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see calculus
Reply 2
Original post by Big-Daddy
Which SUVAT equations continue to apply in some form or other for non-constant acceleration? And how do I modify them to reach expressions for non-constant acceleration (where a is a function of t)?

e.g. v=u+at modified a=6t2+13t-7.


You can't really. They are derived from constant accelerations. If you have a = a(t), then one can find v(t) and hence x(t), via integration.
well if you have an equation like a=6t^2+13t-7 then you need to integrate to find velocity (in other words find the area under the graph), so you equation for velocity would be v=2t^3 + 6.5t^2 - 7t (+c)
and to find displacement you integrate velocity do s=0.5t^4 +(13/6)t^3 -3.5t^2 +ct (+d)

where c and d are constants :smile:

To find the velocity from displacement, just differentiate, and to find the acceleration from velocity, just differentiation.

My integration might be a bit iffy there btw, I was only using it as an example :wink:

It's nto really SUVAT but equations of motion :smile:
Reply 4
There aren't really any SUVAT equations for non-constant acceleration, you need to use calculus.
Reply 5
Original post by AspiringGenius
well if you have an equation like a=6t^2+13t-7 then you need to integrate to find velocity (in other words find the area under the graph), so you equation for velocity would be v=2t^3 + 6.5t^2 - 7t (+c)
and to find displacement you integrate velocity do s=0.5t^4 +(13/6)t^3 -3.5t^2 +ct (+d)

where c and d are constants :smile:

To find the velocity from displacement, just differentiate, and to find the acceleration from velocity, just differentiation.

My integration might be a bit iffy there btw, I was only using it as an example :wink:

It's nto really SUVAT but equations of motion :smile:


OK your constant c is v0 i.e. u. What's d? s0?

Let's say I want to find the stationary points of s: set v(t)=0, solve for t, then plug this into s(t)? Stationary points of v: set a(t)=0, solve for t, plug into v(t)?

What about v2=u2+2as? I'm guessing there's no equivalent (since t is inherently needed for a).
Original post by Big-Daddy
OK your constant c is v0 i.e. u. What's d? s0?

Let's say I want to find the stationary points of s: set v(t)=0, solve for t, then plug this into s(t)? Stationary points of v: set a(t)=0, solve for t, plug into v(t)?

What about v2=u2+2as? I'm guessing there's no equivalent (since t is inherently needed for a).


it's just a constant used in integration for when it's just a general formula- when finding the velocity between 2 and 5 seconds, you don't need to use it.

When you get an equation like a=6t^2+4t etc, you generally don't have to use SUVAT in that instance so it's best to put it out of mine. SUVAT is for constant acceleration :wink:
Original post by Big-Daddy
Which SUVAT equations continue to apply in some form or other for non-constant acceleration? And how do I modify them to reach expressions for non-constant acceleration (where a is a function of t)?

e.g. v=u+at modified a=6t2+13t-7.


If acceleration is changing uniformly then the rate of change of acceleration is often called jerk, jolt or surge and is given by
J = (a2 - a1) / t
This gives
a2 = a1 + jt
This is similar to the suvat v = u + at

Other equations here

http://en.wikipedia.org/wiki/Jerk_(physics)#Equations
(edited 11 years ago)
Reply 8
Original post by AspiringGenius
it's just a constant used in integration for when it's just a general formula- when finding the velocity between 2 and 5 seconds, you don't need to use it.


This is because you're talking about definite integration and thus the v0 which describes your constant is cancelled out, right?

This is necessary - just integrating on its own won't be enough, you will need to provide a value for your constant, and to me that seems to be v0 and s0 respectively. If you are given the equation for acceleration in terms of time and then integrate to find velocity in terms of t, then the constant must be v0, lest you make the assumption that at t=0, v=0, which may be very wrong.

A more interesting scenario, though - what if you are given a in terms of v? Or s in terms of v or v in terms of s? (Only a in terms of v should be common really but it's still an interesting idea to me.)
(edited 11 years ago)
Original post by Big-Daddy
This is because you're talking about definite integration and thus the v0 which describes your constant is cancelled out, right?

This is necessary - just integrating on its own won't be enough, you will need to provide a value for your constant, and to me that seems to be v0 and s0 respectively. If you are given the equation for acceleration in terms of time and then integrate to find velocity in terms of t, then the constant must be v0, lest you make the assumption that at t=0, v=0, which may be very wrong.

A more interesting scenario, though - what if you are given a in terms of v? Or s in terms of v or v in terms of s? (Only a in terms of v should be common really but it's still an interesting idea to me.)


I didn't know if you knew differentiation/integration, sorry :smile: If they require you to write an equation, for an indefinite integral, they will usually give you some information (for example, at t=0, the object is at rest) which you can substitute to find the constants. If it is definite integrals you don't need to do this.

they can give you a funky equation for a in terms of t and v in terms of t, but as a=v/t I don't think they can give you a crazy euqtion with a in terms of v and if they did it'd probably be a rearra nge and substittute questino. In any case I haven't come across this yet :smile:
Reply 10
Original post by AspiringGenius
I didn't know if you knew differentiation/integration, sorry :smile: If they require you to write an equation, for an indefinite integral, they will usually give you some information (for example, at t=0, the object is at rest) which you can substitute to find the constants. If it is definite integrals you don't need to do this.

they can give you a funky equation for a in terms of t and v in terms of t, but as a=v/t I don't think they can give you a crazy euqtion with a in terms of v and if they did it'd probably be a rearra nge and substittute questino. In any case I haven't come across this yet :smile:


I see. Thank you very much for the help. :smile:
Original post by Stonebridge
If acceleration is changing uniformly then the rate of change of acceleration is often called jerk, jolt or surge and is given by
J = (a2 - a1) / t
This gives
a2 = a1 + jt
This is similar to the suvat v = u + at

Other equations here

http://en.wikipedia.org/wiki/Jerk_(physics)#Equations



Hi Stonebridge,


I have a question which asks how do the three main suvat equations change if the acceleration is non constant? It doesn't give more information than that, so I was wondering whether I could change to the suvat equations using calculus to include jerk?
Original post by LiamEagle
Hi Stonebridge,


I have a question which asks how do the three main suvat equations change if the acceleration is non constant? It doesn't give more information than that, so I was wondering whether I could change to the suvat equations using calculus to include jerk?


It depends what level you are working at, but from what has been said above, I would guess they are asking you to integrate the defining formula for j (jerk) which is j = (a2 - a1) / t
jerk = rate of change of acceleration, and is constant.
As I have already said, this gives
a2 = a1 + jt

Integrate this wrt time and you get the equation for v (which looks like the suvat for s = ut + ½at²)

Integrate again and you get s
Original post by Stonebridge
It depends what level you are working at, but from what has been said above, I would guess they are asking you to integrate the defining formula for j (jerk) which is j = (a2 - a1) / t
jerk = rate of change of acceleration, and is constant.
As I have already said, this gives
a2 = a1 + jt

Integrate this wrt time and you get the equation for v (which looks like the suvat for s = ut + ½at²)

Integrate again and you get s


Makes sense now, thank you for your help!
v2=u2+2*integral of a wrt s from s initial to s final

You can derive this either by using the fact that a =vdv/dx, rearranging and integrating both sides or by using the work-energy theorem and rearranging

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