0
$\begingroup$

Question:

1) Suppose you have a street light at a height 10 meters. You drop a rock vertically so that it hits the ground at a distance 5 meters from the street light. The shadow of the rock moves along the ground. Find the speed of the shadow of the rock at the moment that the rock is 3 meters above the ground and moving with the speed of 5 m/s.

2) You say goodbye to your friend at the intersection of two perpendicular roads. At time t = 0 you drive off North at a (constant) speed v and your friend drives West at a (constant) speed w. How fast is the distance between you and your friend increasing at time t?

What should start with these two?

$\endgroup$
1
$\begingroup$

1) Let $x$ be the distance of the shadow from the spot of landing and $h$ be the height of the stone. From similar triangles,

$$\frac xh = \frac{x+5}{10}, \implies x=\frac{5h}{10-h}$$

Take the time derivatives,

$$x' = \frac{50}{(10-h)^2}h'$$

Given that $h=3m$ and $h'=-5m/s$, the speed of the shadow is $x'=-\frac{250}{49}m/s$.

2) Let $d$ be the distance. From the right triangle,

$$d(t)=\sqrt{(vt)^2+(wt)^2}$$

Take the time derivatives to get

$$d'=\frac{v^2+w^2}{\sqrt{v^2+w^2}}=\sqrt{v^2+w^2}$$

$\endgroup$
  • $\begingroup$ Thanks Quanto!! You help me alot! $\endgroup$ – DZD Nov 14 '19 at 5:48
  • $\begingroup$ @Donald Zou - Glad to be helpful $\endgroup$ – Quanto Nov 14 '19 at 6:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.