MHB Velocity/acceleration/speed of a particle

  • Thread starter Thread starter ineedhelpnow
  • Start date Start date
  • Tags Tags
    Particle
ineedhelpnow
Messages
649
Reaction score
0
i made a calculator thread regarding this same question.

its number 7. ignore the graph on the first picture. its from the last question. the graph for this one is on the last picture. and excuse my sloppy handwriting. i was in a rush. since a(t)=2j meaning $\left\langle 0,2,0 \right\rangle$ i figured a(1) would be $\left\langle 0,2,0 \right\rangle$ but I am looking at how they did it on chegg and they did $a(1)= \left\langle 1,2,0 \right\rangle$ can someone please explain this?

View attachment 3200View attachment 3201View attachment 3202View attachment 3203
 

Attachments

  • DSC00004.JPG
    DSC00004.JPG
    20.5 KB · Views: 90
  • DSC00005.JPG
    DSC00005.JPG
    18.8 KB · Views: 76
  • DSC00006.JPG
    DSC00006.JPG
    16.6 KB · Views: 87
  • DSC00007.JPG
    DSC00007.JPG
    17.2 KB · Views: 86
Last edited:
Physics news on Phys.org
ineedhelpnow said:
i made a calculator thread regarding this same question.

its number 7. ignore the graph on the first picture. its from the last question. the graph for this one is on the last picture. and excuse my sloppy handwriting. i was in a rush. since a(t)=2j meaning $\left\langle 0,2,0 \right\rangle$ i figured a(1) would be $\left\langle 0,2,0 \right\rangle$ but I am looking at how they did it on chegg and they did $a(1)= \left\langle 1,2,0 \right\rangle$ can someone please explain this?

View attachment 3200https://www.physicsforums.com/attachments/3201View attachment 3202https://www.physicsforums.com/attachments/3203

Next time, please crop your picture in MS Paint or Gimp or Picasa down to only what you want people to look at. Increases the signal-to-noise ratio (always a good thing).

Anyway. You are correct, and chegg is wrong. If you differentiate $t$ twice, as you need to do for the $\mathbf{i}$ component, you get zero.
 
Ackbach said:
Next time, please crop your picture in MS Paint or Gimp or Picasa down to only what you want people to look at. Increases the signal-to-noise ratio (always a good thing).
yes sir (Nod)

Ackbach said:
Anyway. You are correct, and chegg is wrong. If you differentiate $t$ twice, as you need to do for the $\mathbf{i}$ component, you get zero.
im kind of confused. can you draw both vectors v(1) and a(1) so i can see what they look like?
 
So, we have
\begin{align*}
\mathbf{r}(t)&=\langle t,t^2,2 \rangle \\
\mathbf{v}(t)&=\langle 1,2t,0 \rangle \\
\mathbf{a}(t)&=\langle 0,2,0 \rangle \\
\mathbf{v}(1)&=\langle 1,2,0 \rangle \\
\mathbf{a}(1)&=\langle 0,2,0 \rangle.
\end{align*}

Here's the WA plot. See if you can figure out which axis is which.
 
the one on top is the y. the one coming down is x and the one at the bottom z?

- - - Updated - - -

i can't seem to remember how to draw vectors again. i can't even remember how i drew v(1) on there earlier.
 
Fine!

http://3-ps.googleusercontent.com/x/www.intmath.com/intmstat.com/vectors/313x311x235-3D-vector.png.pagespeed.ic.ZP7PZ_6kVk.png

That's how you draw a vector.

From Ackbach's plot, we see that both vectors have $z=0$, so the plane that both vectors lie on is the z axis. Thus, z is the axis going up and down, x-axis is going left and right, and y-axis is going in and out...of course if you look at it slanted. :D
 
Last edited by a moderator:
thanks rido. i understand the simple ones like that. i just can't remember how to draw one on the the surface of another graph/curve from a certain point.
 
ineedhelpnow said:
i just can't remember how to draw one on the the surface of another graph/curve from a certain point.

Are you saying you still can't remember how to draw such a vector, or that you used to not know, but you do now? Vectors are directed line segments (just arrows), and they're straight. So if you know where a vector's origin is, and where it needs to point to, then you just draw a straight line from point A to point B, and put an arrowhead on the tip to where it's pointing.
 
is the vector i drew in my drawing (v(1)) correct?
 
  • #10
ineedhelpnow said:
is the vector i drew in my drawing (v(1)) correct?

I see no $\vec{v}(1)$ drawings in any of your posts that are correct for Problem 7. Now it seems to me that in one pic, you were working on 7, but the drawing got clipped. If you were to re-scan that work, I could evaluate it for you.

But you haven't answered my question: Do you, or do you not currently know how to draw vectors in 3D space?
 
  • #11
I do know. I just got confused but I figured it out
 
Back
Top