Question

Particles A and B are free to move along parallel paths. Particle B is moving at a constant velocity of 10 m/s and particle A starts ( from rest) moving at the moment particle B passes it. If particle A accelerates at 1 m/s^2 for 5 seconds and then travels at a constant velocity, the distance between particles A and B 10 seconds after particle B passes particle A is

Answer 1

Distance covered by particle B in 10 second

S=ut+1/2 at^2
u=10 a =0 t=10
so S = 10*10=100m

distance covered by A in 10 seconds
= distance in first 5 seconds + distance in second 5 seconds

distance in first five seconds
s = ut +1/2 at^2
u = 0
a=1 t = 5
s= 1/2 (1*5*5) = 25/2 m

distance in last 5 seconds
v= u+at
u = 0 a =1 t=5
so v=5
now s = ut+1/2 at^2 u =5 a =0: t=5
= 5*5 =25
so total diatance
​​​​​25 +25/2 =75/2 m =37.5m

distance between A and B = 100-37.5=62.5m