Is there Instantaneous Acceleration?

In summary, the instantaneous acceleration is just the acceleration at a particular time. The acceleration is a constant (it doesn't even depend on the value of t) and it is 17.
  • #1
brandtw
4
0
Is there such a thing as instantaneous acceleration or is it always zero? Problem gives: x(t)=8.5t^2-2t+6 and asks for instantaneous acceleration at t=3

Any help would be greatly appreciated
 
Physics news on Phys.org
  • #2
Originally posted by brandtw
Is there such a thing as instantaneous acceleration or is it always zero?
The instantaneous acceleration is just the acceleration at a particular time. In this case, it is certainly not zero. (Acceleration is the second derivative of distance with respect to time.)
 
  • #3
thank you so with the distance equation x(t)=8.5t^2-2t+6 the instantaneous acceleration is 17?
 
  • #4
Originally posted by brandtw
thank you so with the distance equation x(t)=8.5t^2-2t+6 the instantaneous acceleration is 17?
Yep. And in this case, the acceleration is a constant: it doesn't even depend on the value of t.
 
  • #5
thank you so much
 
  • #6
how do you know it's a constant, and when would it not be constant?
 
  • #7
Originally posted by NanoTech
how do you know it's a constant, and when would it not be constant?
If you are given the position function x(t), take the second derivative and see what happens.
 
  • #8
i've learned today of a third derivative, which is a change or "jerk" in the acceleration. (a rate of change in a change of position) wouldn't that be changing always and never a constant?
 
  • #9
Originally posted by NanoTech
i've learned today of a third derivative, which is a change or "jerk" in the acceleration. (a rate of change in a change of position) wouldn't that be changing always and never a constant?
Not at all. It depends. In the example that started this thread, the "jerk" is zero at all times.

Think of these simple examples:
- a body at rest = v=0, a=0, jerk=0
- a body moving with constant speed (in straight line): v=constant, a=0, jerk=0
- a falling body (no air resistance, etc.): v= changing, a=constant, jerk=0
 
  • #10
Let us review first the definition of acceleration, and the question asked.

given a position [tex] x [/tex] which varies with time.
We know that the derivitve of that [tex] \frac{dx}{dt} [/tex] is the instantious time rate of change (derivitive) of the position, which is given the name velocity. From that we take the derivitive of that, so you have [tex] \frac{d^2x}{dt^2} [/tex] or the acceleration at an instant in time (given position)

Now, we are asked what the acceleration at [tex] t = 3 [/tex] is. We are given the position function [tex] x(t)=8.5t^2-2t+6 [/tex]. Therefore taking the second derivitive [tex] \frac{d^2x(t)}{dt^2} [/tex] we find the derivitive to be [tex] 17 [/tex]

Now consider if [tex] x(t) = t - 2 [/tex] in which [tex] \frac{d^2x(t)}{dt^2} [/tex] would be 0. As you can see in the position function, no acceleration occurs, it is simply a constant position change based on time (indicating a constant velocity).
 
Last edited:
  • #11
forstajh, the derivative is simply 17, not 17 m/s^2. No units were implied.
 
  • #12
Originally posted by notmuch
forstajh, the derivative is simply 17, not 17 m/s^2. No units were implied.

Correct notmuch, I apologize for any confusion this may have caused to anyone. I am so used to using m/s^2 that I accidently put them even though no units were provided.

Thanks for pointing out my error.

-Jacob
 
  • #13
How does one find the rate of change of acceleration?

i.e. two planets moving toward each other.
 
  • #14
Silverious said:
How does one find the rate of change of acceleration?

i.e. two planets moving toward each other.

Silverious, I am quite interested if someone has offered you any answers. please let me know.
Thank you
 
  • #15
If you are given the position as a function of time for each object, say x1(t) and x2(t), then the distance between them is |x1(t)- x2(t)|. The rate at which the two objects are moving toward each other is the derivative of that. The acceleration of each object, relative to the other, is the second derivative.

Now, if you are actually referring to two object, planets, you know that F= ma and that, for gravity, F= GmM/r2 so you can find the acceleration of either one directly from that.
 
  • #16
That is a good explanation. Thanks.
 

1. What is instantaneous acceleration?

Instantaneous acceleration is the rate at which an object's velocity changes at a specific moment in time. It is a measure of how quickly an object is speeding up or slowing down at a given point.

2. How is instantaneous acceleration different from average acceleration?

While average acceleration is calculated over a period of time, instantaneous acceleration is measured at a specific instant. This means that instantaneous acceleration can change rapidly, while average acceleration may remain constant over a longer period of time.

3. Can instantaneous acceleration be negative?

Yes, instantaneous acceleration can be negative. This occurs when an object is slowing down, or when its velocity is decreasing. Negative acceleration is also known as deceleration or retardation.

4. How is instantaneous acceleration measured?

Instantaneous acceleration is measured using calculus, specifically by taking the derivative of the velocity function with respect to time. It can also be calculated by measuring an object's change in velocity over a very small time interval.

5. What factors affect instantaneous acceleration?

The factors that affect instantaneous acceleration include the initial velocity of the object, the force acting on the object, and the mass of the object. The direction of the force also plays a role in determining the direction of the instantaneous acceleration.

Similar threads

Replies
18
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
208
  • Special and General Relativity
Replies
11
Views
940
  • Classical Physics
Replies
1
Views
600
  • Calculus
Replies
1
Views
2K
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
25
Views
1K
  • Introductory Physics Homework Help
Replies
11
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
828
Back
Top