A police cruiser, approaching a right angled intersection from north, is chasing a speeding car that has turned the corner and is now moving straight east. When the cruiser is 0.6 km. north of the intersection and the car is 0.8 km to the east, the police determine with radar that the distance then and the car is increasing at 20 km/h. Suppose that the cruiser is moving at 60 km/h at the instant of measurement.
(i) If s is the distance between car and cruiser at time t, x = position of cr at time t and y position of cruiser at time t then 
is equal to:
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Ans.
(i)
Sol. At the instant in question x = 0.8, y = 0.6,
= – 60 km/h
= 20 km/h, s 2 = x 2 + y 2 2s 
= 2x 
+ 2y 
= 
... (i)
(ii)
Sol. Substituting the given values, in eqn. (i), we get
20 = 0.8 
– 36 ⇒ 
= 70
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