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 0.6 km north of the intersection and the car is 0.8 km to the east, the police determine with radar that the distance between them and the car is increasing at 20 km/h . If the jeep is moving at 60 km/h at the instànt of measurement, what is the speed of the car?



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Answer 1

$\large \underline{ \pmb{ \orange{{ \frak{Supposition : - }}}}}$

• Let the distance of jeep with respect to the intersection at a time interval be (a) = 0.6 km
• Let the distance of car with respect to the intersection be at the same time interval (b) = 0.8 km
• Let the distance between them at the time interval be (c)

As we know that :-

• the jeep is moving towards -ve y axis hence, velocity will be negative = -60km/hr

$:\implies$ $\sf\dfrac{da}{dt}=-60km/hr$

Also,

$:\implies$ $\sf\dfrac{dc}{dt}= 20km/hr$

We are asked to find :-

$:\implies$ $\sf\dfrac{db}{dt}= \:?$

$\large \underline{ \pmb{ \orange{{ \frak{Required\: Solution : - }}}}}$

According to the figure :-

$\leadsto \large \underline{ \boxed {\pink{{ \frak{ {c}^{2} = {a}^{2} + {b}^{2} }}}}} \bigstar$

$\dag\:\frak{Differentiating\: this\: equation\: with \:respect\: to\: t:- }$

$\\ : \implies \sf \dfrac{d }{dt} {c}^{2} = \dfrac{d}{dt} {a}^{2} + \dfrac{d}{dt} {b}^{2}$

$\\ : \implies \sf 2c \dfrac{dc}{dt} = 2a \dfrac{da}{dt} + 2b \dfrac{db}{dt}$

$\\ : \implies \sf \dfrac{dc}{dt} = \dfrac{1}{c} \bigg(a \dfrac{da}{dt} + b \dfrac{db}{dt} \bigg)$

$\\ : \implies\boxed{ \sf \frac{dc}{dt} = \frac{1}{ \sqrt{ {a}^{2} + {b}^{2} } } \bigg(a \frac{da}{dt} + b \frac{db}{dt} \bigg) }$

$\dag\:\frak{Substituting \:the\: given\: values\: in\: this \:equation :-}$

$\\ : \implies \sf 20 = \frac{1}{ \sqrt{ {(0.6)}^{2} + {(0.8)}^{2} } } \bigg[0.6 ( - 60) + 0.8 \dfrac{db}{dt} \bigg]$

$\\ : \implies \sf 20 = - 36 + 0.8 \dfrac{db}{dt}$

$\\ : \implies \sf 56 = 0.8 \dfrac{db}{dt}$

$\\ : \implies \boxed{ \pink{{ \sf \dfrac{db}{dt} = 70 \: km {hr}^{ - 1} }}} \bigstar$

$\therefore \underline{ \sf our \: required \: answer \: is \: 70 \: km {hr}^{ - 1} }$

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And we are done :"D

Answer 2

I will give photo's of Answer.

Answer is 70 km h ¹

if its perfect so mark as brainlist