What is the Equation for Calculating Uncertainty in Velocity and Acceleration?

[Raza]

Homework Statement

Hi. I need to prove that these 3 eqns are the same.

p \rightarrow q \vee r

p \wedge \neg q \rightarrow r

p \wedge \neg r \rightarrow q

Homework Equations

p \rightarrow q \equiv \neg p \wedge q

The Attempt at a Solution

p \rightarrow q \vee r

p \rightarrow \neg q \rightarrow r

\neg p \wedge \neg q \rightarrow r

That's all I got. Please help me.

 

[HallsofIvy]

I'm not sure what you are doing. I don't see how you got p \rightarrow \neg q \wedge r from any of those!

Using your "relevant equation", the first becomes \not p\wedge (q\vee r), the second \not (p\wedge \not q)\vee r which is itself equivalent ot (\not p \vee q)\vee r.<br /> <br /> Frankly I would use truth tables!

 

[Raza]

I had made a mistake, now I have corrected it. and I am not allowed to use the truth table.

Thank you

 

[Raza]

just ignore this, I need latex for MS word.
\sqrt{2 \times (\frac{0.1}{3.2})^2 + (\frac{0.2}{3.0})^2}

=0.079984804cm^3

=0.08cm^3

 

[Raza]

Again, please ignore this
just ignore this, I need latex for MS word.
\frac{\Delta v}{v} = \sqrt{(\frac{\Delta \ell}{\ell})^2 + (\frac{\Delta t}{t})^2}

\frac{\Delta v}{v} = \sqrt{(\frac{0.001}{0.101})^2 + (\frac{0.00001}{0.3950})^2}

\frac{\Delta v}{v} = 0.0099m/s

\frac{\Delta a}{a} = \sqrt{(\frac{\Delta v_1}{v_1})^2 + (\frac{\Delta v_2}{v_2})^2} + (\frac{\Delta d}{d})^2}

\frac{\Delta a}{a} = \sqrt{(\frac{0.0099}{0.256})^2 + (\frac{0.0099}{0.620})^2 + (\frac{0.01}{0.60})^2}

\frac{\Delta a}{a} = \sqrt{(0.00149) + (0.000254) + (0.000277)}

\frac{\Delta a}{a} = \sqrt{0.002021}

\frac{\Delta a}{a} = 0.041945

\frac{\Delta a}{a} = 0.042 m/s^2