Ace school

with brainly

  • Get help from millions of students
  • Learn from experts with step-by-step explanations
  • Level-up by helping others

A community for students.

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?"

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SEE EXPERT ANSWER

To see the expert answer you'll need to create a free account at Brainly

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?
@hartnn , can you help me please?
I just need to know what I should be calculating for acceleration, and why. thanks!

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

i would have calculated d^2y/dx^2 because i didn't know the sqrt formula :P
But, d^2y/dx^2 is the movement of y with respect to x - it doesn't say the particle is accelerating does it?
if x is position, dy/dx gives velocity and d2y/dx2 gives the acceleration. so, it does.
no, dx/dt gives velocity - velocity is the derivative of distance with respect to time right?
oh, yes, sorry....
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 >.<
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
i am sure , just realised , that d2y/dx2 is NOT acceleration
Okay - how should I prove to my teacher that d^2y/dx^2 is NOT acceleration?
units.....x and y are distances..... d2y/dx2 is dimensionless. it doesn't have the unit of acceleration.
Okay - any other ways? I tried emailing her saying units don't add up but I'm not sure if she read. >.<
d2y/dx2 gives change in dy/dx w.r.t x which is in no way, acceleration
because the change in y with respect to x isn't velocity right?
yeah, it isn't velocity , its just the slope
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 >.<
that equality doesn't hold!
it does, but it has nothing to do with the definition of acceleration in two dimensions
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.
you know the right thing,thats enough
my grade is hurt though >.<
Ah well I'll think of something - thank you very much!
hmm...welcome ^_^
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
product rule for y'' ?

Not the answer you are looking for?

Search for more explanations.

Ask your own question