How to derive this equation of motion: s = 1/2 (v+u)t?

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Kindred.Spirit
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Need to learn this for my exam, but my own notes on this equation aren't clear.

Please explain how you can derive this equation?
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Mr M
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Do you know the formula for the area of a trapezium?

If so, sketch a velocity time graph with initial velocity u and final velocity v and a straight line joining these values (remember the acceleration is constant so the gradient is constant too).

All done.
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the bear
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or you could say that the distance travelled is

average speed X time

the average speed is ( u + v )/2
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electriic_ink
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s=ut+0.5at^2
at=v-u
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Mr M
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(Original post by electriic_ink)
s=ut+0.5at^2
at=v-u
Isn't that circular reasoning?

The standard approach is to start with v=u+at and s=\frac{1}{2}(u+v)t from a velocity-time graph and to then derive the others by substitution.
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Kindred.Spirit
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(Original post by Mr M)
Do you know the formula for the area of a trapezium?

If so, sketch a velocity time graph with initial velocity u and final velocity v and a straight line joining these values (remember the acceleration is constant so the gradient is constant too).

All done.
Yep - The area of a trapezium is 1/2 × h × (a + b)

Ah, so Displacement would be the area under the graph = ut + 1/2 (v-u)t

Then if if you cancel the brackets out you get = ut +1/2vt - 1/2ut

Simplify to:

s=1/2ut +1/2vt

so then, s = 1/2 (v+u)t

Is this correct?
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Mr M
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(Original post by Kindred.Spirit)
Yep - The area of a trapezium is 1/2 × h × (a + b)

Ah, so Displacement would be the area under the graph = ut + 1/2 (v-u)t

Then if if you cancel the brackets out you get = ut +1/2vt - 1/2ut

Simplify to:

s=1/2ut +1/2vt

so then, s = 1/2 (v+u)t

Is this correct?
Rather complicated.

a = u, b = v, h = t, A = s
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Kindred.Spirit
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(Original post by Mr M)
Rather complicated.

a = u, b = v, h = t, A = s
Oh right - I see!

But why is it similar to the shape of a trapezium?
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Mr M
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(Original post by Kindred.Spirit)
Oh right - I see!

But why is it similar to the shape of a trapezium?
Private message me with an email address and I will send you a Powerpoint about this.
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electriic_ink
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(Original post by Mr M)
Isn't that circular reasoning?

The standard approach is to start with v=u+at and s=\frac{1}{2}(u+v)t from a velocity-time graph and to then derive the others by substitution.
Not at all. s=ut+0.5at^2 follows immediately from v=u+at since you just integrate once with respect to t.
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Mr M
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(Original post by electriic_ink)
Not at all. s=ut+0.5at^2 follows immediately from v=u+at since you just integrate once with respect to t.
Many students derive the SUVAT equations before they first encounter calculus but fair enough.

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Milouw
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How do you derive it using calculus? Please I need help asap
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Milouw
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(Original post by electriic_ink)
Not at all. s=ut+0.5at^2 follows immediately from v=u+at since you just integrate once with respect to t.
How do you derive it using calculus?
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Sinnoh
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It's just taking the average velocity and multiplying it by time.
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Somil Kaushik
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v²-u²=2as
(v-u)(v+u)=2as
at(v+u)=2as
{v-u=at}
t(v+u)=2s
s=(v+u)t/2
Last edited by Somil Kaushik; 11 months ago
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Deaths ray
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Let us take the third equation of motion.....i.e. v²-u²=2aS . Here, v = final velocity, u = initial velocity, a= acceleration and S= distance coveredNow,v²-u²=2aS=> (v u)(v-u) = 2aS=> (v u) = 2aS/ (v-u)=> (v u) = 2aS/at ...........1 [ Since By formula of acceleration, a=(v-u)/t => at = (v-u)]Again, putting this in 1,=> (v u) = 2S/t .... (Since 'a' gets cancelled)=> (v u)t = 2S=> S = 1/2t (v u)
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