Question

Derive third equation of motion by calculus method

Answer 1

Newton's 3rd law of motion: Every action in the universe have their equall and opposite reaction..

Newton's 3rd equation of motion consists of final velocity (v), initial velocity(u), displacement (s) and constant acceleration (a). The equation is as follows-

v²= u²+2as

Actually I don't know how to derive it. But here is some hints are provided.

You want it by integration.

(v² - u²) /2 indicates the integration of v dv limit u to v.

Here is the proof -

a= dv/dt……1

v=ds/dt……….2

from 1 & 2:

a ds/dt = v dv/ dt

implies a ds =v dv

after integration you will get 3rd equation. (pic of integration is shown above the matter).

Hope it will help...

Answer 2

Answer:

Acceleration (a) of a body is defined the rate of change of its velocity (v) with time (t). Mathematically, we can write,

$a=\frac{dv}{dt}\\adt=dv$

Integrating both sides of the above equation,

$\int\limits^t_0 adt=\int\limits^v_u dv\\ a\int\limits^t_0 {x} dt=\int\limits^v_u dv\\a[t]^t_0=[v]^v_u\\at=v-u\\v=u+at$                                                                         .................(1)

Equation (1) is the 1st equation of motion of a body.

where v is the final velocity, u is the initial velocity and t is the time of motion.

To derive the 2nd equation of motion: The velocity of the body is defined as the rate of change of its displacement.

$v=ds/dt\\ds=vdt$

Substitute the value of the final velocity from the 1st equation of motion

ds = (u + at) dt

⇒ ds = udt + atdt

On Integrating both sides, we get

$\int\limits^S_0 ds=\int\limits^t_0 {u} \, dt + \int\limits^t_0atdt$

$[S]^S_0=u[t]^t_0+\frac{a}{2} [t^2]^t_0\\\\S=ut+\frac{1}{2} at^2$                                                                 ..................(2)

Equation (2) is the second equation of motion.

To derive the third equation of motion,

Use both the expression of the first and the second equations:

$a=\frac{dv}{dt}\; and \;v=\frac{ds}{dt}$

$dt=\frac{dv}{a}=\frac{dS}{v}$

On integrating both sides of the expression, we will get;

$\int\limits^S_0 adS=\int\limits^v_u vdv\\a[S]^S_0=\frac{1}{2}[v^2]^v_u$

$aS= \frac{v^2-u^2}{2}$

$v^2-u^2=2aS$                                                                      ...................(3)

Equation (3) is the 3rd equation of motion.