A particles start from rest and acceleration passing 2 point (A and B)which are 100cm apart in 15seconds if the velocity of the particle at B is three times it velocity at A,find i)the velocity of the particles at A. ii)the accelation. iii) the distance of A from the starting piont.

Answer 1

The language is a bit unclear, but it appears that it takes 15 seconds to travel the 100 cm from A to B. If so,

since velocity
v = at

3at = a(t+15)
t = 15/2 s

Since distance
s = 1/2 at^2,

100 + 1/2 at^2 = 1/2 a(t+15)^2
a = 4/9 cm/s^2

So, we have
v@A = (4/9)(15/2) = 10/3
v@B = (4/9)(45/2) = 10
s@A = 1/2 (4/9)(15/2)^2 = 25/2
s@B = 1/2 (4/9)(45/2)^2 = 225/2
Note that 225/2 = 25/2 + 100 as required