Question

Derive the expression for Newton's second equation and third equation of motion using Calculus.

Answer 1

Dear Student,

​Considering a particle of mass 'm' having initial velocity 'u', Let after time 't' its final velocity becomes 'v' with a uniform acceleration 'a'.
We know the first equation of motion, v = u + at .......(1)
Derivation of eqn (2): -
Let the distance travelled by particle be 's'.
we know the velocity,
$$\begin{gathered} v=\frac{ds}{dt}=u+at[fromeqn(1\left)\right] \\ \Rightarrow ds=udt+atdt \\ \Rightarrow \int ds=u\int dt+a\int tdt\Rightarrow s=ut+\frac{1}{2}a{t}^2+C \\ Att=0\to s=0 \\ Soweget,C=0\Rightarrow \mathbf{}\mathbf{}\mathit{s}\mathbf{}\mathbf{=}\mathbf{}\mathit{u}\mathit{t}\mathbf{}\mathbf{}\mathbf{+}\mathbf{}\mathbf{}\frac{\mathbf{1}}{\mathbf{2}}\mathit{a}{\mathit{t}}^{\mathbf{2}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{2}\mathbf{\right)} \end{gathered}$$

Derivation of eqn (3):-
As we know acceleration,
$$\begin{gathered} a=\frac{dv}{dt}=\frac{dv}{ds}.\frac{ds}{dt}=\frac{dv}{ds}.v \\ \Rightarrow a\int ds=\int vdv\Rightarrow a.s=\frac{v^2}{2}+C \\ Nowatt=0\to s=0\to v=u. \\ Soweget,C=\frac{u^2}{2}. \\ hence,v^2=u^2+2as......\left(3\right) \end{gathered}$$

So eqn (2) & (3) are required equations.


Regards..