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Question
A police jeep, approaching an orthogonal intersection from the northern direction, is chasing a speeding car that has turned and moving straight east. When the jeep is 0.6 km north of the intersection and the car is 0.8 km to the east. The police determine with a radar that the distance between them and the car is increasing at 20 km/hr. If the jeep is moving at 60 km/hr at the instant of measurement, what is the speed of the car?
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Solution
Given x = 0.8
y = 0.6
`("d"y)/"dt"` = – 60
`"ds"/"dtt"` = 20
From the figure
S2 = x2 + y2
S2 = (0.8)2 + (0.6)2
= 0.64 + 0.36
= 1
S2 = 1
⇒ S = 1
S2 = x2 + y2
Differentiating w.r.t. ‘t’
`2"S" "dS"/"dt" = 2x ("d"x)/"dt" + 2y ("d"y)/"dt"` .....(÷ 2)
`"S" "dS"/"dt" = x ("d"x)/"dt" + y ("d"y)/"dt"`
1(20) = `(0.8) ("d"x)/"dt" + (0.6)(- 60)`
20 = `0.8 ("d"x)/"dt" 36`
∴ `("d"x)/"dt" = (20 + 36)/0.8`
= `56/0.8`
= 70
∴ Speed of the car is 70 km/hr.
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