Rectilinear motion is the motion of an object in a straight line. It is a linear motion as it occurs along a single dimension or axis. A **uniform** rectilinear motion refers to the motion of an object in a straight line at a constant velocity. A **uniformly accelerated** rectilinear motion refers to the motion of an object in a straight line at a constant acceleration. The motion of a vertically falling object is uniformly accelerated.

- Displacement, Velocity, & Acceleration
- Use of Graphs & Calculus
- Uniformly Accelerated Motion
- Problems

Let a particle moves on a straight line. Consider a point O on this line as the origin and a direction along this line as $+x$ axis. The location of the particle on the straight line is specified by its *position*, which consists of (1) its distance from O and (2) its direction relative to $+x$. For example, if particle is at the point P which is 1 m distance from O and its direction is along $+x$ (when looking from O) then its position is $x=1$ m. If particle is at the point Q which is 1 m distance from O but its direction is opposite to $+x$ (when looking from O) then its position is $x=-1$ m.

Let position of the particle at time $t_1$ be $x_1$ and the position at time $t_2$ be $x_2$. The *displacement* of the particle in the time interval $t_1$ to $t_2$ is $\Delta x=x_2-x_1$. The displacement can be positive or negative.

The path length $d$ is the total distance traveled by the particle when it travels from the position $x_1$ to the position $x_2$. It is always positive and $d\geq |\Delta x|$.

A particle is said to be in motion if its position changes with time. Let its position at time $t_1$ be $x_1$ which changes to $x_2$ at time $t_2$. The displacement of the particle in the time interval $\Delta t=t_2-t_1$ is $\Delta x=x_2-x_1$. The average velocity of the particle in this time interval is defined as the ratio of displacement to the time interval i.e., \begin{align} v_\text{av} &=\frac{\Delta x}{\Delta t} =\frac{x_2-x_1}{t_2-t_1}.\nonumber \end{align} On an $x\text{-}t$ graph, the average velocity over a time interval is the slope of the line connecting the initial and final positions corresponding to that interval.

The instantaneous velocity (or velocity) is defined as the limit of the average velocity as the time interval $\Delta t$ becomes infinitesimally small i.e., \begin{align} v &=\lim_{\Delta t\to 0} \frac{\Delta x}{\Delta t} =\frac{\mathrm{d}x}{\mathrm{d}t}.\nonumber \end{align}

Average speed is the ratio of total path length traversed and the corresponding time interval. The speed of the particle at a time instant $t$ is equal to the magnitude of its velocity at that instant.

In uniform motion, the velocity $v$ is constant. In non-uniform motion, the velocity changes with time. Let the velocity of the particle be $v_1$ at time $t_1$ which changes to $v_2$ at time $t_2$. The average acceleration is the change in velocity divided by the time interval during which the change occurs i.e., \begin{align} a_\text{av} &=\frac{\Delta v}{\Delta t} =\frac{v_2-v_1}{t_2-t_1}.\nonumber \end{align} The instantaneous acceleration is defined as the limit of average acceleration as the time interval $\Delta t$ goes to zero i.e., \begin{align} a &=\lim_{\Delta t\to 0}\frac{\Delta v}{\Delta t} =\frac{\mathrm{d}v}{\mathrm{d}t}.\nonumber \end{align}

The slope of the tangent drawn on $x\text{-}t$ graph at a time $t$ gives velocity at time $t$. The slope of the tangent drawn on $v\text{-}t$ graph at a time $t$ gives acceleration at that time.

The area under $v\text{-}t$ graph from time $t_1$ to $t_2$ gives the displacement of the particle in this time interval. The area under $a\text{-}t$ graph from time $t_1$ to $t_2$ gives the change in velocity in this time interval. By convention, the area above the time axis is taken as positive whereas the area below the time axis is taken as negative.

The calculus (differentiation and integration) are very useful to study motion. Note the following relations: \begin{align} v&=\frac{\mathrm{d}x}{\mathrm{d}t}, && \Delta x=\int_{t_1}^{t_2}\! v\,\mathrm{d}t, \nonumber\\ a&=\frac{\mathrm{d}v}{\mathrm{d}t}, &&\Delta v=\int_{t_1}^{t_2}\! a\,\mathrm{d}t. \nonumber \end{align}

One of the useful relation is \begin{align} \frac{\mathrm{d}v}{\mathrm{d}t} & =\frac{\mathrm{d}x}{\mathrm{d}t}\frac{\mathrm{d}v}{\mathrm{d}x}=v\frac{\mathrm{d}v}{\mathrm{d}x}. \end{align} The velocity attains its extreme value (maximum or minimum) when acceleration is zero i.e., $\frac{\mathrm{d}v}{\mathrm{d}t}=0$.

The motion is said to be uniformly accelerated if $a$ is constant. If a particle starts its motion from origin (i.e., $x=0$ at $t=0$) with an initial velocity $u$ then, \begin{align} &v=u+at,\nonumber\\ &x=ut+\tfrac{1}{2}at^2,\nonumber\\ &v^2-u^2=2ax,\nonumber \end{align} where $x$ and $v$ are displacement and velocity at time $t$.

One example of uniformly accelerated motion is free fall under the gravity. Note that direction of acceleration due to gravity is downward and its magnitude is $g=9.8$ m/s^{2}. Always choose the origin O and a direction ($+x$) before applying equations of motion.

Let $v_A$ and $v_B$ be velocities of two bodies A and B. These velocities are defined in a common reference frame (say frame attached to the ground). The velocity of B relative to A (i.e., velocity of the body B as seen by an observer on the body A) is given by \begin{align} v_{B/A}=v_B-v_A.\nonumber \end{align}

**Problem (IIT JEE 1993):**
A small block slides without friction down an inclined plane starting from rest. Let $s_n$ be the distance travelled from $t=n-1$ to $t=n$. Then $\frac{s_n}{s_{n+1}}$ is,

- $\frac{2n-1}{2n}$
- $\frac{2n+1}{2n-1}$
- $\frac{2n-1}{2n+1}$
- $\frac{2n}{2n+1}$

**Solution:**
The downward acceleration of the block on an inclined plane with inclination angle $\theta$ is given by $a=g\sin\theta$. Initial velocity of the block is $u=0$. The distance travelled in the first $n$ seconds is given by,
\begin{alignat}{2}
s(n) &=u n+\tfrac{1}{2}g\sin\theta\, n^2 \\
&=\tfrac{1}{2}\,g\sin\theta \,n^2. \nonumber
\end{alignat}
Thus, the distances travelled in the $n^\mathrm{th}$ and the $(n+1)^\mathrm{th}$ seconds are,
\begin{align}
s_n &=s(n)-s(n-1) \\
&=(2n-1)\, \tfrac{1}{2}\,g\sin\theta, \\
s_{n+1} &=s(n+1)-s(n) \\
&=(2n+1)\, \tfrac{1}{2}\,g\sin\theta.
\end{align}
Divide first equation by the second equation to get
\begin{align}
\frac{s_n}{s_{n+1}}=\frac{2n-1}{2n+1}.
\end{align}

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