Answer to Question #333779 in Calculus for Jane

Question

An automobile traveling at the rate of 20 m/s is approaching an intersection. When the automobile is 100 meters from the intersection, a truck traveling at the rate of 40 m/s crosses the intersection. The automobile and the truck are on perpendicular roads. How fast is the distance between the truck and the automobile changing two seconds after the truck leaves the intersection?

Answer 1

Let the equation of the displacement of an automobile be

$$y(t)=-100+20t$$

Let the equation of the displacement of a track be

$$x(t)=40t$$

Then the distance between the truck and the automobile will be

$$s(t)=\sqrt{[x(t)]^2+[y(t)]^2}$$

$$=\sqrt{(-100+20t)^2+(40t)^2}$$

$$=\sqrt{10000-4000t+2000t^2}$$

$$=20\sqrt{5}\sqrt{t^2-2t+5}$$

Find the first derivative with respect to $t$

$$s'(t)=\dfrac{20\sqrt{5}}{2\sqrt{t^2-2t+5}}(4t-2)$$

$$=\dfrac{20\sqrt{5}(2t-1)}{\sqrt{t^2-2t+5}}, t\ge0$$

$t=2$

$$s'(2)=\dfrac{20\sqrt{5}(2(2)-1)}{\sqrt{(2)^2-2(2)+5}}$$

$$=60(m/s)$$