Question

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

Answer 1

See below.

Step-by-step 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(dv)/(dt)`

From equation (i) we have,

`\displaystyle \longrightarrow ma = m(dv)/(dt)`

If the mass is constant,

`\displaystyle \longrightarrow a =(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</p><p> \underline{\underline{ v =v_0+at}}`

Hence we have derived the First equation of motion.

For second , as we know that ,

`\displaystyle \longrightarrow v =(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[(t^2)/(2)\bigg]^t_0\\`

Simplify,

`\displaystyle \longrightarrow \underline{\underline{ x-x_0= v_0t +(1)/(2)at^2}}`

Therefore we have derived the second equation of motion.

And we are done!