# Van travels over the hill described by y=(-1.5(1/1000)x^2 + 15)ft

1. Sep 6, 2009

### vanquish

1. The problem statement, all variables and given/known data
The van travels over the hill described by y=(-1.5(1/1000)x^2 + 15)ft. If it has a constant speed of 75ft/s, determine the x and y components of the van's velocity and acceleration when x=50ft.

http://img215.imageshack.us/img215/3631/dynamcs.jpg [Broken]

2. Relevant equations

3. The attempt at a solution
I took the derivative of the equation of the hill in order to find the equation for the velocity in the Y direction. The result was:
Vy=(-3x/1000) This bothered me, seeing as it would make the Y velocity always negative.

So when x=50, Vy=-.15ft/s
and since the problem states that the van has a constant velocity of 75m/s I used Pythagorean to solve for the velocity in the x direction

(75^2)-(-.15^2)=Vx^2
This gives me a value of Vx=74.9

When I plug these numbers into the answer (we use an online homework system) It tells me that the first term (74.9) has an incorrect sign, yet when I change the sign the problem is simply wrong. Unless the program is trying to tell me I messed up a sign in my calculations.

So after getting stuck on this problem I moved on to the next part: finding the x,y components of the acceleration at x=50. And it's been a few hours and I'm stumped.

Last edited by a moderator: May 4, 2017
2. Sep 6, 2009

### tiny-tim

Hi vanquish!

(try using the X2 and X2 tags just above the Reply box )
No, it makes the Y component of velocity negative when x is positive, and vice versa.
I think you're getting confused between dy/dx and dy/dt.

Vy = dy/dt, Vx = dx/dt.

So Vy/Vx = dy/dx (and Vy2 + Vx2 = 752).

Try again.

3. Sep 6, 2009

### vanquish

Re: Dynamics

Thanks for the help. So now there is only 1 thing I can think of doing to solve this problem.

At x=50ft, y=11.25, so I created a triangle and found theta, which turned out to be 12.68
then, since it always has a constant speed of 75, I created another triangle with 12.68 as the angle and 75 as the hypotenuse

this gave me Vy=16.4 and Vx=73.17

The numbers looked good, and (in my mind: made sense) But alas, they are still wrong apparently.

4. Sep 6, 2009

### tiny-tim

uhh? wrong triangle!!

that's the slope of the line joining it to the origin.

You need the slope of the tangent … that's dy/dx.

5. Sep 6, 2009

### vanquish

Re: Dynamics

Oh that was silly of me, so dy/dx=-3x/1000 and at 50ft, the slope is -.15 (now that number makes sense)

since slope is rise over run, I can find theta by doing tan-1(-.15)

which gives me: theta=-8.5

That gives me:
Vy=-11.1
Vx=74.17

:uhh:

6. Sep 6, 2009

### tiny-tim

Look at the diagram …

how can Vy be negative?

7. Sep 6, 2009

Re: Dynamics

no idea :?

8. Sep 6, 2009

### tiny-tim

Then why have you made it negative?

9. Sep 7, 2009

### vanquish

Re: Dynamics

because the slope was -3x/1000 and since we want the car at 50 ft, that gives us a -.15 for the slope which gives us a negative theta, which gives us a negative Vy

10. Sep 7, 2009

### tiny-tim

or a negative Vx ?

11. Sep 7, 2009

### vanquish

Re: Dynamics

it doesn't though, theta=-8.5
Vy=75sin$$\Theta$$ and Vx=75cos$$\Theta$$

sin(-8.5)=-.1478
cos(-8.5)=.989

12. Sep 7, 2009

### tiny-tim

oh i see what you're doing …

no, you got it from tan-1(-.15), which is either -8.5º or 171.5º (ie, 180º more), isn't it?

in this case, just by looking at the diagram, you should have seen that Vx is always negative (because the car's going to the left!!), so you have to choose 171.5º

MORAL: this wouldn't have happened if you'd used Pythagoras!

13. Sep 7, 2009

### vanquish

Re: Dynamics

Ah, I see what I did. That was a silly mistake.

So now I have to find the acceleration components

My first instict would be to take the second derivative of the line, to give me the acceleration and then do the same thing i did before, and find theta

Last edited: Sep 7, 2009
14. Sep 8, 2009

### tiny-tim

(just got up :zzz: …)

(oh, and have a theta: θ )

Yup, that should do it!

15. Sep 8, 2009

### vanquish

Re: Dynamics

So i'm stumped again. I took the second derivative which ends up being -3/1000 which would be the slope and I can find θ from that, but im not sure how to get the components of the acceleration because when I did it for velocity, I knew the magnitude of the velocity and with the help of θ I found the components, but I don't know the magnitude of the acceleration.

16. Sep 11, 2009

### nvn

vanquish: Hint 1: Look for a formula to compute normal acceleration. Hint 2: Look in your calculus book for a formula to compute curvature.

17. Sep 11, 2009

### vanquish

Re: Dynamics

could i do:

ρ=[1+(dy/dx)2]3/2/(d2y/dx2)

then

a=(dy/dx)ut+(v2/ρ)un

and then since it asks for Vx and Vy, rather than Vn and Vt, i would use the angle I found earlier from the velocity, and find the components that way

I think this is a viable solution but I'd like someone else to see if this makes sense

Last edited: Sep 11, 2009
18. Sep 11, 2009

### nvn

vanquish: Nice work on the first equation (except I think it should have an absolute value sign in there somewhere, if I recall).

Regarding your second equation, it is starting to look great, except dy/dx is just the hill slope, not velocity, which tiny-tim mentioned in an earlier post. Let's say, e.g., you wanted to call the hill arc length s. Then what is ds/dt? And what is d2s/dt2? See if you can fix your equation now.

19. Sep 12, 2009

### vanquish

Re: Dynamics

ρ=[1+(dy/dx)2]3/2/|(d2y/dx2)|
the absolute value was on the denominator

a=(d2y/dx2)ut+(v2/ρ)un
the forumula in the book had it as the derivative of velocity

So when i calculated everything and used the angle i found before to project this information from the At and An directions into the Ax and Ay directions I get these numbers:
Ax=-2.9*10^-3
Ay=16.14

edit: those answers are incorrect apparently :(

20. Sep 12, 2009

### nvn

That is correct for ay, except for the sign. But ax is currently wrong. d2y/dx2 doesn't go in the "a" equation. Hint 3: Can you think of another name for ds/dt? Try to answer my questions in post 18.