# Derivation of Instantaneous Velocity

1. Nov 30, 2012

### 0x5B

Find the instantaneous velocity where r is the position vector as a function of time:
r(t)=(3.0m/s^2)t$\hat{x}$+(4.0m/s)t$\hat{y}$

I attempted to find the derivative of this to find instantaneous velocity, but the book's solution was different. I think the author of the book may have made a mistake, but if not, I would like to know what I've done wrong.
My answer: v(t)=(6.0m/s)t$\hat{x}$+(4.0m/s)$\hat{y}$
Book's answer: v(t)=(6.0m/s^2)t$\hat{x}$+(4.0m/s)$\hat{y}$

2. Nov 30, 2012

### aralbrec

Should this be:

r(t) = 3t2 $\hat{x}$ + 4t $\hat{y}$

Then

r'(t) = 6t $\hat{x}$ + 4 $\hat{y}$

as you and the book found.

Units for velocity are in m/s so '6' is '6 m/s2' and '4' is '4 m/s'
so I agree with the book :)

When you multiply m/s2 by seconds, the result is m/s

3. Nov 30, 2012

### 0x5B

Ah, thank you, I was unsure of how to handle the units and being in only ninth grade I wasn't so sure about calling out a university level physics book on a mistake.

Last edited: Nov 30, 2012
4. Nov 30, 2012

### aralbrec

Oh, there are mistakes... later editions and more elementary level tends to reduce the number of errors.

At ninth grade the math will probably get in your way but it's good to see some early interest :)

5. Nov 30, 2012

### 0x5B

Hmm... Taking a second look, I'm not sure we're on the same page. The '6t' should represent (6m/s^2), not just the '6', so it should be (6m/s^2) not (6m/s^2)*t, right? Or am I still missing something? I apologize for such a protracted conversation about something as simple as the units.

6. Nov 30, 2012

### aralbrec

The x component of the velocity is x(t) = 6t so '6t' is the horizontal velocity in m/s. It is changing with time that is why the t is there. At time 0, the horizontal velocity is x(0)=0 m/s. At time t=1, the horizontal velocity is x(1)=(6 m/s2)*(1 s) = 6 m/s

The units of x(t) are m/s. 't' is in seconds so that means the units of the constant '6' must be m/s2. m/s2 * s = m/s (cancel out one of the 's')

Last edited: Nov 30, 2012
7. Nov 30, 2012

### 0x5B

Ah, again, thank you. In hindsight, this was such an obvious oversight on my part.