Question

Starting from F=ma, and using calculus, derive v=v₀+at and x=x₀+v₀ t+1/2 at^2.

Answer 1

Answer:

See below.

Explanation:

Here using F = ma , we need to derive the fire and second equation of motion.

$\displaystyle \longrightarrow F = ma \dots(i)$

As we know that the rate of change of velocity is called acceleration. Therefore,

$\displaystyle \longrightarrow F = m\dfrac{dv}{dt}$

From equation (i) we have,

$\displaystyle \longrightarrow ma = m\dfrac{dv}{dt}$

If the mass is constant,

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

$\displaystyle \longrightarrow dv = a.dt$

On integrating both sides,

$\displaystyle \longrightarrow \int dv = \int a .dt$

LHS will be integrated from v₀ to v and RHS will be integrated from 0 to t , as ;

$\displaystyle \longrightarrow\int^v_{v_0} dv =\int^t_0 a .dt$

Here a is constant , so ;

$\displaystyle \longrightarrow v|^v_{v_0}= a \int^t_0 dt$

$\displaystyle \longrightarrow v - v_0= a(t-0)$

Adding v₀ both sides,

$\displaystyle \longrightarrow

\underline{\underline{ v =v_0+at}}$

Hence we have derived the First equation of motion.

For second , as we know that ,

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

$\displaystyle \longrightarrow dx = v.dt$

Integrating both sides, we have;

$\displaystyle \longrightarrow \int dx =\int v.dt$

Putting the limits,

$\displaystyle \longrightarrow \int^x_{x_0} dx =\int_0^t v.dt$

From first equation,

$\displaystyle \longrightarrow x|^x_{x_0}= \int^t_0 ( v_0+at).dt$

Distribute ,

$\displaystyle \longrightarrow x - x_0= \int^t_0 v_0dt +\int^t_0 at.dt$

$\displaystyle \longrightarrow x-x_0= v_0(t-0)+a\bigg[\dfrac{t^2}{2}\bigg]^t_0$

Simplify,

$\displaystyle \longrightarrow \underline{\underline{ x-x_0= v_0t +\dfrac{1}{2}at^2}}$

Therefore we have derived the second equation of motion.

And we are done!