How Do You Calculate Instantaneous Velocity in Vector Animation?

[kk727]

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?

 

[collinsmark]

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!

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

\dot r = \left( 5t \right) \hat \imath + \left( 5 \right) \hat \jmath

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. \hat \imath is the x-component and \hat \jmath 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,
c = \sqrt{a^2 + b^2}

\tan \theta = \left( \frac{\mathrm{opposite}}{\mathrm{adjacent}} \right)

 

[kk727]

Oh! Completely makes sense now...hahaha, I feel stupid! Thank you for your help, you saying that just made it all click!