Calculate Distance Traveled: Instantaneous vs. Average Velocity

[Slimy0233]

Homework Statement

Velocity of a particle is $\vec{v}=V_0\left[-\sin \omega t \hat{i}+\cos \omega t \hat{j}\right]$
$V_0, \omega$ are constants.
Calc dist, displacement during time $t = 0$ to $t = t_{0}$

Relevant Equations

$\vec{v}=V_0\left[-\sin \omega t \hat{i}+\cos \omega t \hat{j} \right]$

edit: I don't know why my latex isn't rendering, any help would be appreciated.

Edit 2: The question was due to a misunderstanding I had, I thought integrating instantaneous velocity would give me average velocity.

I have attached what I have tried so far. I had a doubt. Can you calculate the distance travelled by an object/particle using only the instantaneous velocity?

I mean, is the speed travelled by an object in time interval t = 0 to t = $t_{0}$
S = $V_{inst}*t_{0}$

My professor basically did this [Image with prof in it]

The other one is what I think the answer would be. I calculate the average velocity from instantaneous velocity by integrating $V_{inst}$ and then found out the average speed by finding the magnitude of the average velocity <vec{v}>

So, who is right? (I know I probably am wrong, but I want to know why)

 

[haruspex]

You are asked to find a distance. Integrating a velocity gives displacement.
To find the distance travelled you need to integrate the speed, a magnitude, not a vector.
It just happens that in this case the speed is constant, so using the instantaneous speed is ok.

 

[Slimy0233]

I am really stupid to have asked this question, thank you for your reply!

 

[Slimy0233]

hey @haruspex why is the avg velocity same as instantaneous velocity here?

 

[haruspex]

hey @haruspex why is the avg velocity same as instantaneous velocity here?

No, the average velocity is not the same as the instantaneous velocity. Velocity is a vector, speed is its magnitude.
Look at what your prof did: found the speed by taking the magnitude of the velocity.
$$\left.\vec{v}=V_0[-\sin( \omega t )\hat{i}+\cos( \omega t )\hat{j} \right]$$
$$\vec{v}^2=V_0^2[(-\sin( \omega t ))^2+(\cos( \omega t ))^2]=V_0^2$$
So the speed is constant, and the average speed is the same as the instantaneous speed. The velocity is changing, though, because the direction keeps changing. Over time $\frac{2\pi}{\omega}$, the average velocity is zero.

 

[hutchphd]

Homework Statement: Velocity of a particle is $\left.\vec{v}=V_0[-\sin \omega t \hat{i}+\cos \omega t \hat{j}\right]$ $V_0, \omega$ are constants Calc dist, displacement during time t = 0 to t = $t_{0}$
Relevant Equations: $$\left.\vec{v}=V_0[-\sin \omega t \hat{i}+\cos \omega t \hat{j} \right]$$

I mean, is the speed travelled by an object in time interval t = 0 to t = t0
S = Vinst∗t0

Alarm. Speed has the same units as velocity.