Motion In One Dimension

Motion of an object in a straight line is called one dimensional (1-D) motion. The position of a particle in one dimensional motion can be described by only one variable (say x). For a particle moving along a straight line (1-D motion) all the vector quantities such as position, velocity, displacement and acceleration have only one non-zero component.

1. Displacement: The change in position of a body in a particular direction is known as displacement. It is a vector quantity and its unit is meter in SI. The shortest distance between the initial and final positions of the object in a specified direction.

2. Distance: The total length of actual path traversed by a body in a certain interval of time is called distance. It is the actual path travelled by an object between its initial and final positions. It is a scalar quantity and its unit in SI is meter. Displacement may be positive, negative or zero but distance is always positive. If a particle moves in a straight line without change in its direction, the magnitude of displacement is equal to the distance travelled. Otherwise it is always less than it. Thus, magnitude of Displacement $\le$ distance

Illustration 1: What will be the distance and displacement while  moving in a circle from A to B and then B to A as shown in adjoining figure?

Key concept: Remember difference between distance and displacement.  

Solution:

3. Average Speed and Velocity: The average speed of a particle in a given interval of time is defined as the ratio of the distance travelled to the time taken while, average velocity is defined as the ratio of the displacement to the time taken.

If a particle moves from A to C through a path ABC. Then distance $\left(\Delta s\right)$ travelled is the actual path length ABC, while the displacement is,

$\Delta \overrightarrow{r}=\overrightarrow{r_C}-\overrightarrow{r_A}$            

Thus, if the distance travelled is and displacement of a particle is $\Delta \overrightarrow{r}$ in a given time interval $\Delta t$ then 

$v_{av}=Averagespeed=\frac{\Delta s}{\Delta t}and{\overrightarrow{V}}_{av}=Averagevelocity=\frac{\Delta \overrightarrow{r}}{\Delta t}$

4. Instantaneous Speed and Velocity: Instantaneous speed and velocity are defined at a particular instant and are given by

$v=\underset{\Delta t\to 0}{\lim }\frac{\Delta s}{\Delta t}=\frac{ds}{dt}and\overrightarrow{V}=\underset{\Delta t\to 0}{\lim }\frac{\Delta \overrightarrow{r}}{\Delta t}=\frac{d\overrightarrow{r}}{dt}$

5. Average and Instantaneous Acceleration: Average acceleration is defined as the change in velocity $\Delta \overrightarrow{V}$ over a time interval $\Delta t$. Hence,  $\overrightarrow{a_{av}}=\frac{\Delta \overrightarrow{V}}{\Delta t}$

The instantaneous acceleration of a particle is the rate at which its velocity is changing at that instant i.e.,

         $\overrightarrow{a_{ins}}=$$\underset{\Delta t\to 0}{\lim }\frac{\Delta \overrightarrow{V}}{\Delta t}=\frac{d\overrightarrow{V}}{dt}$Illustration 2. A particle moves along a semi circle path A to B in a time T as shown in the following fig. (a) Determine the average speed of the particle.(b) Determine the average velocity of the particle.Solution: (a) The average speed of the particle = $\frac{distance}{time}=$$\frac{\pi \text{R}}{\text{T}}$

(b) The average velocity of the particle= $\frac{displacement}{time}=$$\frac{2R}{T}$

1. Equations of motion

Following are the three equations of motion for an object with constant acceleration. 

(a)   $v=u+at$

(b)   $s=ut+\frac{1}{2}a{t}^2$

(c)   $v^2=u^2+2as$

where u is the initial velocity of the body (if body start from rest u = 0 ), v is the final velocity, s = displacement travelled by the body in time t seconds and a = acceleration of the body (take + sign for acceleration and – for retardation).

The displacement by the body in n^th second is given by

         $s_n=u+\frac{a}{2}(2n-1)$

2. Equation of motion on an inclined plane

Let a body of mass m slip down a plane, which is inclined at an angle q with the horizontal. If at t = 0, the body is at top of the inclined plane, then in this case u = 0 and a = g sinq 

(i)    In this case the equations of motion are 

(a)   $v=(g\sin \theta )t$                                                                 

(b)   $s=\frac{1}{2}(g\sin \theta )t^2$

(c)   $v^2=2(g\sin \theta )s$

(ii)   If time taken by the body to reach the bottom is t, then 

        $s=\frac{1}{2}(g\sin \theta )t^2$

        $t=\sqrt{\frac{2s}{g\sin \theta }}$

But   $\sin \theta =\frac{h}{s}$ or $s=\frac{h}{\sin \theta }$

         $\therefore t=\frac{1}{\sin \theta }\sqrt{\frac{2h}{g}}$

(iii)   The velocity of the body at the bottom

        $v=g\left(\sin \theta \right)t=\sqrt{2gh}$

(iv)  The velocity of a body moving on an inclined plane does not depend on the inclination of a plane but the time taken to reach the bottom of the plane depends on the inclination of the plane. The velocity and the time taken by the body on an inclined plane depend on the height.

(v)   When a body moves on an inclined plane. it traverses one-fourth of the length of the inclined. Plane in time interval 0 to t/2 and the remaining three fourth in the time interval t/2 to t.

Illustration 3. A ball is projected with a velocity of 20 m/s vertically. Find the distance travelled in first three second. (use g = 10m/sec² )

Solution:   Problem here is to find the distance. We can calculate that the  direction of ball is changes at t = 2s. (From v = u + at, since v = 0 at highest point therefore 0=20 - 10t $\Rightarrow$ t = 2s)

                  Distance traveled in first two second      ($\because$ Distance = Displacement,         because of velocity does not change direction in one dimension)

         $\therefore S_1=20\times 2-\frac{1}{2}10\times 4$

         = 40 – 20 = 20 m (upward)

                  Distance traveled in next one second

         $S_2=-\frac{1}{2}10\times 1=-5mdownward$

         So total distance travelled by the ball in first three seconds

         = 20 +5 = 25m

Illustration 4. A particle moves along the x-axis according to x = 4t – t². Find the distance travelled from t = 0s to t = 3s.

Solution:  The direction of velocity will change when V = 0 at t = 2 sec.

                  Distance =  $\underset{0}{\overset{3}{\int }}vdt$  = $\left|\underset{0}{\overset{1}{\int }}(4-2t)dt\right|+\left|\underset{1}{\overset{2}{\int }}(4-2t)dt\right|+\left|\underset{2}{\overset{3}{\int }}(4-2t)dt\right|$

                  = 3+1+ 1 =5 meter

3. x – t, v – t & a – t GRAPHS FOR MOTION IN ONE DIMENSION 

(i)    Variation of displacement (x), velocity (v) and acceleration (a) with respect to time for different types of motion.

(ii)   Displacement calculation from Velocity - Time Graphs

The displacement during an interval between time t_i and t_f is the area bounded by the velocity curve and the two vertical lines t = t_i and t = t_f, as shown in figures (a) and (b).

(a) For each segment of motion, the velocity is constant. The displacement $\Delta$x₁ during the i ^th interval is the area v_1$\Delta$t₁. So the total displacement is $\Delta$x

$\Delta$x = v_1$\Delta$t₁ + v_2$\Delta$t₂ + v_3$\Delta$t₃ + v_4$\Delta$t₄

$\left(\Delta x=\sum v_i\Delta t_i\right)$

(b) When v vs. t graph is a smooth complex curve. Area bounded by the curve and time axis between t=t_i and t=t_f is the displacement. The area under the curve may be obtained by using integration.

(iii)  Velocity calculation from Acceleration - time Graphs

Given an acceleration–versus–time graph, the change in velocity between t = t_i and t = t_f is the area bounded by the acceleration curve and the vertical lines t = t_i and t = t_f

(a)The area under the a vs t smooth curve is the change in velocity $v=a_0\Delta t$

(b) When a vs t graph is a complex curve, the area under the curve may be obtained by using integration

Illustration 5.    The velocity-time graph of a particle moving along a straight line is shown in following figure. 

(i)    If the particle starts its motion from x = –4m, then draw the (a–t) and (x–t) graphs.

(ii)   Find the displacement of the particle at t = 3 s

Key concept: Use uniform acceleration concept. 

Solution:   (i)

(ii) $x=x_0+[\text{Area}\text{of}\text{v-t}\text{graph}{]}_{t=0}^{t=3}$

$\begin{array}{r}x=\left[-+\frac{\text{1}}{\text{2}}(2)(4)+\left(\frac{4+2}{2}\right)(3-2)\right]\\ \text{or}x=3m\end{array}$

4. Relative Velocity

When two objects moves in the same straight line, we compare their motion in terms of their relative velocity. If two objects A and B are moving in a straight line with velocities V_A and V_B respectively, the relative velocity of object A with respect to object B is given by

         $V_{AB}=V_A-V_B$

where V_B is called reference object velocity

It follows that the relative velocity of object B with respect to object A will be

         $V_{BA}=V_B-V_A$

where V_A is called reference object velocity

Key points regarding relative motion while calculating relative velocity:

(i)     Relative velocity of a particle = velocity of a particle - velocity of reference object

(ii)    If the velocity of a particle be V_A and the velocity of a reference object be V_B then the relative velocity of the particle $V_{AB}=V_A-V_B$

(iii)   Relative velocity of a particle while moving in the same direction.

        Relative velocity $=V_A-V_B$

(iv)   Relative velocity of a particle while moving in the opposite direction.

        Relative velocity $=V_A+V_B$

Illustration 6: The position of a particle moving on a straight line path is given by:

         $x=12+18t+9t^2$metre

         Its velocity at t = 2s is :

(A) 84 ms^–1                                   

(B) 72 ms^–1

(C) 54 ms^–1      

(D) 36 ms^–1

Sol:  (C) Velocity = dx/dt = 18 + 18t. It depends upon time. For t = 2s, the velocity = 18 + 18 x 2 ms^–1 = 54 ms^–1.