Two cars A and B are racing along straight line. Car A is leading, such that their relative velocity is directly proportional to the distance between the two cars. When the lead of car A is λ 1 = 10 m, its running 10 m/s faster than car B. If the time car A will take to increase its lead to λ 2 = 20 m from car B is t = (log e n) sec, then find n.
Text Solution
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(2)
Sol.
As given
(V A –V B ) ∞ x A –x B
(V A –V B ) = K(x A –x B )
when x A –x B = 10 We have V A –V B = 10
We get
10 = K10 ⇒ K = 1
⇒ V A –V B = (x A –x B ).........(1)
Now Let
x A –x B = y .................(2)
On differentiating with respect to ‘t’ on both side.
⇒ 
= 
⇒ Using (1) and (2)
= x A – x B
= dt
⇒ 
= t
t = (log e 2) sec
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