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In high school physics, I was taught three different equations related to accelerated motion:

  1. $v=at$
  2. $x=\frac{1}{2}at^2$
  3. $x=vt$

As one does in high school physics, I pretty much just mindlessly used these equations to get an okay-ish passing grade and didn't think about them much. However, I recently noticed that, when using equations 2 and 3 to form $vt=\frac{1}{2}at^2$ and solving for $t$, I got $v=\frac{1}{2}at$. Predictably, using this equation to calculate velocity yields incorrect results. I know acceleration is defined as velocity per time (without any $\frac{1}{2}$ in there), so I think I understand why this equation doesn't work, but I don't understand how deriving from two (as far as I know) correct equations can yield an incorrect one. What's up here (layman's terms would be appreciated)?

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  • $\begingroup$ If you like this question you may also enjoy reading this related Phys.SE post. $\endgroup$ – Qmechanic Aug 17 at 8:27
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Equations 1 and 2 are valid only for constant acceleration. Equation 3 is valid only for constant velocity. Constant acceleration (if nonzero) and constant velocity are incompatible conditions, so combining 2 and 3 produces nonsense.

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  • $\begingroup$ Actually, it is not entirely nonsense. The equation does have a solution for $v=0$ and $a=0$. This isn’t surprising since equation 2 assumes $v(0)=0$ and equation 3 assumes $a=0$. Satisfying both equation’s assumptions gives a valid solution. $\endgroup$ – Dale Aug 17 at 2:43
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First of all, understand that your equations are for special starting conditions e.g. the position, x, and the velocity, v, are 0 at t=0. Second, your third equation relates distance and time for a constant velocity, not for a velocity that is changing due to acceleration.

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the equation x=vt holds only for constant velocity v, or for the mean velocity in time t. if you accelerate v=at so if you start with v=0 and accelerate the time t, the mean velocity will be vm=(0+at)/2 and so x=vmt=a/2t^2 als the equation x=a/2*t^2 holds only if you start with v=0

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Those three formulae are only applyable for very particular cases. I don't know why you didn't even go slightly deep on them.

  1. These are valid for STRAIGHT movement only.

  2. For a Straight uniform movement, that is, constant speed, you have

FINAL POSITION = INITIAL POSITION + SPEED * TIME

$x_F=x_o + v\cdot t$

Here $v$ does not vary. This formula follows from the definition of velocity:

$$v=\frac{distance}{time} \qquad \Rightarrow \qquad distance=v\cdot time $$

(and distance is obviously $x_F-x_o$


  1. for Uniformly accelerated Straight Movement

that is, the object has a constant acceleration, what you have is two formulae. The firsto ne is pretty similar to the previous one:

$v_F=v_o+at$

Obviously, if theire is an acceleration, the speed varies in time. Check that your formula $v=at$ is only a particular case in which the initial speed is 0. But for any other initial speed $v_o$, you can use $v_F=v_o+at$

Again, this follows from the definition of acceleration:

$$ a=\frac{change\ in\ speed}{time}=\frac{v_F-v_o}{t}$$

Finally, the last formula is a bit harder to deduce, but just learn it. The distance for a uniformly accelerated movement is

$$x_F=x_o+v_o\cdot t + \frac{1}{2} at^2 $$


Notice that there are two different situations:

A) Constant velocity: $x_F=x_o+vt$. Only this formula. This formula is useful for ANY exercise.

B) Constant aceleration (velocity aries, but at a constant rate). Then you have two formulae that must be considered together.

$$\begin{cases} v_F=v_o+at \\ x_F=x_o+v_o t + \frac{at^2}{2} \end{cases}$$

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There are 4 standard kinematic equations, which are derived from the definition of velocity and the definition of acceleration. These equations are:

$\Delta x = 1/2(v_i + v_f) \Delta t \qquad(1)$

$\Delta x = v_i \Delta t + 1/2 a (\Delta t)^2 \quad (2)$

$v_f = v_i + a \Delta t \qquad (3)$

$v_f^2 = v_i^2 + 2 a \Delta x \quad (4)$

Note that I am using the $\Delta$ form of these equations because the initial value of the variables in these equations may or may not be zero, so this mathematical form is the most general.

Equation (1) comes from the mathematical definition of "average", and it applies when there is constant acceleration over time interval $\Delta t$. Equation (3) comes from the definition of acceleration, which states that $a = \Delta x / \Delta t$, where $\Delta$ is defined as final condition minus initial condition (e.g., $\Delta v = v_f - v_i$). Equation (2) is derived from the circumstance where you don't know $v_f$, and you substitute equation (3) into equation (1). Equation (4) comes from the circumstance where you don't know $\Delta t$, and you solve equation (3) for $\Delta t$ and substitute it into equation (1).

There are 4 variables in each of these equations, and in some cases, one or more of these variables will have a starting value of zero, which leads to the equations that you stated in your question. Trying to do a derivation with your stated equations is on "shaky ground" because those equations are not actually equivalent to the kinematic equations listed above.

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