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## Equations of Motion

Available from trevorpythag.wordpress.com

The 4 equations of motion deal with an object which is travelling with constant acceleration (which can be 0 and therefore constant speed).

The equations are as follows:

(1) v = u + at
(2) s= t(v+u)/2
(3) v2=u2+2as
(4) s = ut + at2/2

Where
a = acceleration – this must be constant for equations to hold
u = initial velocity, ie at the start of the journey
v = final velocity, ie at the end of the journey
s = displacement which is a vector quantity for the distance of the object from its starting point
t = time taken for journey
Any of these values can be found using the equations if at least 3 of the other values are known.

Equation (1) comes from the definition of acceleration as acceleration is the rate of change of velocity and therefore for constant acceleration
a = (v-u)/t which you can rearrange to make
v = u + at

Equation 2 can be found by either considering a distance time graph or using the average speed. (v+u)/2 gives the average speed during the journey as a is constant and buy multiplying this by t we find the displacement. Or from the graph we can find it as the area under the graph is a trapezium of height t and sides u and v so using the area of a trapezium formulae we find 2.

Equation (4) can be formed by substituting 1 into 2 so
s = t(u+at+u)/2
s = t(2u + at)/2
s= 2ut/2 + at2/2
s = ut + at2/2

Equations 3 can also be found using 1 and 2 by rearranging 2 to get an expression for t we find
t=2s/(u+v)

If we substitute this into v = u+at we get
v=u+2as/(v+u)
And by bringing up the v+u we find
v2+vu = u2+vu + 2as
And because we have vu on both sides we can cancel these to find 3

v2=u2+2as

Categories: maths
1. February 6, 2009 at 2:21 pm

@Dave,

Nice explanation,

But if the acceleration of particle (a) dependent to it’s position (s), how to get both of it’s velocity and position? for example when a=-100s, where at t=0, v(0)=v0, and s(0)=s0.

Sorry @Dave, I read this problem from basic physics’s book my father Mr.Rohedi.

2. February 6, 2009 at 4:17 pm

Acceleration is the rate of change of velocity which in turn is the rate of change of displacement (s) both with respect to time. So if we integrate a we find a function v and if we integrate this function for v we find one for s. The constant added in the integrations is equal to v0 and s0.

As acceleration is dependant on displacement we then get a differential equation:
s”(t) = – 100s(t)
however with my limited knowledge of maths I’m not sure exactly how to solve it to find the required answer (though I think were learning that in school soon) but please reply if you do

dave

3. February 6, 2009 at 7:04 pm

But, I am interested to this form s”(t) = – 100s(t), this is a so-called a linear second order differential equation, isn’t it. Of course this is out of my memory, hehe..but I believe my father expert to solve it.

Thanks you dave.

Denaya Lesa.

• February 11, 2009 at 2:38 pm

I think i now have the correct solution to your problem though check it because i may well be wrong
s(t) = socos10t + vosin10t /10
v(t) = -10sosin10t + vocos10t
It would seem that your object is performing simple harmonic motion as the acceleration is proportional to the displacement but in the opposite direction which makes sense with the solutions i have found which can be written with just sin or just cos
anyway hope that im right and that it helps

dave

4. February 12, 2009 at 11:23 am

Hi Dave,

Of course your above answers of s(t) and v(t) respectively are correct mathematically. Now I would ask you, may Denaya Lesa posts the answers Physically?
Hehehe…this is according my father daddy Rohedi?

If you okay, Denaya as soon as posts the both answers.

5. February 13, 2009 at 12:45 am

%Dave,

I think in posting the physically answer of s”(t)+100s(t)=0 it doesn’t need to permission you, coz here Denaya only purposes to complete your above answer. Okay Dave..

For visitors here.. we must remember again that every mass moved harmonically such as when oscillating or vibrating about it’s equilibrium position, it will reach maximum displacement called amplitude. According to my physics theacher, the position of mass any time satisfies one of sinusoid function s(t)=A*cos(ωt+φ0) or s(t)=A*sin(ωt+θ0), where A is the amplitude, and ω is the anguler frequency, while φ0 and θ0 are the initial phase for cosine and sine functions respectively.

Again, hehehe….., according to mr.Rohedi,
the amplitude can be determined from both initial of its position (s0) and velocity (v0) at t=0 by using the trigonometric and pythagoras relations in yeilding the form

A=√[s0^2 + (v0/ω)^2].

Next,if the harmonic equation would be represented in the cosine function, then the initial phase is of form φ0=-atan([v0/ω]/s0), where atan is arctangent function. Hence for the general form :

s”(t)+(ω^2)s(t)=0,

then the position of mass from it’s equilibrium position can be presented as following

s(t)= √(s0^2+[v0/ω]^2)*cos(ωt-atan(v0/ωs0) )

while its velocity was derived by differentiating s(t) respect to t or v(t)=ds/dt to yeilding

v(t)=-ω√(s0^2+[v0/ω]^2)*sin(ωt-atan(v0/ωs0) )

@Dave,

For Denaya’s case of s’’(t)+100s(t)=0, you’ve found previously that ω=10, so Denaya can write the physically answer as following

s(t)=√[s0^2+(v0/10)^2]*cos(10t-atan(v0/10s0))

and

v(t)=-10√[s0^2+(v0/10)^2]*sin(10t-atan(v0/10s0))

Of course the position and velocity of mass still depends on the values of s0 and v0.

Next, turn the visitors to prove Denaya’s analytic solution of s(t) and v(t) from Dave’s solution in detail derivation.

Oh yeah Denaya forgets to ask visitors here, how to get solution of s”(t)=-(ω^2)s(t) by integration, so it still appropriates with your advise in deriving the equation of motion.

See you later Dave..

Denaya Lesa.

• February 24, 2009 at 7:50 pm

We can prvoe your solutions for s(t) and v(t) from my soltuion if we let s(t) = rsin(10t + c) where we can work out c and r by using the identity for compound sin (see my post on this) so
s(t) = rsin10tcosc + rcos10tsinc
and by comparing co-efficents
rcosc = v0/10 and rsinc = s0
so c = atan(v0/10s0) as you showed.
we can then find r by squaring the 2 equations to get

This can also be done in a similar way for v(t)
r^2 (cosc^2 + sinc^2) = (v0/10)^2+s0^2

so we know can write s(t) as
s(t)=√[s0^2+(v0/10)^2]*cos(10t-atan(v0/10s0))
as you showed using the SHM equations.

6. February 14, 2009 at 11:30 pm

Hi @dave,

See you later Dave..

Denaya Lesa.

7. February 15, 2009 at 12:20 pm

Hello,

I think that you explanation is excellent :):) you obviously put a lot of effort into it.

Well done

8. February 16, 2009 at 5:32 am

Thank @Nicola for your nice comment. Important to be known that all of post’s Rohedi’s Family at so many math blogs would be dedicated for our nation Republic of Indonesia. Thank’s for @trevorpythag at al who accept Denaya’s comments at this math blog as well.

Again, thank you @Nicola,
my best regards,

Denaya Lesa.

9. February 25, 2009 at 1:53 pm

@trevorpythag,

Thanks for your help in proving my solution posted previously. From your proof above, Denaya can take a conclusion that the pythagoras relation and all of your trigonometric relation posted previously are very required in solving physics’s problems, isn’t it.
Good Luck..@trevorpythag.

See you later,
Denaya Lesa.

10. March 1, 2009 at 12:30 am

Sorry dave, whether you miss in writing
c=atan(v0.10s0). I see on Denaya’s solution the initial phase in form c=atan(v0/10s0), or which one the right answer?thx.

11. March 1, 2009 at 12:17 pm

sorry that was a typo,
c=atan(v0/10s0)

well done in spotting it 🙂

12. March 4, 2009 at 4:08 am

Dave, I always follow your discussion with miss Denaya. I think this is very educational discussion. Really, until now I try to solve challenge from miss Denaya to solve s”(t)+100s(t)=0 by integration, but sorry I’ve no idea to do it. Can you or all visitors here help me…

• March 8, 2009 at 5:01 pm

we can let m2= -100
so that m = 10i
then
s=e0(Ecos10t + iFsin10t)
so s = (Ecos10t + iFsin10t)
and subsubsiting in the conditions given we can find the values of E and F

13. March 6, 2009 at 6:05 pm

Me too Brandon, I need it for preparing mechanics exam

14. March 18, 2009 at 12:57 am

Thanks Dave, I think you’ve solved by D=d/dt operator, isn’t it?