1057SA=3t+t2/2
SB=t+t2
Snet=10+(SB-SA)
=10-3t-t2/2+t+t2
this function is minimum when it's derivative is 0...
or -3-t+1+2t=0
or t=2 secs...
Snet=8 m
Two particles A and B start moving simultaneously along the line joining them in the same direction with acceleration of 1 m/s2 and 2 m/s2 and speeds 3 m/s and 1 m/s respectively.Initially A is 10m behind B.What is the minimum distance between them?
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5 Answers
49As you can visualise if you can, that A first moves towards B and then away from B.
initial separation = 10 m
A is trying to reduce the distance and B is trying to increase it.
So after time t, the separation S=d_1-(u_At+\frac{1}{2}a_At^2) + (u_Bt+\frac{1}{2}a_Bt^2)
So, you can do this now.
7ans 8???
@Subhomoy: wont the distance be min when the relative velocities be 0..(since acc of B is more which wil result in greater velocity than A after some time.)
let after time t,
Va=3+t
Vb= 1+2t
t=2
now Sa =8
Sb=6
1057
1you can also do it by the concept of relative acceleration
think about it its simple