Instantaneous Velocity

  • Thread starter kk727
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Question: A web page designer creates an animation in which a dot on a computer screen has a position of r=[4.0 cm + (2.5 cm/s2)t2)i + [(5.0 cm/s)t]j.

a. Find the magnitude and direction of the dot's average velocity between t=0 and t=2.0s.
b. Find the magnitude and direction of the instantaneous velocity at t=0, t=1.0s, and t=2.0s.


Homework Equations


r=[4.0 cm + (2.5 cm/s2)t2)i + [(5.0 cm/s)t]j.


The Attempt at a Solution


For part A, I had no problem. I found the position at t=0 and t=2 and used the distance formula to find the resultant. I divided that by the time interval, 2 seconds, and got an average velocity of 7.07 m/s at a 45 degree angle. When I checked this with the back of my book, my answer was correct.

For part B, I figured that I would take the derivative of the given equation, which is what I did.

r=(4.0 + 2.5t2)+5t
r'=5t + 5

I'm already unsure about this step; I'm not quite sure what the I and J indicate in the original equation. I thought they meant X and Y, so I also tried splitting up the formula into an x-component and a y-component, but I still did not get the right answer.
For t=0, the answer was 5.0 cm/s, which I think I got just by sheer luck and coincidence. So really, I don't even know how to approach part B of this problem...the answer for t=1.0s is supposed to be 7.1 cm/s...help?
 

Answers and Replies

  • #2
collinsmark
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For part B, I figured that I would take the derivative of the given equation, which is what I did.

r=(4.0 + 2.5t2)+5t
r'=5t + 5
Don't forget your unit vectors! :smile:

[tex] r = \left(4.0 + (2.5t^2) \right) \hat \imath + (5 t) \hat \jmath [/tex]

[tex] \dot r = \left( 5t \right) \hat \imath + \left( 5 \right) \hat \jmath [/tex]
I'm already unsure about this step; I'm not quite sure what the I and J indicate in the original equation. I thought they meant X and Y,
Essentially, yes that's right. [itex] \hat \imath [/itex] is the x-component and [itex] \hat \jmath [/itex] is the y-component.
so I also tried splitting up the formula into an x-component and a y-component, but I still did not get the right answer.
Did you remember to use the Pythagorean theorem? For a right triangle with sides a and b, with hypotenuse c,
[tex] c = \sqrt{a^2 + b^2} [/tex]

[tex] \tan \theta = \left( \frac{\mathrm{opposite}}{\mathrm{adjacent}} \right) [/tex]
 
  • #3
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Oh! Completely makes sense now...hahaha, I feel stupid! Thank you for your help, you saying that just made it all click!
 

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