Physics > Motion in One Dimension > 4.0 Derivation of the kinematics equation
Motion in One Dimension
1.0 Introduction
2.0 Kinematic variables
2.1 Distance and displacement
2.2 Average speed and velocity
2.3 Instantaneous speed and velocity
2.4 Average and instantaneous acceleration
3.0 Motion in one dimension
3.1 Motion in a straight line with uniform velocity
3.2 Motion in a straight line with uniform acceleration
3.3 Motion in a straight line with non-uniform acceleration
4.0 Derivation of the kinematics equation
5.0 Vertical motion under gravity
5.1 Basic terminologies for motion under gravity
5.2 Detailed concept of motion under gravity
5.3 Solved examples
6.0 Analysis of motion through graph
6.1 Displacement - time graph
6.2 Velocity - time graph
6.3 Area under the graph
6.4 Solved examples
7.0 Relative motion
7.1 Relative displacement
7.2 Relative velocity
7.3 Relative acceleration
7.4 Illustration of relative motion
7.5 Application of relative motion
8.0 Simultaneous motion of two bodies
9.0 River boat problem
9.1 Downstream
9.2 Upstream
9.3 Crosses the river in shortest interval of time
9.4 Reaches the point just opposite from where he started
9.5 River-man problem
9.6 Solved examples
10.0 Aircraft-wind problem
11.0 Rain problem
4.1 Graphical method
2.2 Average speed and velocity
2.3 Instantaneous speed and velocity
2.4 Average and instantaneous acceleration
3.2 Motion in a straight line with uniform acceleration
3.3 Motion in a straight line with non-uniform acceleration
5.2 Detailed concept of motion under gravity
5.3 Solved examples
6.2 Velocity - time graph
6.3 Area under the graph
6.4 Solved examples
7.2 Relative velocity
7.3 Relative acceleration
7.4 Illustration of relative motion
7.5 Application of relative motion
9.2 Upstream
9.3 Crosses the river in shortest interval of time
9.4 Reaches the point just opposite from where he started
9.5 River-man problem
9.6 Solved examples
Consider a body having initial velocity $u$ at time $t=0$ and moves with constant acceleration $a$. It moves with velocity $v$ at any time $t$.
For $\overrightarrow v = \overrightarrow u + \overrightarrow a t$,
The slope of the graph gives us acceleration.
$$\tan \theta = \frac{{v - u}}{{t - 0}}$$$$a = \frac{{v - u}}{t}$$$$v = u + at$$ or $$\overrightarrow v = \overrightarrow u + \overrightarrow a t \quad ...(i)$$
For $\overrightarrow s = \overrightarrow u t + \frac{1}{2}\overrightarrow a {t^2}$,
Area under the velocity-time $(v-t)$ graph gives us displacement.
Area $=$ Area of $\Delta ABC$ + Area of $\Delta OACD$$$s = ut + \frac{1}{2}\left( {v - u} \right)t$$
As we know, $$v - u = at$$ So, $$s = ut + \frac{1}{2}a{t^2}$$ or $$\overrightarrow v = \overrightarrow u + \overrightarrow a t \quad ...(ii)$$
For ${v^2} = {u^2} + 2\overrightarrow a .\overrightarrow s $,
Eliminating $t$ from equation $(i)$ in equation $(ii)$ we get,
$$\overrightarrow s = \overrightarrow u \left[ {\frac{{\overrightarrow v - \overrightarrow u }}{a}} \right] + \frac{1}{2}\overrightarrow a {\left[ {\frac{{\overrightarrow v - \overrightarrow u }}{a}} \right]^2}$$$$\overrightarrow s = \overrightarrow u \left[ {\frac{{\overrightarrow v - \overrightarrow u }}{a}} \right] + \frac{1}{2}\overrightarrow a \left[ {\frac{{{v^2} + {u^2} - 2\overrightarrow v .\,\overrightarrow u }}{{{a^2}}}} \right]$$$$\overrightarrow s = \overrightarrow u \left[ {\frac{{\overrightarrow v - \overrightarrow u }}{a}} \right] + \left[ {\frac{{{v^2} + {u^2} - 2\overrightarrow v .\,\overrightarrow u }}{{2\overrightarrow a }}} \right]$$$$\overrightarrow s = \left[ {\frac{{2\overrightarrow u .\,\overrightarrow v - 2{u^2} + {v^2} + {u^2} - 2\overrightarrow v .\,\overrightarrow u }}{{2\overrightarrow a }}} \right]$$$$2\overrightarrow a .\overrightarrow s = {v^2} - {u^2}$$$${v^2} = {u^2} + 2\overrightarrow a .\overrightarrow s $$
For ${\vec s_{{n^{th}}}} = \vec u + \frac{1}{2}\vec a\left( {2n - 1} \right)$
Displacement in $n^{th}$ second means the displacement of a body between $(n-1)$ to $n$ seconds.
$${{\vec s}_n} = \vec un + \frac{1}{2}\vec a{n^2}$$$${{\vec s}_{n - 1}} = \vec u\left( {n - 1} \right) + \frac{1}{2}\vec a{\left( {n - 1} \right)^2}$$ Then, $${{\vec s}_n} - {{\vec s}_{n - 1}} = \vec u\left( {n - n + 1} \right) + \frac{1}{2}\vec a\left[ {{n^2} - {{\left( {n - 1} \right)}^2}} \right]$$$${{\vec s}_{{n^{th}}}} = \vec u + \frac{1}{2}\overrightarrow a \left[ {{n^2} - {n^2} - 1 + 2n} \right]$$$${{\vec s}_{{n^{th}}}} = \vec u + \frac{1}{2}\overrightarrow a \left[ {2n - 1} \right]$$
