Question

An automobile travels past the farmhouse at a speed of v = 45 km/h. How fast is the distance between the automobile and the farmhouse increasing when the automobile is 3.7 km past the intersection of the highway and the road?

Answer 1

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

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

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