Change in Displacement Formulation

[microdosemishief]

Homework Statement

Change in displacement

Relevant Equations

Assuming initial displacement is zero, if x = v₁(t) + ½a(t)², why isnt Δx = x2-x1 = (v₁(t2) + ½a(t2)²) - (v₁(t1) + ½a(t1)²) = v₁(t₂ - t₁) + ½a(t₂² - t₁²)

Assuming initial displacement is zero, if x = v₁(t) + ½a(t)², why isnt Δx = x2-x1 = (v₁(t2) + ½a(t2)²) - (v₁(t1) + ½a(t1)²) = v₁(t₂ - t₁) + ½a(t₂² - t₁²)

Why is it instead v₁(t₂ - t₁) + ½a(t₂ - t₁)²

 

[PeroK]

The two calculations have a different initial velocity. Your initial formula should be:
$$x = (v_0(t_2) + \frac 1 2 at_2^2) - (v_0(t_1) + \frac 1 2 at_1^2)$$And you also have$$v_1 = v_0+ at_1$$If you use those, you'll see that everything works out with $v_0$ in one case and $v_1$ in the other.

 

[kuruman]

Assuming initial displacement is zero, if x = v₁(t) + ½a(t)², why isnt Δx = x2-x1 = (v₁(t2) + ½a(t2)²) - (v₁(t1) + ½a(t1)²) = v₁(t₂ - t₁) + ½a(t₂² - t₁²)

Why is it instead v₁(t₂ - t₁) + ½a(t₂ - t₁)²

What exactly are you asking? In the second paragraph above Isn't "it" the displacement? Are the expressions outlined in red in the two paragraphs different? If so how?

As @PeroK pointed out, your use of v₁ is confusing. In your equation for the position it stands for the velocity at time t = 0. Since you are using subscripts 1 and 2 to denote specific times, v₁ could also stand for "the velocity at time t₁". Using ##v₀## for the initial velocity makes it clear what's what.

 

[Herman Trivilino]

Homework Statement: Change in displacement

Who wrote that homework statement? First of all, it doesn't ask for a response. Secondly, displacement equals change in position. Change in displacement is encountered usually when you're comparing the displacements of two different objects.

Relevant Equations: Assuming initial displacement is zero, if x = v₁(t) + ½a(t)²,

That's assuming initial position is zero. Also, the notation is rather strange. If there's a subscript "1" on $v$ why is the same subscript not also on $x$ and $t$?

why isnt Δx = x2-x1 = (v₁(t2) + ½a(t2)²) - (v₁(t1) + ½a(t1)²) = v₁(t₂ - t₁) + ½a(t₂² - t₁²)

To me, assuming $\Delta x$ is the displacement, or change in position as the object moves from position $x_1$ to position $x_2$, the correct equation would be $\Delta x=x_2-x_1=(v_2 t_2-\frac{1}{2} a t_2^2) -(v_1 t_1-\frac{1}{2} a t_1^2)$.

why isnt Δx = x2-x1 = (v₁(t2) + ½a(t2)²) - (v₁(t1) + ½a(t1)²) = v₁(t₂ - t₁) + ½a(t₂² - t₁²)

Why is it instead v₁(t₂ - t₁) + ½a(t₂ - t₁)²

Who says it is? $\Delta x = (v_2 t_2 - v_1 t_1) - \frac{1}{2} a (t_2^2-t_1^2)$.

It seems that besides the confusion between position and displacement, there is also some confusion about the positions of the object. My understanding is that there are three positions, $x_o=0$, $x_1$, and $x_2$.

 

[kuruman]

To me, assuming $\Delta x$ is the displacement, or change in position as the object moves from position $x_1$ to position $x_2$, the correct equation would be $\Delta x=x_2-x_1=(v_2 t_2-\frac{1}{2} a t_2^2) -(v_1 t_1-\frac{1}{2} a t_1^2)$.

Sure, but if the velocities and the times are known, it would be more sensible (and easier to remember) to write this equation in the equivalent form $$\Delta x= \frac{v_2+v_1}{2}\left(t_2-t_1\right)=v_{\text{avg.}}\Delta t.$$

 

[Gavran]

Both expressions for Δx

v₁(t₂ - t₁) + ½a(t₂² - t₁²)

and

v₁(t₂ - t₁) + ½a(t₂ - t₁)²

are only valid for the case where t₁=0.

 

[kuruman]

Just to tidy this up. One way to do this formally is to write the position $x(t)$ at any time $t$ then write expressions for the position at specific times 1 and 2 and subtract. It is implicitly understood that time $t=0$ is when motions starts, i.e. the clock starts when the object starts moving. Also, numerical subscripts denote values of quantities at specific times, e.g. $v_0$ is the velocity when the motion (and the clock) starts, $v_1$ is the velocity when the clock reads $t_1$, etc. Then,
$x(t)=x_0+v_0t+\frac{1}{2}at^2$
$x_1=x_0+v_0t_1+\frac{1}{2}at_1^2$
$x_2=x_0+v_0t_2+\frac{1}{2}at_2^2$
Thus, the displacement from clock time $t_1$ to clock time $t_2$ is $$\Delta x=x_2-x_1=v_0(t_2-t_1)+\frac{1}{2}a\left(t_2^2-t_1^2\right).$$ A second way to do this formally is to start a second clock when the object is at position $x_1$ and has velocity $v_1$. This clock reads time $\tau.$ The position is, of course, relative to the same origin and is written as
$x(\tau)=x_1+v_1\tau+\frac{1}{2}a\tau^2.$ Clearly, the second clock runs behind the first and the times displayed by the clocks are related by $\tau=t-t_1$. The object reaches $x_2$ at time $\tau_2=t_2-t_1$ so that
$$\begin{align} & x(\tau_2)=x_2=x_1+v_1\tau_2+\frac{1}{2}a\tau_2^2 \nonumber \\
&\implies x_2-x_1=\Delta x=v_1(t_2-t_1)+\frac{1}{2}a\left(t_2-t_1\right)^2.
\end{align}$$ The expressions for the displacement derived by the two methods are the same. This can be verified by substituting $v_1=v_0+at_1$ in equation (1).