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

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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?

Please explain how you can derive this equation?

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#2

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.

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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#3

or you could say that the distance travelled is

average speed X time

the average speed is ( u + v )/2

average speed X time

the average speed is ( u + v )/2

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#5

The standard approach is to start with and from a velocity-time graph and to then derive the others by substitution.

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(Original post by

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.

**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.

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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#7

(Original post by

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?

**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?

a = u, b = v, h = t, A = s

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But why is it similar to the shape of a trapezium?

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#9

(Original post by

Oh right - I see!

But why is it similar to the shape of a trapezium?

**Kindred.Spirit**)Oh right - I see!

But why is it similar to the shape of a trapezium?

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#10

(Original post by

Isn't that circular reasoning?

The standard approach is to start with and from a velocity-time graph and to then derive the others by substitution.

**Mr M**)Isn't that circular reasoning?

The standard approach is to start with and from a velocity-time graph and to then derive the others by substitution.

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#11

(Original post by

Not at all. s=ut+0.5at^2 follows immediately from v=u+at since you just integrate once with respect to t.

**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.

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#13

**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.

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#15

v²-u²=2as

(v-u)(v+u)=2as

at(v+u)=2as

{v-u=at}

t(v+u)=2s

s=(v+u)t/2

(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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#16

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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