In the figure shown A and B are two particles which start from rest. A has constant acceleration ' a ' in the direction shown. B also increases its speed at a constant rate ' b ', but the direction of velocity is always towards A. Find the time after which B meets A. Also find the total distance travelled by B. (b > a)
Text Solution
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t = 
, 
Sol. Let after time t , A is at P and B is at Q. Let T = Total time . Their velocities after time t
V A = at ....(1)
V B = bt ...(2)
Let distance PQ = x.
Velocity of approch along PQ
= V B – V A cos α
⇒ 
= V B – V A cos α = bt – at cos α
⇒ 
= 
⇒ λ = 
– a 
....... (3)
For motion along x-axis :
= 
aT 2
= 
aT 2
⇒ 
= 
........ (4)
Put into (3)
λ = 
– a × 
⇒ T = 
Now , distance travelled by B
s = 
= 
= 
bT 2
⇒ s = 
b. 
= 
Ans.
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