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

`(ds)/(dt) = 39.586 km/h`

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

`(d)/(dt) s^2 = (d)/(dt) (4 + p^2)`

`2s (ds)/(dt)  =2p (dp)/(dt)`

`(ds)/(dt) = (p)/(s)(dp)/(dt)`

`(ds)/(dt) = (p)/(√(p^2 +4)) (dp)/(dt)`

putting p = 3.7 km

`(ds)/(dt) = (3.7)/(√(3.7^2 +4)) 45`

`(ds)/(dt) = 39.586 km/h`