Question

Define acceleration and derive an uniform equation v=u+at​

Answer 1

Explanation:-

Definition of acceleration

When ∆t approaches zero this average acceleration becomes the instantaneous acceleration. instantaneous acceleration is also called acceleration

The dimensions of acceleration is

$\to \sf \: LT {}^{ - 2}$

Its SI unit is

$\sf \to \: ms {}^{ - 2}$

Acceleration are two types

i) Average acceleration

=> Average acceleration is define as the ratio of change in velocity, i.e. ∆v to the time interval ∆t in which this change occurs. Hence

$\implies \boxed{ \rm \: a_{av} = \frac{ \Delta v}{\Delta t} = \frac{v_i - v_f}{\Delta t} }$

ii) Instantaneous Acceleration

The instantaneous acceleration is defined at a particular instant and is given by

$\implies\boxed{ \rm \: a = \displaystyle \lim_{\Delta t \to0} \rm\frac{\Delta v}{\Delta t} = \frac{dv}{dt} }$

Derivation of an uniform acceleration

For one dimensional motion with a = constant

we can write

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \rm \: dv = a \: \: dt$

Integrating both side , we have

$\rm \int dv = a \int dt$

At t=0 , velocity is u and at t= t velocity is v .Hence

$\int \limits_ { {}^{u} }^{v} \: dv \: = a \: \int \limits_ {0}^{t} \: dt$

$\rm \big[v \big] {}^{v} _u = a \big[t \big] {}^{t} _0 \: \: \: or \: \: \: v - u = at$

$\boxed{ \rm \: v = u + at}$

Hence proved