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Definition of acceleration (calculus)
"The position of an object is described by the parametric equations x=lnt and y=5t^2. What is the acceleration of the object in m/sec2 when t=2?"
 one year ago
 one year ago
Definition of acceleration (calculus) "The position of an object is described by the parametric equations x=lnt and y=5t^2. What is the acceleration of the object in m/sec2 when t=2?"
 one year ago
 one year ago

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inkyvoydBest ResponseYou've already chosen the best response.1
I have gotten this problem wrong for calculating \(\sqrt{\frac{d^2y}{dt^2}+\frac{d^2x}{dt^2}}\) instead of \(\frac{d^2y}{dx^2}\) What am I doing wrong?
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.1
@hartnn , can you help me please?
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.1
I just need to know what I should be calculating for acceleration, and why. thanks!
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
i would have calculated d^2y/dx^2 because i didn't know the sqrt formula :P
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.1
But, d^2y/dx^2 is the movement of y with respect to x  it doesn't say the particle is accelerating does it?
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
if x is position, dy/dx gives velocity and d2y/dx2 gives the acceleration. so, it does.
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.1
no, dx/dt gives velocity  velocity is the derivative of distance with respect to time right?
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.1
I mean, the units don't add up when you do d^2y/dx^2 if you have y and x in parametric terms of t >.<
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.1
I've tried to email my teacher about it like 3 times but I don't think she's read what I've said  she just keeps repeating that the acceleration is d^2y/dx^2
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
i am sure , just realised , that d2y/dx2 is NOT acceleration
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.1
Okay  how should I prove to my teacher that d^2y/dx^2 is NOT acceleration?
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
units.....x and y are distances..... d2y/dx2 is dimensionless. it doesn't have the unit of acceleration.
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.1
Okay  any other ways? I tried emailing her saying units don't add up but I'm not sure if she read. >.<
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
d2y/dx2 gives change in dy/dx w.r.t x which is in no way, acceleration
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.1
because the change in y with respect to x isn't velocity right?
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
yeah, it isn't velocity , its just the slope
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.1
So then my teacher argues that d^2y/dx^2=\(\frac{\frac{d}{dt}(\frac{dx}{dt})}{\frac{dx}{dt}}\) Which unfortunately has a d/t in it >.<
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
that equality doesn't hold!
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.1
it does, but it has nothing to do with the definition of acceleration in two dimensions
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.1
yet I'm afraid my teacher will attempt to end the discussion and get back to doing other stuff (it's an online course) by simply telling me to ignore units.
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
you know the right thing,thats enough
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.1
my grade is hurt though >.<
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.1
Ah well I'll think of something  thank you very much!
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
spose we equate t in terms of x x=lnt ; e^x = t sub in e^x into the t for y y = 5e^(x^2) y' = 10x e^(x^2) y'' = 20x^2 e^(x^2) when t=2, x=ln2 y'' = 20(ln2)^2 e^(ln2^2) = 15.536 or so ....if my idea is correct
 one year ago
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