Direct proportionality equations

  • Thread starter Thread starter member 731016
  • Start date Start date
  • Tags Tags
    equations Physics
AI Thread Summary
The equation x = vt + x_0 does not represent direct proportionality between x and v due to the constant x_0. Instead, direct proportionality exists between (x - x_0) and t, as doubling v results in doubling (x - x_0) for a fixed t. Similarly, (x - x_0) is directly proportional to v when t is constant. The relationship can be described as linear, but it is not proportional since it does not pass through the origin. Understanding this distinction is crucial for interpreting the relationship correctly.
member 731016
Homework Statement
Please see below
Relevant Equations
Please see below
For this,
1685668525439.png

Can a the equation ##x = vt + x_0## not be considered a direct proportionality between ##x## and ##v##? If so, is it because there is a constant ##x_0##?

Many thanks!
 
Physics news on Phys.org
  • Like
Likes malawi_glenn
ChiralSuperfields said:
Homework Statement: Please see below
Relevant Equations: Please see below

For this,
View attachment 327344
Can a the equation ##x = vt + x_0## not be considered a direct proportionality between ##x## and ##v##? If so, is it because there is a constant ##x_0##?

Many thanks!
The direct proportionality is between ##(x-x_0)## and ##t##. If you double ##v##, you double ##(x-x_0)## for the same ##t##.
You can also say that there is direct proportionality between ##(x-x_0)## and ##v##. If you double ##t##, you double ##(x-x_0)## for the same ##v##. Makes intuitive sense, no?
 
  • Like
  • Informative
Likes Steve4Physics, hmmm27 and member 731016
kuruman said:
The direct proportionality is between ##(x-x_0)## and ##t##. If you double ##v##, you double ##(x-x_0)## for the same ##t##.
You can also say that there is direct proportionality between ##(x-x_0)## and ##v##. If you double ##t##, you double ##(x-x_0)## for the same ##v##. Makes intuitive sense, no?
Thank you for your reply @kuruman! Yeah that is interesting, and yeah I think it makes sense :)

Many thanks!
 
ChiralSuperfields said:
Thank you for your reply @kuruman! Yeah that is interesting, and yeah I think it makes sense :)
@kuruman's test (Post #3) for proportionality is straightforward. Ask yourself: 'If I double one quantity, does the other quantity always get doubled?'. If the answer is 'yes' the quantities are proportional.

Of course, there's nothing special about doubling. It works for any factor. E.g. if ##y## is proportional to ##x##, then tripling ##x## also triples ##y##. This should be clear if you thnk about the equation ##y=kx##.

It’s also worth thinking graphically. If two quantities are directly proportional, a graph of one quantity against the other is a straight line through the origin.

If you get a straight line which doesn't pass through the origin, the quantities are not proportional;. In this case, the relationship is called 'linearity'. E.g. for the equation ##x=vt + x_0## there are various ways to describe the relationship between ##x## and ##t##: e.g. '##x## is linearly dependent on ##t##'; or 'there is a linear releationship between ##x## and ##t##'.

Edit: typo' corrected.
 
Last edited:
  • Like
  • Informative
Likes member 731016 and kuruman
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top