Question

question for calculus: A rock is dropped straight off a bridge that is 50 meters above the ground. Another person is 7 meters away on the same bridge. At what rate is the distance between the rock and the second person increasing just as the rock hits the ground?

Answer 1

Let the vertical position of the rock be s(t). Then the distance from the other person by Pythagorus' theorem is
\[ f(t) = \sqrt{7^2 + s(t)^2} \]

Now, what is the function s(t)?

Answer 2

Given the way we're measuring distance s(0) = 0.

What is s(t) in general? At what rate is the rock accelerating? Hence what is its velocity and hence what is its position?

Answer 3

would the function s(t)= 0 as well?

Answer 4

clearly not because the rock changes position.

Answer 5

I think the rate of acceleration is -9.8 m/sec2 but I could be wrong

Answer 6

yes, call that g. So at time t, where is the rock? How far has it fallen?

Answer 7

so, at t(0), has the rock fallen yet, or is it still at zero?

Answer 8

At t = 0 is the exact moment the rock is released. At the exact moment, it has not travelled any distance from the person who just released it. But there is now a force acting on the rock--gravity--and it is accelerating at a rate of g.

Let's break it down then. Given that the acceleration is constant, what is the rock's speed at time t, given that its velocity at time t=0 is also zero.

Answer 9

so, the equation for velocity would be v(t)=-9.8t+c, but since c=0, would v(t) just equal -9.8t?

Answer 10

yes. Now, what then is s(t)?

Answer 11

s(t) would equal \[-9.8t ^{2} /2 +c\] ?

Answer 12

Yes and c equals what in this case?

Answer 13

would that equal 50, since that's the distance is travelled?

Answer 14

No. At t = 0, s(t) = 0. i.e., s(0) = 0. Hence c = ...

Answer 15

0?

Answer 16

yes. We're measuring vertical distance from the bridge. That is why the function f(t) that shows the distance the rock has travelled has the form it has. Draw a diagram and make sure you understand where f(t) comes from.

Answer 17

So what you now have is that the distance of the rock from the other person is
\[ f(t) = \sqrt{7^2 + s(t)^2} = \sqrt{49 + (9.8^2/2^2)t^4} = \sqrt{49 + at^4} \]
where a = 9.8^2/2^2 = 24.01

Answer 18

The question now asks what's the rate of change of this function when the rock hits the ground.

To answer now you need to
1. Find the time T when s(t=T) = -50, i.e., the ground.
2. Find the derivative of f, f'(t)
3. Evaluate the derivative at t=T, f'(T)

Answer 19

I'll leave you to it.

Answer 20

okay. For 1, we would set the s(t)=-50, and when we figure it out, I think the answer is closest to t=3.19

Answer 21

as for f'(t), I think it equals \[2\sqrt{49+at ^{4)}}^{1/2} \times 4t ^{3} \] ?

Answer 22

and the last part, i'm not too sure about

Answer 23

For the derivative, use the chain rule. You haven't got it right.

Answer 24

\[1/2 \sqrt{49+at ^{4}}^{-1/2} \times 4t ^{3}\]

Answer 25

You dropped an "a"

Answer 26

\[ f'(t) = \frac{4at^3}{2\sqrt{49+at^4}} \]

Answer 27

oh okay. Now, would this function be used as dz/dt in the equation?

Answer 28

What is z? Ask yourself, what is f(t)?

Answer 29

z would be the original f(t) \[\sqrt{49+at ^{4}}\]

Answer 30

If you insist. In which case, if z = f(t), then clearly dz/dt = df/dt = f'(t)

Answer 31

Okay, so then when we plug that back into the original equation, we're left to solve for dy/dt ?

Answer 32

What is it we're trying to measure, i.e., what is the question asking?
"At what rate is the distance between the rock and the second person increasing just as the rock hits the ground?"

Now, as f(t) is the distance between the rock and the second person, we want to know
f'(t)
when the rock hits the ground.

Answer 33

So you want to know the value of f'(t) for the appropriate value of t.

Answer 34

Make sense?

Answer 35

Yes, now it does

Answer 36

Would we use the same value of for t what we found before, 3.19?

Answer 37

for what value of t does the rock hit the ground?

Answer 38

I think 3.19?

Answer 39

How did you derive t = 3.19?

Answer 40

well, when we figured out s(t)=-50, the answer I got was approximately equal to 3.19

Answer 41

Right. So that must be the right value of t.

Answer 42

okay, so then we plug that into the f'(t) equation and solve for it?

Answer 43

solve it, no. But evaluate it, yes.

Answer 44

The final answer to the problem is the value of f'(3.19)

Answer 45

So do yourself a favor now. Go back to the beginning of the problem, and a blank piece of paper, and write out all of the steps yourself and make sure you understand where they come from.

When you can do that without looking at this page on OpenStudy or your earlier versions, then you really understand that problem.

I strongly recommend doing that. You'll really learn that way.

Answer 46

Okay, thank you for your help