Finding the initial velocity, .

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megkirch
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Homework Statement


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.


Homework Equations


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


The Attempt at a Solution


the problem is confusing me becuase I am not sure what numbers should go where in the equation. i know the initial velocity is v0.
 
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So can you compare the equation you wrote in (2) with the one you wrote in (1)?
 
2.257 would be the x0, 4.894 would be the t, and 16.5 would go where the t^2 is. I am not sure what to put as the a.
 
welcome to pf!

hi megkirch! welcome to pf! :wink:
megkirch said:
2.257 would be the x0, 4.894 would be the t

yes :smile:
… , and 16.5 would go where the t^2 is. I am not sure what to put as the a.

16.5 is the ½a :wink:
 
ohh ok thank you! So it would be, x=2.257+V0(4.894)+1/2(16.5)?
 
megkirch said:
ohh ok thank you! So it would be, x=2.257+V0(4.894)+1/2(16.5)?

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

would you like to rephrase it? :smile:
 
do i have that equation set up right?
 
i have been stuck on this one problem all day, please help.
 
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)
 
megkirch said:
do i have that equation set up right?

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

you have to find v0 and a …

so what is a?​
 
sorry, I am completely lost
 
do you know how to find a derivative?
 
yes the derivative of the equations is. 2.257=0 4.894t=4.894 16.5t^2=33.
 
thank you so much! i finally got the answer its 4.894
 
thats what I think :) want a second opinion? I mean I'm sure that's right.
 
my homework is online it said its right. thank you again