Why is the s = vt - (1/2)at2 formula not included in exams?

[influx]

In both my Maths and Physics exams they give these equations of motion:

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


However the following equation is never included (even though it can simplify some problems significantly)

s = vt - (1/2)at²

Why is this deliberately omitted by the exam boards? Is it because this formula can only be applied in certain situations ?

Thanks

 

[gneill]

In both my Maths and Physics exams they give these equations of motion:

v = u + at
v² = u² + 2as
s = ut + (1/2)at² [itex]\color{red}\Leftarrow[/itex]
s = (u+v/2)t


However the following equation is never included (even though it can simplify some problems significantly)

s = vt - (1/2)at²

Why is this deliberately omitted by the exam boards? Is it because this formula can only be applied in certain situations ?

Thanks

What happens if the acceleration a is negative in your third equation?

 

[andrien]

from v=u+at, put u=v-at in third.what will you get

 

[influx]

What happens if the acceleration a is negative in your third equation?

Oh then it becomes s = vt + 1/2(at²)? but this is still different to s = ut+1/2(at²) right? Because in the latter equation you would substitute a value for u, but if you do not have u given, then you would use the other one where you substitute a value for v?

 

[azizlwl]

It just abbreviation. Different books different abbreviation for initial and final velocity

v_i, v_f
v₀,v₁
u, v

Once you adopt u=initial and v=final, be consistent.

 

[gneill]

Oh then it becomes s = vt + 1/2(at²)? but this is still different to s = ut+1/2(at²) right? Because in the latter equation you would substitute a value for u, but if you do not have u given, then you would use the other one where you substitute a value for v?

No, the acceleration is not squared. Only the time is squared. What happens to the sign of the whole term?

 

[influx]

No, the acceleration is not squared. Only the time is squared. What happens to the sign of the whole term?

Oh sorry, I meant to say :

If a is negative, the third equation becomes s = ut - (1/2)at²

but s = ut - (1/2)at² is not the same as s = vt - (1/2)at² is it? Because the latter requires you to substitute a value for 'v' (final velocity), whilst the first one require you to substitute a value for 'u' (initial velocity)..

 

[Genericcoder]

The negative is included when acceleration is negative like gravity which will adjust itself to get positive time since if ur going from high elevation to low elevation you will get positive time when you solve for it. Remember those equations are only used when acceleration is constant not changing.

 

[meldraft]

Your velocity multiplied by "t" is always the initial velocity no matter which letter is used (u or v). Different books (and often different chapters in the same book) use different symbols.