Calculate the instantaneous acceleration

[Heat]

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.

 

[Feldoh]

Lets go over a few easy rules for calculus:

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

\frac{d}{dx}x^n = nx^{n-1} 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.

 

[Heat]

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?

 

[rocomath]

what is the derivative of your velocity equation?

 

[Heat]

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...

 

[rocomath]

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

a_x(t)=2\beta{t}

 

[Heat]

I'm taking Physics with Calc, Calc being taken at the same time, now about the problem.

 

[rocomath]

I'm taking Physics with Calc, Calc being taken at the same time, now about the problem.

if you want some examples i'll be more than happy to show you.

 

[Feldoh]

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

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

where h is very very small...

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

 

[Heat]

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

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

where h is very very small...

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

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

= .00005/.00005 = 1?

nvm, I think I got it, I am working on it, I forgot to plug into the original equation provided.

 

[Heat]

Update:

5.5 @ 5s
5.50001 @ 5.00001s

5.5 - 5 / 5.00001 - 5 = 50000m/s^2 ? Is this right, seems large

 

[learningphysics]

1 is the right answer.

 

[Heat]

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

 

[learningphysics]

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...

 

[Heat]

lol, ok, (I was marking it on the wrong question)

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 .

 

[learningphysics]

lol, ok, (I was marking it on the wrong question)

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 .

Yes, you can do it that way... that will give you a good approximation... for exact results you need derivatives.

 

[Heat]

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.

 

[learningphysics]

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.

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

[V(0.0001) - V(0)]/(0.0001 - 0) = 0.000000001/0.0001 = 0.00001 not 1.

 

[learningphysics]

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.

No at t=0, it is not 1. It is only 1 at t = 5.

 

[learningphysics]

From rocophysics' post

a_x(t)=2\beta{t}

where \beta = 0.1

 

[Heat]

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?

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

 

[learningphysics]

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?

lol. I think 0 will be fine.

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

no prob.