Why do kinematic equations only work with constant acceleration?

Question

People say that the equations are derived assuming a constant acceleration. I just don't see how this is the case. (I am new to calculus.)

Answer 1

Kinematic equations, one of which is $v=u+at$, are derived making the assumption that the acceleration $a$ is a constant.

Why this is so can be seen by inspecting a graph of velocity against time for constant acceleration (gradient of the straight line).

Gradient, $a= \frac {v-u}{t} \Rightarrow v = u+at$ If the acceleration varied the gradient of the graph would not be constant and then this derivation would be invalid.

The area under a velocity time graph is equal to the displacement $s = \left (\frac{v+u}{2}\right ) t$ (area of a trapezium) and from these two equation one can generate the other two constant acceleration kinematic equations.

Because the gradient is constant it is not necessary in this case to use the calculus notation , $a = \frac {dv}{dt}$ and $s = \int ^t_0 v \,dt$ which you would need to use if the acceleration depended on time.

Answer 2

Why do kinematic equations only work with constant acceleration?

In general, in one dimension: $$ \frac{dv}{dt} = a(t)\;, $$ where the acceleration $a(t)$ is a function of time.

Integrating both sides, we find: $$ v(t) - v(t_0) = \int_{t_0}^{t}a(t')dt' \;, $$ where, for simplicity, I will set $t_0=0$ in the remainder of this post so that I can write: $$ v(t) = v_0 + \int_{0}^{t}a(t')dt' \;. $$

We also have: $$ \frac{dx}{dt} = v(t)\;, $$ so that: $$ x(t) - x_0 =\int_{0}^{t}v(t')dt' = \int_{0}^{t}\left(v_0 + \int_{0}^{t'}a(t'')dt''\right)dt' $$ $$ =v_0t + \int_0^t dt'\int_0^{t'}dt''a(t'')\;. $$

If the acceleration is a constant value, $a$, we can take it out of the integrals and then the last integrals are easy to do. We find: $$ x(t) = x_0 + v_0t + a\int_0^tdt't' $$ $$ = x_0 + v_0t + a\frac{t^2}{2}\;, $$ where is the usual constant-acceleration kinematic equation.

Answer 3

It's pretty simple.

The formal definition of velocity is the derivative of position with respect to time. So in one dimension:

$v = \frac{dx}{dt}$

And therefore

$x(t) = \int_{0}^{t}v(t')dt'$

If you assume $v(t)$ is constant, then that equation becomes $x = vt$ (hence the kinematic equation). If $v(t)$ isn't constant, then $x=vt$ is almost certainly not correct.

Edit: As @hft pointed out, the more central assumption to kinematics is that $a(t)$ is constant, so that:

$v(t) = \int_{0}^{t}a(t')dt' = at$

Now if we substitute $at$ in for $v(t)$, we get:

$x(t) = \int_{0}^{t}v(t')dt' = \int_{0}^{t}at' + v_0dt' = \frac{1}{2}at^2 + v_0t$

Which should look like the form we know and love.

Answer 4

The fundamental equations describing Newtonian mechanics are differential equations. Newton's Second Law (for the position $\vec{x}$ of single mass) takes the form $$m\frac{d^{2}\vec{x}}{dt^{2}}=\vec{F},$$ where $\vec{F}$ is whatever the force is. You cannot say anything more about the motion of the mass without knowing the details of the force $\vec{F}$ that it is exposed to. Different force laws are going to give different solutions.

The kinematic equations are, essentially, the solution of the differential equation when the force law takes the simple form of being a constant: $\vec{F}=-m\vec{g}$ (for a gravitational acceleration $-\vec{g}$, but the same math applies to any force that produces a constant acceleration). The differential equation (Newton's Second Law) is second order in time, so it requires two initial conditions, the position at the initial time (taken to be $t=0$), $\vec{x}(t=0)=\vec{x}_{0}$ and the first derivative of the position at the initial time, $\frac{d\vec{x}}{dt}(0)=\vec{v}(0)=\vec{v}_{0}$. Once these initial data are specified, there is a unique solution, $$\vec{x}(t)=\vec{x}_{0}+\vec{v}_{0}t-\frac{1}{2}\vec{g}t^{2}.$$

From this solution for $\vec{x}(t)$, the other kinematical equations follow. Taking a time derivative gives the behavior of the velocity, $$\vec{v}(t)=\vec{v}_{0}-\vec{g}t.$$ Combining the solutions for position and velocity (with a bit of algebra) gives the third equation, $$v^{2}=v_{0}^{2}-2\vec{g}\cdot(\vec{x}-\vec{x}_{0}).$$

Note that this last equation may be reorganized to say $$\frac{1}{2}mv^{2}+m\vec{g}\cdot\vec{x}=\frac{1}{2}mv^{2}_{0}+m\vec{g}\cdot\vec{x}_{0},$$ which is just conservation of energy (constancy of $K+U$), with a potential energy $U(\vec{x})=m\vec{g}\cdot\vec{x}$. With a different force law (that is, a different $U$ and thus different $\vec{F}=-\vec{\nabla}U$), the explicit form of the energy conservation equation would have to be different.

Answer 5

You can have kinematic equations for different types of variable acceleration. It is just that the equations derived for the case of constant acceleration are not valid when the acceleration is not constant. Which is to be expected. The equations treated in the beginning of an introductory physics course are limited to the case of constant acceleration because this is the simplest case of accelerated motion. It does not mean that this is the only case and these are not "THE" kinematic equations. Just one set of equations, valid for a specififc case - constant acceleration. If you study simple harmonic oscillator you learn of another set of kinematic equations which express the position, velocity and acceleration as functions of time (with sin and cosine functions). This is an example of kinematic equations for non-constant acceleration.