Competition - post a photo you've taken of nature >>

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

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?

Reply 1

Mr M

Study Forum Helper

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.

Reply 2

the bear

or you could say that the distance travelled is

average speed X time

the average speed is ( u + v )/2

s=ut+0.5at^2
at=v-u

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.

Reply 5

Kindred.Spirit

Original post by Mr M

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?

Reply 6

Mr M

Study Forum Helper

Original post by Kindred.Spirit

Rather complicated.

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

Reply 7

Kindred.Spirit

Original post by Mr M

Rather complicated.

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

Oh right - I see! :biggrin:

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

Reply 8

Mr M

Study Forum Helper

Original post by Kindred.Spirit

Oh right - I see! :biggrin:

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.

Reply 9

electriic_ink

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.

Reply 10

Mr M

Study Forum Helper

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.

:smile:

Reply 11

Milouw

How do you derive it using calculus? Please I need help asap

Reply 12

Milouw

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?

Reply 13

Sinnoh

It's just taking the average velocity and multiplying it by time.

Reply 14

Somil Kaushik

v²-u²=2as
(v-u)(v+u)=2as
at(v+u)=2as
{v-u=at}
t(v+u)=2s
s=(v+u)t/2

(edited 4 years ago)

Reply 15

Deaths ray

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)