Deriving the formula for distance

In summary, the equation for velocity is v=dx/dt, and the equation for acceleration is a=dv/dt. The equation for displacement is d=1/2 at^2. If you know the equation for velocity, you can find the equation for acceleration.
  • #1
DarthRoni
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I am currently redoing grade 11 physics and they have introduced the following formula,
[tex]\Delta d = v_1\Delta t + \frac{1}{2}a\Delta t^2[/tex]
I am trying to find this equation using calculus.
So from what I understand, [itex]\Delta d[/itex] is the area under the curve for [itex]v(\Delta t)[/itex].
We can define [itex]v(\Delta t) = v_1 + a\Delta t[/itex] where [itex]\Delta t = t_2 - t_1[/itex]
I can take the integral of [itex]v(\Delta t)[/itex] to find [itex]\Delta d[/itex] call it [itex]D(\Delta t)[/itex]. Where, [itex]D'(\Delta t) = v(\Delta t)[/itex].
[tex]D(\Delta t) = \int v(\Delta t)\ d\Delta t \implies D(\Delta t) = \int (v_1 + a\Delta t)d\Delta t\implies D(\Delta t) = v_1\Delta t + \frac{1}{2}a\Delta t^2[/tex]

and suppose I know [itex]v(t)[/itex] could I say:
[tex]\Delta d = \int_{t_1}^{t_2}v(t)dt = P(t_2) - P(t_1)\ where\ P'(t) = v(t)[/tex]
[tex]\Delta d = D(\Delta t) \implies D(\Delta t) = P(t_2) - P(t_1)[/tex]

Is this all correct ? Any feedback would be appreciated.
 
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  • #2
We define velocity as v=dx/dt, and acceleration as a=dv/dt.

Start with the constant acceleration, a. One integration gives the change in velocity as v=at; the second gives the change in distance as d=1/2 at^2.

Then add in the expressions for an initial velocity, and an initial displacement - this gives the general result for constant acceleration.
 
  • #3
So are you saying since [itex] a = \frac{\Delta v}{\Delta t}\ and\ v = \frac{\Delta d}{\Delta t}[/itex]
[tex]\Delta v = \int (a)dt = at[/tex]
[tex]\int (\Delta v) dt = \int (at)dt = \frac{1}{2}at^2[/tex]
and since [itex]v_1[/itex] is constant
[tex]\int (v_1)dt = v_1t[/tex]
and since [itex]v(t) = v_1[/itex] when [itex]t = 0[/itex], we have
[tex]\int v(t)dt = \int (v_1)dt + \int (\Delta v)dt[/tex]
[tex]\implies \int v(t)dt = v_1t + \frac{1}{2}at^2[/tex]

Is this what you mean? Where does initial displacement come into play? How would I proceed to use definite integrals ?
[tex]\int_{t_1}^{t_2}v(t)dt = (v_1t_2 + \frac{1}{2}at_2^2) -(v_1t_1 + \frac{1}{2}at_1^2)[/tex]
does this work ?
 
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  • #4
well I suppose all we did was define v(t) and took the integral. I'm still not sure what you mean by adding initial displacement.
 
  • #6
Here's another way to do it besides the one in the link UltrafastPED gave. Integrate the acceleration as an indefinite integral, so you add a constant of integration:
$$v=\int {a dt}\\
v = at + C$$

To find the constant of integration, you have to specify an "initial condition", namely that ##v = v_0## at t = 0:
$$v_0 = a \cdot 0 + C\\
C = v_0$$

Therefore ##v = at + v_0##

You can probably do the second integration to get x, with the initial condition ##x = x_0## at t = 0.
 
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  • #7
Thanks guys !
 

1. How is the formula for distance derived?

The formula for distance, d = rt (distance = rate x time), is derived from the definition of distance as the amount of space between two points. It is also derived from the formula for speed, s = d/t (speed = distance/time), and rearranging it to solve for distance.

2. What is the significance of the formula for distance?

The formula for distance is significant because it allows us to calculate the distance traveled by an object given its speed and time. This is useful in many real-world applications, such as calculating travel time or determining the distance between two locations.

3. How does the formula for distance relate to other physics concepts?

The formula for distance is closely related to the concepts of speed, time, and displacement. It also has a direct relationship with the formula for velocity, which includes the direction of movement in addition to speed.

4. Can the formula for distance be used for moving objects with varying speeds?

Yes, the formula for distance can be used for moving objects with varying speeds. However, in this case, the average speed would need to be used in the calculation, rather than the instantaneous speed at any given moment.

5. Are there any limitations to the formula for distance?

The formula for distance has limitations when applied to objects with changing acceleration or when considering the effects of external forces such as air resistance. It also assumes that the object is moving in a straight line, and does not account for changes in direction. In these cases, more complex formulas may be needed to accurately calculate distance traveled.

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