1. Sep 2, 2011

### megkirch

1. The problem statement, all variables and given/known data
The position x of an object as a function of time is given as x = 2.257 + 4.894 t + 16.5 t2. All quantities are understood to be quoted in SI units. What is the initial velocity ? Indicate with a negative (positive) sign whether the acceleration is constant (not constant) for the given expression.

2. Relevant equations
i think i have to use. x=x0+v0(t)+1/2(a)t^2)

3. The attempt at a solution
the problem is confusing me becuase im not sure what numbers should go where in the equation. i know the initial velocity is v0.

2. Sep 2, 2011

### olivermsun

So can you compare the equation you wrote in (2) with the one you wrote in (1)?

3. Sep 2, 2011

### megkirch

2.257 would be the x0, 4.894 would be the t, and 16.5 would go where the t^2 is. im not sure what to put as the a.

4. Sep 2, 2011

### tiny-tim

welcome to pf!

hi megkirch! welcome to pf!
yes
16.5 is the ½a

5. Sep 2, 2011

### megkirch

ohh ok thank you! So it would be, x=2.257+V0(4.894)+1/2(16.5)?

6. Sep 2, 2011

### tiny-tim

i hesitate to dampen your enthusiasm, but that doesn't really make any sense :tongue2:

would you like to rephrase it?

7. Sep 2, 2011

### megkirch

do i have that equation set up right?

8. Sep 2, 2011

9. Sep 2, 2011

### Rayquesto

What I would do is derive the equation using nx^n-1 where n represents the power in terms of time and you will get the equation that represents velocity.

from this equation, v= v0 + at, what can you say about v0 in the equation that you derived. (hint: the derived equation is identical to the velocity equation)

10. Sep 2, 2011

### tiny-tim

the equation is x = 2.257 + 4.894 t + 16.5 t2

you have to find v0 and a …

so what is a?​

11. Sep 2, 2011

### megkirch

sorry, im completely lost

12. Sep 2, 2011

### Rayquesto

do you know how to find a derivative?

13. Sep 2, 2011

### megkirch

yes the derivative of the equations is. 2.257=0 4.894t=4.894 16.5t^2=33.

14. Sep 2, 2011

### megkirch

thank you so much! i finally got the answer its 4.894

15. Sep 2, 2011

### Rayquesto

thats what I think :) want a second opinion? I mean I'm sure that's right.

16. Sep 2, 2011

### megkirch

my hw is online it said its right. thank you again