Two cars starts from same point on a straight road due north .While car A starts with an initial velocity of 10 m/s and an acceleration of 1 m/s2 ,car B starts from rest with an acceleration of 2 m/s^2 .The maximum separation between the two cars is .

							
Since the one car has initial velocity and another car is starting from rest the separation will starts increasing first. Now after some time second car will have more velocity than other car since it has higher acceleration. So the separation will starts decreasing. Then they will come to same point at some time and then their separation will starts increasing again.
						

							
Dear student 
For car A 
u=10m/s. a= 1 m/s² 
s1 = 10t+ t²/2 
For car B 
u = 0 m/s 
a = 2 m/ s² 
s2 = 2t²/2= t² 
A/Q 
s1 = s2 
from this t= 20 sec 
After 20 sec the separation will be zero again
						

							
initially the seperation will increase till the velocty of both becomes same... then the velocity of car B will increase and hence the separation between both will start decreasing. 
 
So if we find the time, (say time = t) at which the velocity of car B  and A becomes equal..
 
Vb = Va
0+2*t =  10 + 1*t
or t = 10 seconds.
 
means after 10 sec of start up the velocity of car B(Vb) becomes euqal to the velocity of car A (Va)
The displacement of car A in 10 seconds , 
 
Sa = 10*10 + 0.5*1*10*10
      = 100 + 50
    = 150 meters
Similary 
 
he displacement of car B in 10 seconds , 
Sa = 0*10 + 0.5*2*10*10
      = 0 + 100
    = 100 meters
 
 
So the max separation between A and B will be 100 meters. 
 
 
 
 
						

							
 
nitially the seperation will increase till the velocty of both becomes same... then the velocity of car B will increase and hence the separation between both will start decreasing. 
 
So if we find the time, (say time = t) at which the velocity of car B  and A becomes equal..
 
Vb = Va
0+2*t =  10 + 1*t
or t = 10 seconds.
 
means after 10 sec of start up the velocity of car B(Vb) becomes euqal to the velocity of car A (Va)
The displacement of car A in 10 seconds , 
 
Sa = 10*10 + 0.5*1*10*10
      = 100 + 50
    = 150 meters
Similary 
 
he displacement of car B in 10 seconds , 
Sa = 0*10 + 0.5*2*10*10
      = 0 + 100
    = 100 meters
 
 
So the max separation between A and B will be 150-100 = 50 meters. 
 
						

							
Dear student,
 
Since the one car has initial velocity and another car is starting from rest the separation will starts increasing first.
 
For car A 
u=10m/s. a= 1 m/s² 
s1 = 10t+ t²/2 
For car B 
u = 0 m/s 
a = 2 m/ s² 
s2 = 2t²/2= t² 
A/Q 
s1 = s2 
from this t= 20 sec 
 
Regards