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My teacher wrote the following:

Constant Acceleration

If acceleration is constant, then:

$$\vec{v}(t) = \int_0^t \vec{a}(t')dt'\ + \vec{v_0}$$

and

$$\vec{x}(t) = \int_0^t \vec{v}(t')dt'\ + \vec{x_0}$$

Why does acceleration need to be constant? I can't see why integration would need a constant acceleration as such.

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    $\begingroup$ If does not imply constant is a requirement $\endgroup$ – paparazzo Aug 13 '16 at 10:13
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    $\begingroup$ Your teacher wanted to make life easier for you by explaining the easy case and omitting the hard one(s). $\endgroup$ – Guntram Blohm supports Monica Aug 13 '16 at 10:52
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    $\begingroup$ It looks like they may have cut pieces out of a previous document they wrote. What you have are the generic equations. However, they may have pulled them from a document that was talking about the special case where acceleration is constant. In that case, they would have several more lines showing how this equation simplifies in that special case. They may have simply removed the later equations, but forgotten to change the text to reflect that they were now showing the general case, not the specific one. $\endgroup$ – Cort Ammon - Reinstate Monica Aug 13 '16 at 15:29
  • $\begingroup$ We are considering this equation only when the acceleration remain constant. I mean to say that equations of kinematics only applicable when a=c $\endgroup$ – Prabhat Aug 14 '16 at 1:37
  • $\begingroup$ @dr.honey This is not correct. Some equations of kinematics only apply at constant acceleration. But not these ones in the question above - they are the pure definitions and work in any case. $\endgroup$ – Steeven Aug 16 '16 at 12:49
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Acceleration does not need to be constant. By definition, $a=dv/dt$. You can still solve for $v(t)$ by integrating $\int a(t) dt$.

If acceleration is constant, you will arrive at the common situation of $v(t)=v_0 +at$. If acceleration is not constant, you will have some other (more interesting) result for $v(t)$ since you are now integrating over a function that includes $t$.

For example, if $a(t)=\frac{a_0}{t^2}$, then $v(t)=-\frac{a_0}{t}+v_0$.

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Why does acceleration need to be constant?

The equations you give don't require constant acceleration, they are true regardless:

\begin{align} v(t) &= \int^t_0 a(t')dt' + v_0 \\ &= \int^t_0 \frac{dv(t')}{dt'}dt' + v_0 \\ &= \int^t_0 dv(t') + v_0 \\ &= v(t) - v(0) + v_0 \\ &= v(t) \end{align}

where I used the definition of acceleration, $a(t) = \frac{dv(t)}{dt}$, on line 2. And similarly for position

$$ x(t) = \int^t_0 v(t')dt' + x_0 = \int^t_0 \frac{dx(t')}{dt'}dt' + x_0 = \int^t_0 dx(t') + x_0 = x(t) - x(0) + x_0 = x(t) $$

using the definition of velocity, $v(t) = \frac{dx(t)}{dt}$, in the $2^{nd}$ step. Crucially, constant acceleration wasn't assumed: $a(t)$ could be anything differential-able.

I can't see why integration would need a constant acceleration as such.

In general it doesn't, but if you do assume it then the equations simplify massively :)

If acceleration is constant, then $a(t) \rightarrow a$ as it does not depend on time. This lets you pull it out of the integral, which makes the integral solvable. Starting from the definition of acceleration in integral form

\begin{align} v(t) &= \int a(t)dt \\ &\rightarrow a\int dt \quad\text{ (!!)} \\ &= at+c \end{align}

where $c$ is a constant and $(!!)$ means that I used the fact that acceleration is constant. If you consider $t=0$: $v(t=0) = c$ then it becomes obvious that $c$ is the initial velocity, $v(t=0)$, while I will rename as $v_0$.

\begin{equation} v(t) = at+v_0 \tag{1} \end{equation}

You can then repeat this process with position, $x$, given the definition of velocity

\begin{align} x(t) &= \int v(t)dt \\ &= \int (at+v_0)dt \quad\text{ (!!)} \\\\ &= \int at dt + \int v_0 dt \\ &\rightarrow a\int t dt + v_0\int dt \quad\text{ (!!)} \\\\ &= \frac{1}{2}at^2 + v_0t + c \end{align}

Considering $t=0$ again: $x(t=0) = c$, so we have

\begin{equation} x(t) = \frac{1}{2}at^2 + v_0t + x_0 \tag{2} \end{equation}

Equations 1 and 2 are used extensively throughout classical Physics, because we often consider simple cases with constant forces (e.g. gravity and electrostatics), which produce constant accelerations. There are also some more equations for constant acceleration, more on them here.


If acceleration isn't constant, then you have some function of $t$. For example, \begin{align} a(t) &= xt^2 + yt + z \\ v(t) &= \int a(t) dt \\ &= \int (xt^2 + yt + z) dt \\ &= \int xt^2 dt + \int yt dt + \int z dt \\ &= x\int t^2 dt + y\int t dt + z\int dt \\ &= \frac{1}{3}xt^3 + \frac{1}{2}yt^2 + zt + c \end{align}

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