Instantaneous velocity animation

[Whatupdoc]

A web page designer creates an animation in which a dot on a computer screen has a position of r^\rightarrow [4.40cm + (2.20cm/s^2)t^2]\underlinei + (5.00 cm/s)t\underlinej

okay i already have the correct answer, but i would like to know how the author got it. i came close to getting the correct answer.

Question.) Find the instantaneous velocity at t=1.0. Give your answer as a pair of components separated by a comma(x,y).

ok to find the instantaneous velocity, i need to find the derivative of the function...

d/dt r = 2(2.20cm/s)*2(t)i + (5.00cm/s)
plug in 1.0 for t and got...

(8.8cm/s,5.00cm/s) <--- my answer

(4.40cm/s, 5.00 cm/s) <--- correct answer

it looks like the correct answer divided my x-component by 2, but why? can someone explain?

 

[Hurkyl]

<br /> \vec{r} = [4.40cm + (2.20 cm/s^2)t^2] \hat{i} + (5.00 cm/s)t \hat{j}<br />

That what you were trying to write?



Anyways, you've identified that your answer is twice the alledgedly correct answer. I notice that your answer has several factors of two in it, so the first thought that springs to my mind is: "Can I find a reason why one of those 2's shouldn't be there?"

 

[SpatialVacancy]

I think...

I think you are working the problem wrong. Take the derivative of the acceleration ONLY, we do not care about the initial position, we only want the derivative of a.

\dfrac{d}{dt} 2.20t^2

You do the math.

Hope this helps.

 

[ExtravagantDreams]

"2(2.20cm/s)*2(t)i"
too many twos

 

[Whatupdoc]

I think you are working the problem wrong. Take the derivative of the acceleration ONLY, we do not care about the initial position, we only want the derivative of a.

\dfrac{d}{dt} 2.20t^2

You do the math.

Hope this helps.

so...(2.20cm/s^2)t^2 is the acceleration of i?

wait...

i got two 2's because you see 2.20cm/s^2? the derivative of that is 2(2.20cm/s) right? and the derivative of t^2 is equal to 2t. so...
2(2.20cm/s) * 2t

am i not suppose to care about the square root on cm/s?

 

[thermodynamicaldude]

No...s^2 is simply a unit. It is NOT a variable. You are taking the derivative of the function with respect to the variable t, so try to envision that particular term as 2.20t^2...and the derivative of that would be 4.40t.

 

[Hurkyl]

Doesn't matter if s is a unit or a varaible: it's a constant with respect to t, so its derivative (WRT t) is zero.