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Question

Two cars A and B are moving west to east and south to north respectively along crossroads. A moves with a speed of 72 kmh1 and is 500 m away from point of intersection of crossroads and B moves with a speed of 54 kmh1 and is 400 m away from point of intersection of crossroads. Find the shortest distance between them in metres?

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Solution

Method-I (Using the concept of relative velocity)
In this method we watch the velocity of A w.r.t. B. To do this we plot the resultant velocity Vr. Since the accelerations of both the bodies is zero, so the relative acceleration between them is also zero. Hence the relative velocity will remain constant. So, the path of A with respect to B will be straight line and along the direction of relative velocity of A with respect to B. The shortest distance between A & B is when A is at point F (i.e. when we drop a perpendicular from B on the line of motion of A with respect to B).

tanθ=VBVA=5472=34(i)
This θ is the angle made by the resultant velocity vector with the x axis.
From the figure, we can say
tanθ=OEOA=OE500(ii)
From equation (i) and (ii) we get
OE=375 m
BE=OBOE=400375=25 m
But the shortest distance is BF.
From the magnified figure we see that BF=BEcosθ=25×45
BF=20 m
Method II (using the concept of relative velocity of approach)
After time t let us plot the components of velocity of A and B in the direction along AB.

When the distance between the two is minimum, the relative velocity of approach is zero.
VAcosαf+VBsinαf=0
(Where αf is the angle made by the line AB with the x axis)
20cosαf=15sinαf
tanαf=2015=43
(Here do not confuse this angle with the angle θ in method (l) because that θ is the angle made by the net relative velocity with x axis, but αf is the angle made by the line joining the two particles with x axis when velocity of approach is zero.)
Distance of particle B from the origin after time t=40015t
Distance of particle A from the origin after time t=50020t
tanαf=40015t50020t=43
t=1285
So, OB=16 m and OA=12 m
AB=162+(12)2=20 m

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