# 2.6 One dimensional motion with constant acceleration  (Page 3/6)

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The acceleration due to gravity near the earth surface is nearly constant and equal to 9.8 $m/{s}^{2}$ . Value of ‘g’ is taken as 10 $m/{s}^{2}$ as an approximation to facilitate ease of calculation.

When only acceleration due to gravity is considered, neglecting other forces, each of the bodies (feather and iron ball) starting from rest is accelerated at the same rate. Velocity of each bodies increases by 9.8 m/s at the end of every second. As such, the feather and the iron ball reach the surface at the same time and at the same velocity.

A close scrutiny of three equations of motion derived so far reveals that they relate specific quantities, which define the motion. There is possibility that we may encounter problems where inputs are not provided in the manner required by equation of motion.

For example, a problem involvoing calaculation of displacement may identify initial velocity, final velocity and acceleration as input. Now, the equation of motion for displacement is expressed in terms of initial velocity, time and acceleration. Evidently, there is a mis-match between what is given and what is required. We can, no doubt, find out time from the set of given inputs, using equation for velocity and then we can solve equation for the displacement. But what if we develop a relation-ship which relates the quantities as given in the input set! This would certainly help.

From first equation :

$\begin{array}{l}v=u+at\\ ⇒v-u=at\\ ⇒t=\frac{\left(v-u\right)}{a}=at\end{array}$

From second equation :

$\begin{array}{l}\frac{\left(u+v\right)}{2}=\frac{\left({x}_{2}-{x}_{1}\right)}{t}\\ ⇒\left(u+v\right)=\frac{2\left({x}_{2}-{x}_{1}\right)}{t}\end{array}$

Eliminating ‘t’,

$\begin{array}{l}⇒\left(u+v\right)=\frac{2a\left({x}_{2}-{x}_{1}\right)}{\left(v-u\right)}\\ ⇒{v}^{2}-{u}^{2}=2a\left({x}_{2}-{x}_{1}\right)\end{array}$

$\begin{array}{l}⇒{v}^{2}-{u}^{2}=2a\left({x}_{2}-{x}_{1}\right)\end{array}$

This equation relates initial velocity, final velocity, acceleration and displacement.

Also, we observe that equation for displacement calculates displacement when initial velocity, acceleration and time are given. If final velocity - instead of initial velocity - is given, then displacement can be obtained with slight modification.

$\begin{array}{l}\Delta x={x}_{2}-{x}_{1}=ut=\frac{1}{2}a{t}^{2}\end{array}$

Using v = u + at,

$\begin{array}{l}⇒\Delta x={x}_{2}-{x}_{1}=\left(v-at\right)t+\frac{1}{2}a{t}^{2}\end{array}$

$\begin{array}{l}⇒\Delta x={x}_{2}-{x}_{1}=vt-\frac{1}{2}a{t}^{2}\end{array}$

## Constant acceleration in one dimensional motion

Problem : A train traveling on a straight track moves with a speed of 20 m/s. Brake is applied uniformly such that its speed is reduced to 10 m/s, while covering a distance of 200 m. With the same rate of deceleration, how far will the train go before coming to rest.

Solution : To know the distance before train stops, we need to know the deceleration. We can find out deceleration from the first set of data. Here, u = 20 m/s ; v = 10 m/s ; x = 200 m ; a = ? An inspection of above data reveals that none of the first three equations of motion fits the requirement in hand, while the additional form of equation ${v}^{2}-{u}^{2}=2ax$ serves the purpose :

$\begin{array}{l}a=\frac{{v}^{2}-{u}^{2}}{2x}=\frac{{10}^{2}-{20}^{2}}{2x200}=-\frac{3}{4}\phantom{\rule{2pt}{0ex}}m/{s}^{2}\end{array}$

For the train to stops, we have

$\begin{array}{l}u=20\phantom{\rule{2pt}{0ex}}m/s;\phantom{\rule{2pt}{0ex}}v=0\phantom{\rule{2pt}{0ex}}m/s;\phantom{\rule{2pt}{0ex}}a=-\frac{3}{4}\phantom{\rule{2pt}{0ex}}m/{s}^{2};\phantom{\rule{2pt}{0ex}}\phantom{\rule{2pt}{0ex}}x=?\end{array}$

and,

$\begin{array}{l}x=\frac{{v}^{2}-{u}^{2}}{2a}=\frac{{0}^{2}-{20}^{2}}{2x\left(-\frac{3}{4}\right)}=266.67\phantom{\rule{2pt}{0ex}}m\end{array}$

## Displacement in a particular second

The displacement in a second is obtained by subtracting two displacements in successive seconds. We calculate displacements in ${n}^{\mathrm{th}}$ second and ${\mathrm{\left(n-1\right)}}^{\mathrm{th}}$ seconds. The difference of two displacements is the displacement in the ${n}^{\mathrm{th}}$ second. Now, the displacements at the end of n and (n-1) seconds as measured from origin are given by :

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