Question

A police jeep, approaching a right-angled intersection from the north, is chasing a speeding car that has turned the corner and is now moving straight east. When the jeep is 0.8 km north of the intersection and the car is 0.6 km to the east, the police determine with radar that the distance between them and the car is increasing at 30 km/h . If the jeep is moving at 50 km/h at the ínstant of measurement, what is the speed of the car?​

Answer 1

Explanation:

We draw a diagram of the car and jeep in the coordinate plane, using the positive x-axis as the eastbound highway and the positive y-axis as the northbound highway. Let x be the position of car at time t.

y= position of jeep at time t.

s=distance between can and jeep at time t.

We assume x,y, and s to be differentiable functions of t.

x=0.8 km,y=0.6 km,

dt

dy

=−60 km h −1

dtds

=20 km h −1

Hope its help..

Answer 2

$\bold{s= \sqrt{ {x}^{2} + {y}^{2}}}$

$\bold{= \sqrt{ {0.8}^{2} + {0.6}^{2}}}$

$\bold{= \sqrt{1} = 1}$

$\bold{Given \: x = 0.8, \: y = 0.6,}$

$\bold{\dfrac{dy}{dt} = - 60, \: \dfrac{ds}{dt}}$

$\bold{From \: the \: figure \: {s}^{2} = {x}^{2} + {y}^{2}}$

$\bold{differentiate \: w.r \: to \: t}$

$\bold{2s \frac{ds}{dt} = 2x \frac{dx}{dt} + 2y \frac{dy}{dt} ( \div \:by \: 2)}$

$\bold{ \implies \: s \frac{ds}{dt} = x\frac{dx}{dt} + y \frac{dy}{dt} }$

$\bold{1(20) = (0.8) \frac{dx}{dt} + (0.6)( - 60)}$

$\bold{ \frac{dx}{dt} = \frac{20 + 36}{0.8} = \frac{56}{0.8} = 70}$

$\bold{ \frac{dx}{dt} = 70 \: km/hr}$