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katrin2010 [14]
3 years ago
8

An automobile travels past the farmhouse at a speed of v = 45 km/h. How fast is the distance between the automobile and the farm

house increasing when the automobile is 3.7 km past the intersection of the highway and the road?
Mathematics
1 answer:
yulyashka [42]3 years ago
3 0

Answer:

\frac{ds}{dt} = 39.586 km/h

Step-by-step explanation:

let distance between farmhouse and road is 2 km

From diagram given

p is the distance between road and past the intersection of highway

By using Pythagoras theorem

s^2 = 2^2 +p^2

differentiate wrt t

we get

\frac{d}{dt} s^2 = \frac{d}{dt} (4 + p^2)

2s \frac{ds}{dt}  =2p \frac{dp}{dt}

\frac{ds}{dt} = \frac{p}{s}\frac{dp}{dt}

\frac{ds}{dt} = \frac{p}{\sqrt{p^2 +4}} \frac{dp}{dt}

putting p = 3.7 km

\frac{ds}{dt} = \frac{3.7}{\sqrt{3.7^2 +4}} 45

\frac{ds}{dt} = 39.586 km/h

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