Question

A road perpendicular to a highway leads to a farmhouse located 7 mile away. An automobile traveling on the highway passes through this intersection at a speed of 55mph. How fast is the distance between the automobile and the farmhouse increasing when the automobile is 4 miles past the intersection of the highway and the road?

Answer 1

Answer:

The rate at which the automobile is moving away from the farmhouse is 27.29 m/h.

Explanation:

As shown in the figure, A denotes the position of farmhouse, B be the location of highway intersection and C be the direction along which automobile is moving.

Consider s be the distance between farmhouse and automobile which is represent by AC, x is the distance between intersection and automobile which is represent by BC and the distance between intersection of highway and automobile is represent by AB.

Applying Pythagoras Theorem to the figure,

(AB)² + (BC)² = (AC)²

Since, AB = 7 miles, BC = x and AC = s.

7² + x² = s²

Differentiating both sides of the above equation with respect to time :

$\frac{d}{dt}(7^{2} +x^{2} )=\frac{d}{dt}s^{2}$

$2x\frac{dx}{dt} = 2s\frac{ds}{dt}$

$\frac{x}{s}\frac{dx}{dt}=\frac{ds}{dt}$

$\frac{x}{\sqrt{7^{2}+x^{2} } }\frac{dx}{dt}=\frac{ds}{dt}$

When the automobile is 4 miles past the intersection, i.e.

x = 4 miles and $\frac{dx}{dt}$ = 55 m/h, then

$\frac{4}{\sqrt{7^{2}+4^{2} } }55=\frac{ds}{dt}$

$\frac{ds}{dt}=27.29$ m/h