Instantaneous velocity and acceleration as functions of time

In summary, the position of an object changes as a function of time according to x = At^2 + Bt + C. A = 8 m/s^2, B = 6 m/s, and C = 4 m.
  • #1
jensson
25
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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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  • #2
how does velocity relate to position?
how does acceleration relate to velocity?
 
  • #3
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.
 
  • #4
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')?
 
  • #5
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
 
  • #6
You have the right idea, you're just not taking the derivative correctly.

[tex]x = At^2 + Bt + C[/tex]
[tex]v = 2At^1 + Bt^0 + 0[/tex] 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). ====>
[tex]v = 2At + B[/tex]
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?
 
  • #7
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!
 
  • #8
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?
 
  • #9
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
 
  • #10

FAQ: Instantaneous velocity and acceleration as functions of time

1. What is instantaneous velocity?

Instantaneous velocity is the rate of change of an object's position with respect to time at a specific moment. It is also known as the object's velocity at a specific point in time.

2. How is instantaneous velocity calculated?

Instantaneous velocity can be calculated by finding the slope of the tangent line to the position-time graph at the specific point in time. It can also be calculated by taking the derivative of the position function with respect to time.

3. What is the difference between instantaneous velocity and average velocity?

Instantaneous velocity is the velocity of an object at a specific point in time, while average velocity is the total displacement of an object over a period of time. Average velocity is calculated by dividing the change in position by the change in time.

4. How is acceleration related to instantaneous velocity?

Acceleration is the rate of change of an object's velocity with respect to time. In other words, it is the change in velocity over a specific period of time. Instantaneous velocity is used to calculate acceleration by finding the slope of the tangent line to the velocity-time graph.

5. Can instantaneous velocity and acceleration be negative?

Yes, instantaneous velocity and acceleration can be negative. This means that the object is moving in the opposite direction of the positive direction, as defined by the chosen coordinate system. Negative acceleration, also known as deceleration, means that the object is slowing down.

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