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

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The discussion revolves around calculating the x and y components of velocity and acceleration for a van traveling over a hill described by the equation y=(-1.5(1/1000)x^2 + 15) ft at x=50 ft. The initial calculations for the y-component of velocity (Vy) were found to be negative, which led to confusion, as the van's constant speed is 75 ft/s. Participants clarified that the slope of the hill affects the direction of velocity components, and the correct angles and projections were discussed to resolve the signs and magnitudes of the components. Ultimately, the correct values were determined to be Ax = -2.42 ft/s² and Ay = -16.14 ft/s², emphasizing the importance of maintaining proper units and significant figures in calculations. The thread highlights the collaborative effort to troubleshoot and refine the solution to the problem.
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Homework Statement


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

Homework Equations

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.
 
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Hi vanquish! :smile:

(try using the X2 and X2 tags just above the Reply box :wink:)
vanquish said:
Vy=(-3x/1000) This bothered me, seeing as it would make the Y velocity always negative.

No, it makes the Y component of velocity negative when x is positive, and vice versa. :wink:
… 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

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. :smile:
 


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.
 
vanquish said:
At x=50ft, y=11.25, so I created a triangle and found theta, which turned out to be 12.68

uhh? :redface: wrong triangle!

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

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


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

:rolleyes:
 
Look at the diagram …

how can Vy be negative? :redface:
 


no idea :?
 
Then why have you made it negative? :confused:
 


tiny-tim said:
Then why have you made it negative? :confused:

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
or a negative Vx ? :redface:
 
  • #11


tiny-tim said:
or a negative Vx ? :redface:
it doesn't though, theta=-8.5
Vy=75sin\Theta and Vx=75cos\Theta

sin(-8.5)=-.1478
cos(-8.5)=.989
 
  • #12
vanquish said:
it doesn't though, theta=-8.5
Vy=75sin\Theta and Vx=75cos\Theta

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

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 :wink:

MORAL: this wouldn't have happened if you'd used Pythagoras! o:)
 
  • #13


tiny-tim said:
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 :wink:

MORAL: this wouldn't have happened if you'd used Pythagoras! o:)
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
 
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  • #14
(just got up :zzz: …)

(oh, and have a theta: θ :wink:)
vanquish said:
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

Yup, that should do it! :biggrin:
 
  • #15


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 I am 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
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


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
 
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  • #18
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


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
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.
 
  • #21


nvn said:
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.

nvn said:
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.
if the hill arc is s, then ds/dt would be the tangent which is the velocity, and the second derivative would be a line through that point which is the acceleration
 
  • #22
Very good, except change "arc" to "arc length," change "tangent which is the velocity" to "tangential velocity," change "line" to "vector," and insert "tangential" before "acceleration."
 
  • #23


alright, but I am still not quite sure what the correct equation is supposed to be
a=(d2y/dx2)ut+(v2/ρ)un
 
  • #24
Hint 4: Wouldn't you want to change d2y/dx2 in the above "a" equation to the tangential acceleration?
 
  • #25


nvn said:
Hint 4: Wouldn't you want to change d2y/dx2 in the above "a" equation to the tangential acceleration?
I sure would.

At=dvt/dt=r \alpha
 
  • #26
OK, so what is the value of vt? And then can you find dvt/dt?
 
  • #27


i don't think i can find either of them because i have no time, Vt is d\theta/dt*radius
 
  • #28
Hint 5: But check the problem statement in post 1 to see if you can find vt.
 
  • #29


75ft/s?
 
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  • #30
[strike]Wrong units[/strike]. EDIT: Now that you edited your above post, it is correct. Except always leave a space between a numeric value and its following unit symbol. E.g., 75 ft/s, not 75ft/s. You can see this by looking in any textbook.
 
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  • #31


a=(75ft/s^2)ut+(v2/ρ)un
At=75ft/s^2 ?
An=v^2/ρ=75^2/344.6=16.32
 
  • #32
an looks correct, but at is wrong. Look in your textbook for the definition of acceleration, and compute at.
 
  • #33


the only things my textbook says about this are:
at=\dot{v}
or
atds=vdv
 
  • #34
OK, use the first equation to compute at.
 
  • #35


I think I got it, At=11.25
An=v^2/ρ=75^2/344.6=16.32

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=11.12
Ay=16.14
 
  • #36
an looks correct, but at is wrong. Show your calculations for at.
 
  • #37


according to my book:
atds=vdv
so at=vdv/ds
v=75m/s and dv/ds=-3x/1000
and since we want At at 50ft
dv/ds=-.15
so at=(75)(-.15)
 
  • #38
Statements by vanquish highlighted in blue.
at*ds = vt*dvt[/color] --true.
so at = vt*dvt/ds[/color] = (ds/dt)*dvt/ds = dvt/dt[/color] --true.
vt = 75 m/s[/color] --true, except for units; also, see the note I added to post 30.
and dv/ds = -3x/1000[/color] --false; dy/dx = -0.003*x.
and since we want at at 50 ft[/color]
dv/ds = -0.15[/color] --false.
so at = 75(-0.15)[/color] --false.

By the way, according to the international standard (ISO 31-0), numbers less than 1 must have a zero before the decimal point. E.g., 0.15, not .15.
 
  • #39
vanquish: Hint 6: If s = 75*t, can you compute ds/dt, and d2s/dt2?
 
  • #40


nvn said:
vanquish: Hint 6: If s = 75*t, can you compute ds/dt, and d2s/dt2?

ds/dt=75
d2s/dt2=0
 
  • #41
That's correct. What is another name for ds/dt and d2s/dt2?
 
  • #42


nvn said:
That's correct. What is another name for ds/dt and d2s/dt2?

velocity and acceleration?
 
  • #43
That's correct. I mean, can you think of the variable name for these quantities, in post 40?
 
  • #44


vt and at
 
  • #45
That's correct. Want to try to solve the problem now?
 
  • #46


well if at=0At=0
An=v^2/ρ=75^2/344.6=16.32

Ax=0
Ay=16.14
 
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  • #47
That is correct for at and an. That is correct for ay, except for the sign. But ax is currently wrong. Try again.
 
  • #48


if at=0 then when i take the pojection of At along Ax i get
Ax=At*cos(8.5)
 
  • #49
That's right; but don't forget to also project an onto the x axis.
 
  • #50


nvn said:
That's right; but don't forget to also project an onto the x axis.
oh wow

Ax=At*cos(8.5)+An*sin(8.5)=-2.41 m/s2
Ay=-16.14 m/s2

EDIT: -2.41
 
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