# A Particle's Velocity

1. Jan 24, 2010

### erok81

1. The problem statement, all variables and given/known data

A particle's velocity is described by the function $$v_x=kt^2 m/s$$, where k is a constant and t is in s. The particle's position at $$t_0$$=0 is $$x_0$$ = -6.90 . At $$t_1$$= 1.00, the particle is at $$x_0$$= 8.70 .

Determine the units of k in terms of m and s.
2. Relevant equations

None

3. The attempt at a solution

So I really have no idea what this question is asking. I tried integrating the velocity equation then solving for c and eventually k getting a value of 46.8. It's online homework and it wouldn't take that, even if I added m/s to the end.

I am pretty sure what I tried isn't remotely close to what I am supposed to be doing. Any ideas?

2. Jan 24, 2010

### ideasrule

Aren't you supposed to be finding the units instead of calculating any numerical values? What unit do you multiply by s^2, the unit of t^2, to get m/s, the unit of v?

3. Jan 24, 2010

### erok81

Yeah, you are right. Although I am almost more confused now. I seriously have no idea what I am supposed to do or what the answer is supposed to look like. It's a first for me so I am completely dumbfounded. :rofl:

Am I just supposed to get rid of the t^2 so k is alone with m/s?

For some reason I am always stumped on these extremely easy problems. I don't get them. It's pretty sad.

4. Jan 24, 2010

### ideasrule

Here's an example: if k has the unit "s", kt^2 would have the unit s^3 (s * s^2 = s^3). If k has the unit "m", kt^2 would have the unit ms^2 (m * s^2 = ms^3). What unit, multiplied by s^2, gives m/s? You can treat the units the same way you treat algebraic variables.

5. Jan 24, 2010

### erok81

m/s^3?

I think why I am so confused it because it already has m/s in the first part. Does that even matter?

6. Jan 24, 2010

### ideasrule

That's just to tell you that v is in m/s. It really shouldn't be written like that; I can see why it would confuse the hell out of a lot of students.

But yes, m/s^3 is right.

7. Jan 24, 2010

### erok81

Thank you, thank you, thank you. There is no way I would have figured that out on my own.

That original m/s was definately throwing me off. Since it almost seemed like it was already in m/s terms...almost.

Thanks again.

8. Jan 24, 2010

### erok81

On a side note the next question was to figure out the value for k. Which I had correct from my original incorrect answer. So that makes me feel a bit better about it.:tongue2: