
Put the value of a=2 in equation 1, we get ⇒u=2m/s
Hence, Initial velocity =2m/s
Acceleration =2m/s2