# SUVAT Equations for Non-Constant Acceleration

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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=6t

e.g. v=u+at modified a=6t

^{2}+13t-7.
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#3

(Original post by

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=6t

**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=6t

^{2}+13t-7.
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#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

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

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#5

There aren't really any SUVAT equations for non-constant acceleration, you need to use calculus.

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(Original post by

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

**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

_{0}i.e. u. What's d? s

_{0}?

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 v

^{2}=u

^{2}+2as? I'm guessing there's no equivalent (since t is inherently needed for a).

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#7

(Original post by

OK your constant c is v

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 v

**Big-Daddy**)OK your constant c is v

_{0}i.e. u. What's d? s_{0}?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 v

^{2}=u^{2}+2as? I'm guessing there's no equivalent (since t is inherently needed for a).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

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#8

**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=6t

^{2}+13t-7.

J = (a

_{2}- a

_{1}) / t

This gives

a

_{2}= a

_{1}+ jt

This is similar to the suvat v = u + at

Other equations here

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

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(Original post by

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.

**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.

_{0}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 v

_{0}and s

_{0}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 v

_{0}, 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.)

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#10

(Original post by

This is because you're talking about definite integration and thus the v

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 v

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.)

**Big-Daddy**)This is because you're talking about definite integration and thus the v

_{0}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 v

_{0}and s_{0}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 v_{0}, 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.)

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

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(Original post by

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

**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

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#12

(Original post by

If acceleration is changing uniformly then the rate of change of acceleration is often called jerk, jolt or surge and is given by

J = (a

This gives

a

This is similar to the suvat v = u + at

Other equations here

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

**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 = (a

_{2}- a_{1}) / tThis gives

a

_{2}= a_{1}+ jtThis 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?

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#13

(Original post by

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?

**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?

_{2}- a

_{1}) / t

jerk = rate of change of acceleration, and is constant.

As I have already said, this gives

a

_{2}= a

_{1}+ 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**

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#14

(Original post by

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 = (a

jerk = rate of change of acceleration, and is constant.

As I have already said, this gives

a

Integrate this wrt time and you get the equation for

Integrate again and you get

**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 = (a

_{2}- a_{1}) / tjerk = rate of change of acceleration, and is constant.

As I have already said, this gives

a

_{2}= a_{1}+ jtIntegrate 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**
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