Uniform Motion in a Straight Line

An object is said to be in uniform motion, if it covers equal displacement in equal interval of time however, small these time intervals maybe.

The study of the uniform motion is the simplest one. And accelerated motion requires a cause or an effort. Uniform motion in a straight line is physically important from the point of view that it does not require any such cause or effort.

Consider that an object is in uniform motion along a straight line OX. Let point O be the origin for position measurement. Suppose that the origin for measurement of time is also taken at the instant, when the object is at point O.

Further consider that at any time t the object is at point A and at time t’, it reaches point B such that displacement $ \displaystyle \overrightarrow{{OA}}=\overrightarrow{x}$ and $ \displaystyle \overrightarrow{{OB}}=\overrightarrow{{{x}’}}$. Obviously, the displacement of the object in the time interval t and t’ i.e. t’ – t is $ \displaystyle \overrightarrow{{AB}}=\overrightarrow{{{x}’}}-\overrightarrow{x}$.

Now the velocity of the uniform motion $ \displaystyle \text{= }\frac{{\text{displacement}}}{{\text{time interval}}}$

$ \displaystyle \therefore \overrightarrow{v}=\dfrac{{\overrightarrow{{{x}’}}-\overrightarrow{x}}}{{\overrightarrow{{{t}’}}-\overrightarrow{t}}}$

The following points are true for uniform motion:

1. Generally, the displacement may or may not be equal to the actual distance covered by an object. However, when the uniform motion takes place along a straight line in a given direction, the magnitude of the displacement is equal to the actual distance covered by the object.
2. The velocity of uniform motion is same for different choices of t and t’.
3. The velocity of uniform motion is not affected due to the shift of origin.
4. The positive value of velocity means the object is moving towards right of the origin while the negative velocity means the motion is towards the left of the origin.
5. As said earlier, for an object to be in uniform motion no cause or effort i.e no forces required.
6. The average and instantaneous velocity in a uniform motion or always equal, as the velocity during uniform motion is same at each point of the path or at each instant.

Formulae for Uniform Motion

When an object is in uniform motion, its velocity always remains constant. Mathematically, for uniform motion;

$ \displaystyle \overrightarrow{v}=\text{ constant vector }…\text{ (1)}$

Suppose that the origin of the position access is point O and the origin for the time measurement is taken as the instant, when the object is at point A, such that $ \displaystyle \overrightarrow{{OA}}=\overrightarrow{{{{x}_{\circ }}}}$. If at time t, the object moving with constant velocity $ \displaystyle \overrightarrow{v}$ is at point B, such that $ \displaystyle \overrightarrow{{OB}}=\overrightarrow{x}$, then

$ \displaystyle \overrightarrow{x}=\overrightarrow{{{{x}_{\circ }}}}+\overrightarrow{v}t\text{ }…\text{ (2)}$

Similarly, if at time t’, the object reaches point C, such that $ \displaystyle \overrightarrow{{OC}}=\overrightarrow{{{x}’}}$,

Then,

$ \displaystyle \overrightarrow{{{x}’}}=\overrightarrow{{{{x}_{\circ }}}}+\overrightarrow{v}{t}’\text{ }…\text{ (3)}$

Subtracting equation 2 from 3, we have

$ \displaystyle \begin{array}{l}\overrightarrow{{{x}’}}-\overrightarrow{x}=\left( {\overrightarrow{{{{x}_{\circ }}}}+\overrightarrow{v}{t}’} \right)-\left( {\overrightarrow{{{{x}_{\circ }}}}+\overrightarrow{v}t} \right)\\\Rightarrow \overrightarrow{{{x}’}}=\overrightarrow{x}+\overrightarrow{v}({t}’-t)\text{ }…\text{(4)}\end{array}$

The equations 2, 3 and 5 represent the kinematics of uniform motion along a straight line. The equation 4 may be obtained directly by expanding equation $ \displaystyle \overrightarrow{v}=\frac{{\overrightarrow{{{x}’}}-\overrightarrow{x}}}{{\overrightarrow{{{t}’}}-\overrightarrow{t}}}$ for $ \displaystyle {\overrightarrow{{{x}’}}}$ or the equation 4 may be used to obtain expression for velocity of the uniform motion.

If we call $ \displaystyle \overrightarrow{x}-\overrightarrow{{{{x}_{\circ }}}}=\overrightarrow{S}$ the displacement of the object in time t, then equation 2 becomes

$ \displaystyle \overrightarrow{S}=\overrightarrow{v}t$

In the for going study of motion along a straight line, the various displacements of the object such as $ \displaystyle \overrightarrow{{{{x}_{\circ }}}},\overrightarrow{x}\text{ or }\overrightarrow{{{x}’}}$ represent vectors having the same direction and consequently velocity vector is also in the same direction. Just for the sake of convenience, the arrowheads over the displacement and the velocity vectors may be dropped in the equations from 1 to 5.

Velocity-time graph

Let us considered a train moving with a constant velocity of say 15m/s along a straight track so as to study the velocity time graph of the uniform motion of an object in a straight line. As a train is in uniform motion, the magnitude of its velocity at t = 0, t = 1s, t = 2s,… will always be 15m/s and therefore graph between time and velocity of the train will be shown in figure.

Hence, velocity time graph for an object in uniform motion along a straight path is a straight line parallel to time-axis.

The velocity time graph of an object may be used to calculate it’s displacement in a given time interval geometrically. In the above graph A and B are the two points on the velocity-time graph corresponding to the instants t and t’. If the motion of the object is with a constant velocity $ \displaystyle v$, then AA’ = BB’ = $ \displaystyle v$. Also, A’B’ = t’ – t. Therefore,

area under $ \displaystyle v-t$ graph between t and t’ = $ \displaystyle \text{area }AB{B}’A=A{A}’\times {A}'{B}’=v\left( {{t}’-t} \right)$

But, since $ \displaystyle v\left( {{t}’-t} \right)=\overrightarrow{{{x}’}}-\overrightarrow{x}$

Hence , $ \displaystyle \overrightarrow{{{x}’}}-\overrightarrow{x}=\text{area }AB{B}’A$

i.e. displacement of an object in time interval $ \displaystyle \left( {{t}’-t} \right)$ is numerically equal to the area under velocity-time graph between the instants t and t’.

Note: It may be pointed out that this geometrical method of finding the displacement of an object holds good even in the case, when the body is moving with negative velocity. In such a case, the area below the velocity-time graph is taken negative and corresponding to this the displacement will also be negative.

Position-time graph

To study the position-time graph of the uniform motion, let us again consider the example of a train moving along a straight track with a constant velocity of 15m/s. Suppose that at t = 0, the displacement of the train is 10m from the origin of the position axis. If $ \displaystyle {{{x}_{\circ }}}$ represents the displacement of the train at t = 0, then mathematically, we may say that

when $ \displaystyle t=0,{{x}_{\circ }}=10\text{m}$

The displacement of the train from the origin of the position axis at time t = 1s will be 10 + 15(1) i.e. 25m. Similarly, the displacement or position of the train at $ \displaystyle t=2s,3s,4s\text{ and }5s$ can be calculated.

If we plot a graph between time and position of the train from the data, the position-time graph will be like:

It follows that the position-time graph for an object in uniform motion along a straight path is a straight line inclined to the time-axis.

Let us now find the slope of the position time graph of an object in uniform motion. Since the graph is a straight line, its slope may be found between any two points on the graph. In the figure A and B are two points on position time graph corresponding to time t and t’. The position of the object (i.e. displacement from the origin of position axis) at time t is x (= AA’) and at time t’ is x’ (= BB’), so that $ \displaystyle AC={t}’-t\text{ and }BC={x}’-x$. Now, from right angled triangle ACB,

$ \displaystyle \text{slope of the }x-t\text{ graph }=\frac{{BC}}{{AC}}=\frac{{{x}’-x}}{{{t}’-t}}$

Hence, from the above equations, it follows that the velocity of uniform motion is equal to the slope of position time graph.