# Homework Help: I'm curious what anyone gets for part d

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1. Oct 26, 2015

### Matt_Theta

1. The problem statement, all variables and given/known data
After a wild ride on a windsurfer, you head to a lifeguard tower to fly a kite. You stand on the tower, which is 3 meters above the ground and release your kite. You let out 10 meters of string and the kite begins moving according to the following equation:

y(t) = (0.9 m/s3)t3-(7.5 m/s2)t2+(13 m/s)t+10m

2. Relevant equations
a) How far above the ground is the kite when it begins to dive?
b) How far above the ground is the kite when it begins slowing while descending?
c) Does the kite hit the ground? If not, how close does it get?
d) What is the acceleration of the kite as it passes you while speeding up?

3. The attempt at a solution
Answers: a) 19.43m; b) 10.51m; c) No. It is 1.64m above the ground; d) -17.28 m/s2

I completely agree with the answers for parts a-c. However, no matter what I do, I keep getting -12 m/s2 for part d.

I'm looking for a sanity check.

Thanks!
-Matt

2. Oct 26, 2015

### BvU

Hi Matt, welcome to PF !

No typos ? I don't even get a) right. Could you show your steps ? for all four parts ?

3. Oct 26, 2015

### Ray Vickson

Let's change the definitions of everything to get rid of all those annoying units, so we can write $y = .9 t^3 - 7.5 t^2 + 13 t + 10$. Here, the height is $y$ meters (so that $y$ itself is dimensionless) and the time is $t$ seconds (so that $t$ itself is dimensionless).

Anyway, I cannot make sense of your equation. When $t = 0$ you have $y = 10$, so the kite starts off at height 10 meters (above the ground? above the top of the tower?) Did the kite somehow, magically, transport itself 10 meters up before it started its movement?

4. Oct 29, 2015

### J Hann

If the equation for the acceleration is a = 5.4 t - 15 (taking the second derivative of y)
you can't get a negative value for the acceleration less than -15 for real times.
Don't you need to find the the value of t when y = 3 m and then calculate the acceleration.
Looks like the solution of a cubic equation where you should get 2 real roots and 1 imaginary root.

5. Oct 29, 2015

### haruspex

-12 is not less than -15.

6. Oct 29, 2015

### haruspex

Can't tell where you are going wrong if you don't post your working.
If I substitute that acceleration in the $\ddot y$ equation I get t=1/1.8 s. Plugging that back into the y equation I get 15m.

7. Oct 30, 2015

### J Hann

I used a cubic equation solver on the internet and obtained the following values for t at y = 3.
t = 5
t = 3.9
t = -.569
Both 5 and 3.9 would give positive accelerations at y = 3 and neither agrees with the posted value of -17.28.
Also, if the acceleration were -17.28, how could the kite be speeding up?
The path of the kite appears to rise, dive towards the ground and then accelerate upwards indefinitely.

8. Oct 30, 2015

### Ray Vickson

No: once it hits the ground it stops, and is no longer governed by the equation you wrote initially. That is, assuming your y(t) refers to height above the ground---which I asked you about in my first post, but you ignored the question.

The cubic equation y(t) = 3 has two positive roots, and one negative root. The larger of the positive roots occurs after the kite actually hits the ground (at which time the equation for y(t) no longer applies), and of course, the negative root does not count. So (according to Maple) the relevant root is at t = 3.381463103 ≈ 3.4 (sec), giving an acceleration of a = -15.0 + 5.4*3.381463103 ≈ 3.26 (m/s^2). (This occurs just a bit after the time when acceleration switches from negative to positive.)

9. Oct 30, 2015

### J Hann

I calculated dy/dt = 0 at 4.48 sec at a height of 1.64 m, the lowest point of the trajectory.
This agrees with answer (c) for the lowest point.
I agreed with all answers except (d).
If dy / dt is negative how can you say the speed is increasing.
I guess you can say that if the acceleration is positive a less negative speed is increasing.
That would include the value at t = 3.9 along with t = 5 which were the values
I obtained from the cubic equation calculator for the positive roots.

Last edited: Oct 30, 2015
10. Oct 30, 2015

### Ray Vickson

Because dy/dt is velocity, not speed. Even when velocity is < 0 the speed is still > 0. Speed = |velocity|.

You have still not answered my very simple question: is y(t) = height above the ground, or is it height above the top of the tower? If it is height above the ground, the kite does, indeed, hit the ground and stop. If it is height above the top of the tower, the kite never hits the ground and does, indeed accelerate upward to the end of the 10 meter string (at which time, again, the equation you wrote no longer applies). So, which is it to be?

11. Oct 30, 2015

### J Hann

I can only assume, from the information given, that standing on a 3 ft tower means that the y coordinate
references a point 3 ft above the ground.
Using that assumption, I agree with answers a, b, and c.
I also assume that the string can be let out indefinitely at the fliers discretion.
I didn't write the question.

12. Oct 30, 2015

### Ray Vickson

No, but if you are making an interpretation of it, surely you can tell us what that is. Also, when turning in the work for marks it would be a good idea to state explicitly what you are assuming; after all, how can you be sure what was the intention of the person setting the question?

13. Oct 30, 2015

### haruspex

The OP was posted by Matt_Theta, not J Hann.

14. Oct 30, 2015

### haruspex

So do you now understand Ray's point about speeding up? If so, I assume you get that it passes the kite flier while speeding up at time 5 sec.