Question

Derive three kinematics equations using calculus method. Also tell this

method can be used for non uniform acceleration or not!​

Answer 1

Let a body move with velocity $\displaystyle\sf {u}$ initially (at 0 second) and attain a velocity $\displaystyle\sf {v}$ after $\displaystyle\sf {t}$ seconds travelling a displacement $\displaystyle\sf {s.}$

First we're considering motion with uniform acceleration, i.e., here acceleration is independent of time.

We know that acceleration is first derivative of velocity wrt time.

$\displaystyle\sf{\longrightarrow \dfrac {dv}{dt}=a}$

$\displaystyle\sf{\longrightarrow dv=a\ dt}$

Integrating,

$\displaystyle\sf{\longrightarrow \int\limits_u^v dv=\int_0^t a\ dt}$

$\displaystyle\sf{\longrightarrow \big[v\big]_u^v=a\big [t\big]_0^t}$

$\displaystyle\sf{\longrightarrow v-u=a(t-0)}$

$\displaystyle\sf{\longrightarrow\underline {\underline {v=u+at}}\quad\quad\dots (1)}$

This is the first equation of motion.

We know that velocity is first derivative of displacement wrt time.

$\displaystyle\sf{\longrightarrow \dfrac {dx}{dt}=v}$

So (1) becomes,

$\displaystyle\sf{\longrightarrow\dfrac {dx}{dt}=u+at}$

$\displaystyle\sf{\longrightarrow dx=(u+at)\ dt}$

Integrating,

$\displaystyle\sf{\longrightarrow \int\limits_0^sdx=\int\limits_0^t(u+at)\ dt}$

$\displaystyle\sf{\longrightarrow \big [x\big]_0^s=u\big [t\big]_0^t+a\left [\dfrac {t^2}{2}\right]_0^t}$

$\displaystyle\sf{\longrightarrow s-0=u(t-0)+\dfrac {1}{2}\,a\left(t^2-0^2\right)}$

$\displaystyle\sf{\longrightarrow \underline {\underline {s=ut+\dfrac {1}{2}\,at^2}}}$

This is the second equation of motion.

Again we know that acceleration is first derivative of velocity wrt time.

$\displaystyle\sf{\longrightarrow \dfrac {dv}{dt}=a}$

We can write it as,

$\displaystyle\sf{\longrightarrow \dfrac {dv}{dx}\cdot\dfrac {dx}{dt}=a}$

$\displaystyle\sf{\longrightarrow \dfrac {dv}{dx}\cdot v=a}$

$\displaystyle\sf{\longrightarrow v\ dv=a\ dx}$

Integrating,

$\displaystyle\sf{\longrightarrow \int\limits_u^vv\ dv=\int\limits_0^sa\ dx}$

$\displaystyle\sf{\longrightarrow \left [\dfrac {v^2}{2}\right]_u^v=a\big[x\big]_0^s}$

$\displaystyle\sf{\longrightarrow\dfrac {1}{2}(v^2-u^2)=a(s-0)}$

$\displaystyle\sf {\longrightarrow\underline {\underline {v^2=u^2+2as}}}$

This is the third equation of motion.

These three equations can be used only in a uniform accelerated motion.

But the calculus method can be used to derive the three kinematics equations in non - uniform accelerated motion too!

And those give the general kinematics equations.

Let the acceleration of the body vary with time $\displaystyle\sf {t}$ as,

$\displaystyle\sf{\longrightarrow a=kt^n}$

where $\displaystyle\sf {k}$ is a constant and does not vary with time.

Since acceleration is first derivative of velocity,

$\displaystyle\sf{\longrightarrow \dfrac {dv}{dt}=kt^n}$

$\displaystyle\sf{\longrightarrow dv=kt^n\ dt}$

Integrating,

$\displaystyle\sf{\longrightarrow \int\limits_u^vdv=\int\limits_0^tkt^n\ dt}$

For $\displaystyle\sf {n\neq-1,}$

$\displaystyle\sf{\longrightarrow v-u=k\cdot\dfrac {t^{n+1}-0^{n+1}}{n+1}}$

$\displaystyle\sf{\longrightarrow v=u+\dfrac {1}{n+1}\,kt^{n+1}\quad\quad\dots (2)}$

$\displaystyle\sf {\longrightarrow\underline {\underline {v=u+\dfrac {1}{n+1}\, at}}}$

Since velocity is first derivative of displacement wrt time, (2) becomes,

$\displaystyle\sf{\longrightarrow \dfrac {dx}{dt}=u+\dfrac {1}{n+1}\,kt^{n+1}}$

$\displaystyle\sf{\longrightarrow dx=\left (u+\dfrac {1}{n+1}\,kt^{n+1}\right)\ dt}$

Integrating,

$\displaystyle\sf{\longrightarrow \int\limits_0^sdx=\int\limits_0^t\left (u+\dfrac {1}{n+1}\,kt^{n+1}\right)\ dt}$

For $\displaystyle\sf {n\notin\{-2,\ -1\},}$

$\displaystyle\sf{\longrightarrow s-0=u(t-0)+\dfrac {k}{n+1}\cdot \dfrac {t^{n+2}-0^{n+2}}{n+2}}$

$\displaystyle\sf{\longrightarrow s=ut+\dfrac {1}{(n+1)(n+2)}\,kt^{n+2}}$

$\displaystyle\sf{\longrightarrow \underline{\underline {s=ut+\dfrac {1}{(n+1)(n+2)}\,at^2}}}$

Thus we've derived general kinematics equations by calculus method.

For $\displaystyle\sf{n\in\{-2,\ -1\},}$ logarithmic function wrt time comes on as velocity or displacement, but this logarithmic function is not defined for initial time $\displaystyle\sf {t=0}$ as it tends to infinity. And so an exact expression for them can't be determined for these values of $\displaystyle\sf {n.}$