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

AI Thread Summary
The formula s = vt - (1/2)at² is not included in exams, even though it can simplify problems, likely due to its applicability in specific situations. The discussion highlights the importance of consistency in using symbols for initial and final velocities, with variations like u and v or vi and vf. When acceleration is negative, the equation adjusts but remains distinct from s = ut + (1/2)at², as it requires different values for substitution. The equations of motion are primarily valid under constant acceleration conditions. Overall, the omission may stem from the need for clarity and consistency in teaching foundational concepts.
influx
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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
 
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influx said:
In both my Maths and Physics exams they give these equations of motion:

v = u + at
v2 = u2 + 2as
s = ut + (1/2)at2 \color{red}\Leftarrow
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? :wink:
 
from v=u+at, put u=v-at in third.what will you get
 
gneill said:
What happens if the acceleration a is negative in your third equation? :wink:

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

vi, vf
v0,v1
u, v

Once you adopt u=initial and v=final, be consistent.
 
influx said:
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?
 
gneill said:
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)..
 
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.
 
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.
 
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