Calculate the instantaneous acceleration

In summary, the conversation revolved around calculating the instantaneous acceleration for a car's velocity as a function of time, given by Vx (t) = alpha + beta (t)^2. The solution involved taking the derivative and finding the average over a small interval, or taking the limit as the interval approaches zero. The correct answer was determined to be 1, but there was some confusion about the precision of the answer and how to solve for it at time = 0. Ultimately, it was determined that the instantaneous acceleration would not always be 1, as it depends on the specific values of alpha and beta in the equation.
  • #1
Heat
273
0

Homework Statement


A car's velocity as a function of time is given by Vx (t) = alpha + beta (t)^2, where alpha = 3.00m/s and beta = .100m/s^3.

Calculate the instantaneous acceleration for t= 5seconds

Homework Equations



ax = dv/dt


The Attempt at a Solution



My first attempt is to get various times and get their corresponding acceleration.

vx @ 0s = 3m/s
vx @ 1s = 3.1m/s
vx @ .5s = 3.025 m/s
vx@ 5s = 5.5

I know compared a= 3.1-3/1-0 = .1
3.035-3/.5-0 = .05

the average would end up being around .1

I know this problems involved more calculus math, derivitives etc.

How do I apply this here.
 
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  • #2
Lets go over a few easy rules for calculus:

[tex]\frac{d}{dx}c = 0[/tex] Constants always become zero when take a derivative

[tex]\frac{d}{dx}x^n = nx^{n-1}[/tex] The power rule, if there is a coefficient multiply n by the coefficient.

But if you don't really know calculus you probably shouldn't worry about it...

If you want to approximate the instantaneous acceleration consider this: the idea of an instantaneous rate of change is finding an average between two points secant line, and shrinking the distance between those points until they are infinitely small right? This forms what we call a tangent line. Since you don't have a way to do this without calc you could do the next best thing find really small averages from t=5s. I'm talking .1, .01, .001, .0001 off from 5. You don't want to do 0, 1, .5, and 5 -- None of these are really very close at all.
 
  • #3
yeah but that will give me .00001

is this correct?

also, when I chose close numbers to the time would it not give me an number near one alway?
 
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  • #4
what is the derivative of your velocity equation?
 
  • #5
well the equation would look like this (keep in mind I am taking calc along with physics):

3.00 + .100 t^2

which leads to:

3+.2 = 3.2?

I don't know much about derivitives yet...
 
  • #6
Heat said:
well the equation would look like this (keep in mind I am taking calc along with physics):

3.00 + .100 t^2

which leads to:

3+.2 = 3.2?

I don't know much about derivitives yet...
i'd be real worried then ... :O taking the derivative is real ez, i would go ahead and scan your calculus book on how to do so.

the derivative would be

[tex]a_x(t)=2\beta{t}[/tex]
 
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  • #7
I'm taking Physics with Calc, Calc being taken at the same time, now about the problem.o:)
 
  • #8
Heat said:
I'm taking Physics with Calc, Calc being taken at the same time, now about the problem.o:)
if you want some examples i'll be more than happy to show you.
 
  • #9
Heat said:
yeah but that will give me .00001

is this correct?

also, when I chose close numbers to the time would it not give me an number near one alway?

No, when I mean find the average over a small interval I mean

[tex]m = \frac{f(x_2)-f(x_1)}{x_2-x_1} = \frac{v_{x}(5+h)-v_{x}(5)}{(5+h)-5}[/tex]

where h is very very small...

Or better yet if you know how, take the limit as h -> 0
 
  • #10
Feldoh said:
No, when I mean find the average over a small interval I mean

[tex]m = \frac{f(x_2)-f(x_1)}{x_2-x_1} = \frac{v_{x}(5+h)-v_{x}(5)}{(5+h)-5}[/tex]

where h is very very small...

Or better yet if you know how, take the limit as h -> 0


[tex]m = \frac{v_{x}(5+.00001)-v_{x}(5)}{(5+.00001)-5}[/tex]?

= .00005/.00005 = 1?

nvm, I think I got it, I am working on it, I forgot to plug into the original equation provided.
 
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  • #11
Update:

5.5 @ 5s
5.50001 @ 5.00001s

5.5 - 5 / 5.00001 - 5 = 50000m/s^2 ? Is this right, seems large :redface:
 
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  • #12
1 is the right answer.
 
  • #13
Is is really, I marked 1 again, but it tells me I am incorrect. :(

I got two more tries, I don't know if I shuold mark 1.00 (but isn't that more precise)?

lol
 
  • #14
Heat said:
Is is really, I marked 1 again, but it tells me I am incorrect. :(

I got two more tries, I don't know if I shuold mark 1.00 (but isn't that more precise)?

lol

Make sure you have posted the question right...
 
  • #15
lol, ok, (I was marking it on the wrong question) :blushing:

So the way I did it earlier would be the correct way to solve for it. Let's say time is 0. Then it would be:

(0+.00001) - 0 / (0+.00001 - 0) = 1

If this is right, then all instant acceleration would have to be 1 . :uhh:
 
  • #16
Heat said:
lol, ok, (I was marking it on the wrong question) :blushing:

So the way I did it earlier would be the correct way to solve for it. Let's say time is 0. Then it would be:

(0+.00001) - 0 / (0+.00001 - 0) = 1

If this is right, then all instant acceleration would have to be 1 . :uhh:

Yes, you can do it that way... that will give you a good approximation... for exact results you need derivatives.
 
  • #17
learningphysics said:
Yes, you can do it that way... that will give you a good approximation... for exact results you need derivatives.

The point I am trying to make is that regardless of time, will the instant acceleration alway be one. Because if you add a number to the original time (ex: .0000000001) and follow the equation format, then you will eventually end up with the same thing =1.

But in the question I am trying to solve, when I place 1 as the answer when time = 0 it would mark it as incorrect.

BTW, I just remembered that the top is going to be (3+.00001)- 0 / (0+.00001-0)= 1.

:cry:
 
  • #18
Heat said:
The point I am trying to make is that regardless of time, will the instant acceleration alway be one. Because if you add a number to the original time (ex: .0000000001) and follow the equation format, then you will eventually end up with the same thing =1.

But in the question I am trying to solve, when I place 1 as the answer when time = 0 it would mark it as incorrect.

BTW, I just remembered that the top is going to be (3+.00001)- 0 / (0+.00001-0)= 1.

:cry:

V(0) = 3
V(0.0001) = 3.000000001

[V(0.0001) - V(0)]/(0.0001 - 0) = 0.000000001/0.0001 = 0.00001 not 1.
 
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  • #19
Heat said:
The point I am trying to make is that regardless of time, will the instant acceleration alway be one. Because if you add a number to the original time (ex: .0000000001) and follow the equation format, then you will eventually end up with the same thing =1.

But in the question I am trying to solve, when I place 1 as the answer when time = 0 it would mark it as incorrect.

BTW, I just remembered that the top is going to be (3+.00001)- 0 / (0+.00001-0)= 1.

:cry:

No at t=0, it is not 1. It is only 1 at t = 5.
 
  • #20
From rocophysics' post

[tex]a_x(t)=2\beta{t}[/tex]

where [tex]\beta = 0.1[/tex]
 
  • #21
so it would be 2(.100)(t)

@ time = 0

=0

@ time = 5

=1

@ time = 10

= 2

(I will go into my calculus book and do some reading and work out derivitives)

Also, would it be 0.00 or would it just be 0 at time = 0?:rofl:

and thank you again learningphysics, you and rockophysics and Feldoh are great helpers. :smile:
 
  • #22
Heat said:
so it would be 2(.100)(t)

@ time = 0

=0

@ time = 5

=1

@ time = 10

= 2

Yup.

(I will go into my calculus book and do some reading and work out derivitives)

Also, would it be 0.00 or would it just be 0 at time = 0?:rofl:

lol. I think 0 will be fine.

and thank you again learningphysics, you and rockophysics and Feldoh are great helpers. :smile:

no prob. :smile:
 

1. What is meant by instantaneous acceleration?

Instantaneous acceleration is the rate of change of velocity at a specific moment in time. It is the acceleration at a precise point rather than an average over a period of time.

2. How is instantaneous acceleration calculated?

To calculate instantaneous acceleration, you need to measure the change in velocity over an infinitesimally small time interval. This can be done by taking the derivative of the velocity function with respect to time, or by using the formula a = (vf - vi)/t, where vf is the final velocity, vi is the initial velocity, and t is the time interval.

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

Instantaneous acceleration is the acceleration at a specific moment, while average acceleration is the average rate of change of velocity over a period of time. Instantaneous acceleration is more precise, while average acceleration gives an overall understanding of the acceleration over a certain time frame.

4. Can instantaneous acceleration be negative?

Yes, instantaneous acceleration can be negative. This means that the object is slowing down, since its velocity is decreasing. A positive instantaneous acceleration indicates that the object is speeding up, since its velocity is increasing.

5. How does instantaneous acceleration relate to position and velocity?

Instantaneous acceleration is the second derivative of position with respect to time, and the first derivative of velocity with respect to time. This means that it is a measure of how quickly an object's velocity is changing with respect to time, and how quickly its position is changing with respect to time.

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