Instantaneous velocity and acceleration as functions of time

AI Thread Summary
The discussion focuses on understanding how to derive instantaneous velocity and acceleration from a position function given as x = At^2 + Bt + C. Participants clarify that velocity is the derivative of position with respect to time, while acceleration is the derivative of velocity. The correct derivative process is emphasized, showing that velocity is v = 2At + B and acceleration is the derivative of that result. The conversation highlights the challenge of learning calculus and physics simultaneously, with encouragement to approach problems step by step. Overall, mastering derivatives is crucial for solving these types of physics problems effectively.
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Homework Statement


The position of an object as a function of time is given by x = At^2 + Bt + C
A = 8 m/s^2
B = 6 m/s
C = 4 m

Homework Equations


v = u + at
a = v/t

The Attempt at a Solution



I don't even know how to set this up!
 
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how does velocity relate to position?
how does acceleration relate to velocity?
 
velocity is the change in position with respect to the change in time and acceleration is the change in velocity over time.. is that what you mean?



this question was on a quiz and I got it wrong.. I can't seem to make it make sense in my head. I have the answers and I still can't figure out how to get there.
 
You're exactly right so far! So what mathematical process gives you the change of something (i.e. position 'x') with respect to a variable (i.e. time 't')?
 
I can only think of putting that into equation form

V = dx/dt

but the position is given has a quadratic equation.


V = 8t + 6t + 4/t

but that doesn't seem right
 
You have the right idea, you're just not taking the derivative correctly.

x = At^2 + Bt + C
v = 2At^1 + Bt^0 + 0 You always decrease the power of the variable by one--but if there is no variable, then the derivate is zero---in other words: if there is no variable, it is constant, and thus the slope/change is zero.
Then the 'B' term becomes just B, and the C term drops out (is zero). ====>
v = 2At + B
then you repeat to find the acceleration, and plug into find the values.

Does that make sense? Make sure you review your derivatives :)
What does the acceleration equal?
 
I am in calculus I right now and taking physics. My advisor said I should be fine taking them at the same time, but I seem to be completely lost!
 
Yeah, that's definitely an extra challenge, but its totally doable.
They key is just not getting overwhelmed by it; go one step at a time, one problem at a time, one concept at a time.

Does the derivative I did above make sense?
 
hah I already do feel overwhelmed by it :-/
A 2 problem quiz every thursday and thanks to this problem here I got 1/2 wrong :(

I can see the pattern in your steps, but I have not studied derivatives yet so I don't know why it was done
 
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