How do we find the uncertainty of v in the kinematics eqn?

[minamikaze]

Homework Statement

Given the kinematics equation v² = u² + 2as, find the uncertainty Δv.
Given: Values of a, s, u, and their associated uncertainties.

Homework Equations

v² = u² + 2as

The Attempt at a Solution

I'm aware of the rules for uncertainty propagation, but what do I do in this case where there are powers, sums and products all in one equation?

 

[Delta2]

Just combine the rules of uncertainty propagation. What do you know about the rules for the uncertainty of sum and the uncertainty of product?

 

[minamikaze]

I know that if a = b + c, then Δa = Δb + Δc.
And if a = bc², then Δa/a = Δb/b + 2Δc/c. But I don't understand how the two can come together in the case of v² = u² + 2as.

 

[haruspex]

I know that if a = b + c, then Δa = Δb + Δc.
And if a = bc², then Δa/a = Δb/b + 2Δc/c. But I don't understand how the two can come together in the case of v² = u² + 2as.

Just apply your two rules sequentially. v² = u² + 2as has the overall form of a = b + c, where a is v² etc.

 

[minamikaze]

So, (Δv)² = (Δu)² + 2ΔaΔs ?

 

[haruspex]

So, (Δv)² = (Δu)² + 2ΔaΔs ?

No, I wrote "where a is v² etc." If a represents v² what is Δa?

 

[minamikaze]

Δv² ? What is Δv² and how do you calculate it?

 

[haruspex]

Δv² ? What is Δv² and how do you calculate it?

Use your bc² rule.

 

[minamikaze]

I'm sorry, I already said I know the rules and have listed them down. This isn't really leading me anywhere.
Let me phrase my question very, very clearly.

I know that for a = bc², Δa/a = Δb/b + 2Δc/c.
What about a = b+c², where there is both a product and a sum, and a square? I just do not see how this is coming together - the sum propagation and the product propagation rule.
It cannot be the same, Δa/a = Δb/b + 2Δc/c, isn't it?

So you advised applying the rules sequentially.
Step 1 : Apply the addition rule.
Δa = Δb + Δc²
Step 2: Apply the product rule.
Δa/a = Δb/b + 2Δc²/c

Is this it?

And so for my original question, should it be:
2Δv/v = 2Δu/u + 2ΔaΔs ?
It does not seem right, and I would like to get clear guidance on this, instead of just asking me to apply this rule and that rule, which I already know, and have stated clearly above.

 

[Delta2]

Yes we have to apply the sum and product rules however it is needed, but it seems you doing some mistakes applying the product rule.

By applying the sum rule we get

$\Delta v^2=\Delta u^2+\Delta(2as)$

Now we going to apply the $bc^2$ rule for the first 2 terms of the sum (two different applications of the rule).
So we have applying the $a=bc^2$ rule for $a=v^2 ,b=1, c=v$
$\frac{\Delta v^2}{v^2}=\frac{\Delta 1}{1}+2\frac{\Delta v}{v}\Rightarrow \Delta v^2=v^2(0+2\frac{\Delta v}{v})\Rightarrow \Delta v^2=2v\Delta v$

Similarly applying the rule for $a=u^2,b=1,c=u$ we get $\Delta u^2=2u\Delta u$

Now what rule should we apply to calculate $\Delta (2as)$. We don't have a squared term here to apply the bc^2 rule and what you did is that
$\Delta (2as)=2\Delta a \Delta s$ but this is not correct is it?

 

[minamikaze]

z = 2as
Δz/z = Δa/a + Δs/s
∆z = z(Δa/a + Δs/s)

Is this it?

 

[minamikaze]

So overall, it would be:

2vΔv = 2uΔu + 2as(Δa/a + Δs/s)

Is this correct?

 

[haruspex]

So overall, it would be:

2vΔv = 2uΔu + 2as(Δa/a + Δs/s)

Is this correct?

Yes, but you can simplify that with some cancellation.

 

[minamikaze]

Thank you, I am very clear about it now.
The end result will be:
2vΔv=2uΔu + 2sΔa + 2aΔs

Thanks all for the help.