Question

1. Derive following equations for a uniformly accelerated motion:
V= u + at
S= Ut+ 1/2 at2
Where symbols have their usual meaning.​

Answer 1

Answer:

acceleration=change in velocity/time taken

acceleration =final velocity -initial velocity/time taken

a=v-u/t

at=v- u

v=u+at

average velocity =initial velocity+final velocity/2

average velocity=u+v/2

since:distance travelled=average velocity*time

s=(u+v)/2 *t

from first equation of motion v=u+at

s=(u+u+at)/2*t

s=(2u+at )*t/2

s=2ut+at*t

s=2ut+at*t/2

s=ut+1/2at*t

Answer 2

Let me derive the two equations using calculus method.

But before, the motion is considered with,

• time from 0 to t seconds.

• velocity from u to v in $\displaystyle\sf{m\ s^{-1}.}$

• displacement from 0 to s in metres.

We have, acceleration,

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

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

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

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

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

$\displaystyle\longrightarrow\sf{\underline{\underline{v=u+at}}}$

But,

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

Then,

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

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

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

$\displaystyle\longrightarrow\sf{\big[x\big]_0^s=\int\limits_0^tu\ dt+\int\limits_0^tat\ dt}$

$\displaystyle\longrightarrow\sf{s-0=u\big[t\big]_0^t+a\left[\dfrac{t^2}{2}\right]_0^t}$

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

Hence Derived!