# SUVAT Equations for Non-Constant Acceleration

Watch
Announcements
#1
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.
0
7 years ago
#2
see calculus
1
7 years ago
#3
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.
0
7 years ago
#4
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

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

It's nto really SUVAT but equations of motion
0
7 years ago
#5
There aren't really any SUVAT equations for non-constant acceleration, you need to use calculus.
0
#6
(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

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

It's nto really SUVAT but equations of motion
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).
0
7 years ago
#7
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
0
7 years ago
#8
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
0
#9
(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.)
0
7 years ago
#10
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 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
0
#11
(Original post by AspiringGenius)
I didn't know if you knew differentiation/integration, sorry 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
I see. Thank you very much for the help.
0
6 years ago
#12
(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?
0
6 years ago
#13
(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
0
6 years ago
#14
(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!
0
X

new posts
Back
to top
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### Poll

Join the discussion

#### Are you travelling in the Uni student travel window (3-9 Dec) to go home for Christmas?

Yes (50)
31.85%
No - I have already returned home (19)
12.1%
No - I plan on travelling outside these dates (33)
21.02%
No - I'm staying at my term time address over Christmas (14)
8.92%
No - I live at home during term anyway (41)
26.11%