Motion In One Dimension |

 Introduction

If a body is in uniform motion (velocity = constant or body has acceleration and velocity acting at 0° or 180°, then path followed is straight line and motion is rectilinear motion.

Uniform motion

$\to$ Velocity is constant and acceleration is zero.

$\to$ Body covers equal displacement in equal time interval.

$\to$ Equation for uniform motion s = ut .

$\to$ time-displacement graph is straight line .

$\to$ Time velocity graph is straight line parallel to time axis

$\to$ Time acceleration graph is straight line coincide it to time axis.

Variable motion at constant acceleration

$\to$ Velocity changes at constant rate and acceleration is constant.

$\to$ Body may cover equal distances in equal time intervals but never covers equal displacements in equal time interval (a and v at 180°)

Distance and Displacement

$\to$ The length of actual path followed by a moving body is called distance. It is a scalar quantity.

$\to$ The shortest distance between initial and final position with direction for a moving body is called displacement . It is a vector quantity.

$\to$ Displacement covered by a moving body in certain time interval may be zero, negative or positive but distance never be zero or negative.

$\to$ Distance ≥ Displacement

$\to \dfrac{\text{Distance}}{\text{Displacement}} ≥ 1$

$\to$ For a moving body, the distance travelled always increases with time displacement may increase or decrease with time.

Speed and Velocity

Speed
$\to$ The time rate of change in distance for a moving body is called speed. It is a scalar quantity.
$\to$ Speed =$\dfrac{\text{distance}}{\text{time taken}}$
$\to$ The speed of a body can be zero or positive but never negative.
$\to$ A body is said to be moving with uniform speed if it covers equal distances in equal intervals of time.
$\to$ A body is said to be moving with a variable speed if it covers equal distances in unequal intervals of time or unequal distances in equal intervals of time.
$\to$ Average speed = $\dfrac{\text{distance covered in given time}}{\text{time taken}}$
$V_{av} = \dfrac{(x-x_0)}{(t-t_0)} = \dfrac{∆x}{∆t}$
Instantaneous speed is limiting value of average speed as time interval tends to zero.
$V_{av} = \displaystyle{\lim_{∆t \to 0}} \dfrac{∆x}{∆t} = \dfrac{dx}{dt}$
It is defined for a point of time .

Velocity

$\to$ The time rate of change in displacement tor a moving body is called velocity . It is a vector quantity .

$\to$ Velocity=$\dfrac{\text{displacement}}{\text{time taken}}$

$\to$ Average velocity = $\dfrac{\text{displacement covered in given time}}{\text{time taken}}$

$V_{av} = \dfrac{(s-s_0)}{(t-t_0)} = \dfrac{∆s}{∆t}$

It is defined for period of time.

Instantaneous speed is limiting value of average speed as time interval tends to zero.

$V= \displaystyle{\lim_{∆t \to 0}} \dfrac{∆s}{∆t} = \dfrac{ds}{dt}$

It is defined for a point of time .

Average velocity and Average speed

$\to$ Average velocity for a moving body in certain time interval may be zero, +ve or -ve but average speed can never be zero or –ve .

$\to$ Average speed ≥ Average velocity

$\to \dfrac{\text{Average speed}}{\text{Average velocity}} ≥1$

Average velocity

When a body moving along straight line covers distances $s_1 , s_2 , . . . s_n$ at different velocities $v_1 , v_2, ... v_n$ then

Average velocity = $\dfrac{\text{Total displacement}}{\text{Total time}}$

$V_{av}= \dfrac{(s_1 +s_2+ , . . .+ s_n)}{(t_1 + t_2+ .. . . .+ t_n)}$

$V_{av}= \dfrac{(s_1 +s_2+ , . . .+ s_n)}{(\dfrac{s_1}{v_1} + \dfrac{s_2}{v_2}+ .. . . .+ \dfrac{s_n}{v_n})}$

$\dfrac{(s_1 +s_2+ , . . .+ s_n)}{V_{av}} = \dfrac{s_1}{v_1} + \dfrac{s_2}{v_2}+ .. . . .+ \dfrac{s_n}{v_n}$

Equal Distance

If $s_1=s_2 . . . . . =s_n$ when body covers equal distances then ,

$\dfrac{(ns)}{V_{av}} = \dfrac{s}{v_1} + \dfrac{s}{v_2}+ .. . . .+ \dfrac{s}{v_n}$

$\dfrac{(n)}{V_{av}} = \dfrac{1}{v_1} + \dfrac{1}{v_2}+ .. . . .+ \dfrac{1}{v_n}$

Average velocity is harmonic mean of individual velocities .

For two equal distance

$V_{av} = \dfrac{2v_1v_2}{v_1+v_2}$

Equal Time

If $t_1=t_2 . . . . . =t_n$ when body covers equal distances then ,

$V_{av}= \dfrac{(s_1 +s_2+ , . . .+ s_n)}{(t_1 + t_2+ .. . . .+ t_n)}$

$V_{av}= \dfrac{(v_1t_1 +v_2t_2+ , . . .+ v_nt_n)}{(t_1 + t_2+ .. . . .+ t_n)}$

$V_{av}= \dfrac{(v_1 + v_2 + ...+ v_n)}{(n)}$

For two equal time.

$V_{av} = \dfrac{v_1+v_2}{2}$

When a moving body covers two equal parts of distances with two different constant velocities $v_1$ and $v_2$ in same or any directions, then

Average speed (v) is given by

$= \dfrac{2v_1v_2}{v_1+v_2}$

But average velocity is given by

$0 ≤ \text{average speed} ≤ \dfrac{2v_1v_2}{v_1+v_2}$

When a moving body travels at two different constant velocities $v_1$ and $v_2$ for equal intervals of time, then Average speed =$\dfrac{v_1+v_2}{2}$

$\dfrac{v_1-v_2}{2} ≤ \text{average speed} ≤ \dfrac{v_1+v_2}{2}$

Acceleration

The rate of change of velocity with respect to time is called acceleration. It is a vector quantity.

$\to$ Acceleration=$\dfrac{\text{velocity}}{\text{time taken}}$

$\to$ Average acceleration = $\dfrac{\text{change in velocity}}{\text{time taken}}$

$\overrightarrow{a_{av}} = \dfrac{(\overrightarrow{v}-\overrightarrow{v_0})}{(t-t_0)} = \dfrac{∆\overrightarrow{v}}{∆t}$

It is defined for a period of time.

Instantaneous acceleration is limiting the value of average speed as the time interval tends to zero.

$\overrightarrow{a} = \displaystyle{\lim_{∆t \to 0}} \dfrac{∆\overrightarrow{v}}{∆t} = \dfrac{d\overrightarrow{v}}{dt}$

It is defined for a point of time.

Acceleration

$\to$ If the change in velocity is constant with time. It is either increasing or decreasing by an equal amount in equal interval of time. It is a uniform acceleration.

$\to$ If the change in velocity is different with time. It is either increasing or decreasing by an unequal amount in equal interval of time. It is a non-uniform acceleration.

If $\overrightarrow{s} = \overrightarrow{s}(t)$

$\to$ velocity(v) = $\dfrac{d\overrightarrow{s}}{dt}$

$\to$ acceleration(a) = $\dfrac{d\overrightarrow{v}}{dt} = \dfrac{d^2\overrightarrow{s}}{dt^2}$

$\to s \propto t^0 \to \text{Rest}$

$\to s \propto t^1 \to \text{Uniform velocity}$

$\to s \propto t^2 \to \text{Uniform acceleration}$

$\to s \propto t^3 \to \text{Variable acceleration}$

If $\overrightarrow{v} = \overrightarrow{v}(t)$

$\to$ displacement(s) = $\int \overrightarrow{v} dt$

$\to$ acceleration(a) = $\dfrac{d\overrightarrow{v}(t)}{dt}$

If $\overrightarrow{a} = \overrightarrow{a}(t)$

$\to$ velocity(v) = $\int \overrightarrow{a} dt$

$\to$ displacement(s) = $\int \overrightarrow{v} dt$

Motion with Constant Acceleration

When the motion of the body starts from rest and moves with constant acceleration 'a'. The distance traveled and velocity in nth second is given by:

$s_n = \dfrac{a}{2}(2n-1)$

$v_n = at$

$\to$ The distance traveled in successive equal interval of time are in the ratio of $1:3:5:7 ...: (2n-1)$

$i.e.$ $s_1:s_2:s_3:...:s_n = 1:3:5:...:(2n-1)$

$\to$ The distance after the end of the successive equal interval of time are in the ratio of $1:4:9:16 ...:n^2$

$i.e.$ $s_1:(s_1+s_2):(s_1+s_2+s_3):...(s_1+s_2+...+s_n) = 1:4:9:...:n^2$

$\to$ The velocity after the end of the successive equal interval of time are in the ratio of $1:2:3:4 ...:n$

$i.e.$ $v_1:v_2:v_3:...:v_n= 1:2:3:...:n$

$\to$ The time taken to reach the end of successive equal intervals of time are in the ratio of $1:\sqrt{2}:\sqrt{3}:\sqrt{4} ...:\sqrt{n}$

$i.e.$ $t_1: (t_1+t_2):(t_1+t_2+t_3):...:(t_1+t_2+t_3+...+t_n)=1:\sqrt{2}:\sqrt{3}:\sqrt{4} ...:\sqrt{n}$

$\to$ The time taken to cover successive equal intervals of time are in the ratio of $1:\sqrt{2}-1:\sqrt{3}-\sqrt{2}:\sqrt{4}-\sqrt{3} ...:\sqrt{n}-\sqrt{n-1}$

$i.e.$ $t_1: t_2:t_3:...:t_n=1:\sqrt{2}-1:\sqrt{3}-\sqrt{2}:\sqrt{4}-\sqrt{3} ...:\sqrt{n}-\sqrt{n-1}$

When the motion of the body starts from rest and moves with constant acceleration 'a'. If $v_1,v_2,v_3,...,v_n$ be the velocities at successive points at equal separation x, then

$\to$ $v_2^2 -v_1^2 = v_3^2 - v_2^2 = v_4^2 - v_3^2 =...= v_n^2 - v_{n-1}^2= 2ax$

When the motion of the body starts from rest and moves with constant acceleration 'a'. If $v_1,v_2,v_3,...,v_n$ be the velocities at equal time interval t, then

$\to$ $v_2 -v_1= v_3- v_2 = v_4- v_3 =...= v_n - v_{n-1}= at$

When a particle accelerates from rest for time $t_1$ over distance $x_1$ and at the acceleration $a_1$ and then retards to rest at the rate of $a_2$ over distance $x_2$ in time $t_2$, then

$\to \dfrac{a_1}{a_2} = \dfrac{t_2}{t_1} = \dfrac{x_2}{x_1}$

$\to$ Net acceleration, (a) = $\dfrac{a_1a_2}{a_1+a_2}$

$\to$ Maximum velocity ($v_{max}$) = $a_1t_1= a_2t_2 = \sqrt{2a_1x_1} = \sqrt{2a_2x_2} = \sqrt{\dfrac{2a_1a_2(x_2+x_2)}{a_1+a_2}} = \dfrac{a_1a_2}{a_1+a_2} \times \text{total time}$

$\to$ Average velocity $v_{av} = \dfrac{a_1a_2}{2(a_1+a_2)} \times (t_1+t_2)$

Motion Under Gravity

A body is released from 'H' under the influence of gravity takes time t to reach the ground. Then,

$\to$ The time taken to reach the ground is $t = \sqrt{\dfrac{2H}{g}}$

$\to$ The velocity with which it hits ground is $t = \sqrt{2gH} = gt$

$\to$ The distance covered in last second $h_t = \dfrac{g}{2}(2t-1)$

$\to$ The velocity of body 'h' height above ground $v_h = \sqrt{2g(H-h)}$

$\to$ The average velocity for entire motion $v_{av} = \dfrac{gt}{2} = \sqrt{\dfrac{gh}{2}}$

$\to$ The distance traveled in successive equal interval of time are in the ratio of $1:3:5:7 ...: (2n-1)$

$\to$ The distance after the end of the successive equal interval of time are in the ratio of $1:4:9:16 ...:n^2$

When a body is thrown vertically upward with initial velocity u under gravity and returns to the thrower's hand. then

$\to$ Maximum height attached:

$H = \dfrac{u^2}{2g}$

$\to$ Time taken to reach the maximum height

$t= \dfrac{u}{g} = \sqrt{\dfrac{2H}{g}}$

$\to$ Time of flight

$T= 2t = 2 \times \sqrt{\dfrac{2H}{g}} = \sqrt{\dfrac{8H}{g}}$

$\to$ Velocity of the body at height 'h' above the ground

$v_h = \sqrt{2g(H-h)} = u \sqrt{1 - \dfrac{h}{H}}$

$\to$ Average speed for entire trip in upward motion: $v_{av} = \dfrac{u}{2} = \sqrt{\dfrac{gH}{2}}$

$\to$ Distance covered in last second of upward motion is: $h_t = u- \dfrac{g}{2}(2t-1) = \dfrac{g}{2}$

The body passes a point at height twice after time $t_1$ and $t_2$ from starting then

$\to$ Time to reach maximum height $t = \dfrac{t_1+t_2}{2}$

$\to$ Total time of flight $T =t_1+t_2$

$\to$ Height of that point from the ground, $h = \dfrac{g}{2} t_1t_2$

$\to$ Maximum height attained, $H = \dfrac{g}{8} (t_1+t_2)^2$

$\to$ Initial velocity, $(u) = \dfrac{g}{2}(t_1+t_2)$

When a body of mass 'm' is released from rest from height 'h' along a smooth inclined plane having inclination angle $\theta$ and length 'l', then

$\to$ Velocity of the body at the bottom of inclined plane $v = \sqrt{2gh} = \sqrt{2gl \sin \theta}$

$\to$ Acceleration down the plane $a = g \sin \theta$

$\to$ Time taken to fall down the plane $t = \sqrt{\dfrac{2h}{g}} \cosec \theta$

$\to$ Time taken to slide down along different smooth inclined planes of different inclination from same height are in ratio $t_1:t_2:... = \cosec \theta_1: \cosec \theta_2: ...$

$\to$ Time taken to slide down along different smooth inclined planes of equal length of different inclination are in the ratio $t_1:t_2:... = \sqrt{\cosec \theta_1}: \sqrt{\cosec \theta_2}: ...$

Relative Velocity

Introduction
Let velocity of body A be $\overrightarrow{v_A}$ and the velocity of body B be $\overrightarrow{v_B}$.
$\to$ The relative velocity of A with respect to be is denoted by $\overrightarrow{v_{AB}}$ and given by:
$\overrightarrow{v_{AB}} =\overrightarrow{v_A}-\overrightarrow{v_B}$
$\to$ It is the velocity with which body A appears to be moving while taking the body B at rest.
$\to$ Magnitude: If the angle between $\overrightarrow{v_{A}}$ and $\overrightarrow{v_{B}}$ be $\theta$.
$|\overrightarrow{v_{AB}}| = \sqrt{v_A^2 - 2v_Av_B \cos \theta + v_B^2}$
$\to$ Direction: If the $\alpha$ be angle made by $\overrightarrow{v_{AB}}$ with $\overrightarrow{v_{A}}.$
$\alpha = \tan^{-1} \biggr(\dfrac{v_B \sin \theta}{v_A - v_B \cos \theta}\biggr)$
The velocity of body B with respect to body A is denoted by $\overrightarrow{v_{BA}}$ and given by:
$\overrightarrow{v_{BA}} =\overrightarrow{v_B}-\overrightarrow{v_A}$

Crossing of River with the shortest path:

$\overrightarrow{v_r}$ is the velocity of the river, $\overrightarrow{v_s}$ is the velocity of the swimmer, and d is the width of the river.

$\to$ To reach the exact opposite point of river i.e. cross river with the shortest path, the swimmer should approach with angle $90^{\circ} + \theta = 90^{\circ} + \sin^{-1} \dfrac{v_r}{v_s}$

$\to$ The magnitude of velocity of swimmer with respect to the river is: $\sqrt{v_s^2 - v_r^2}$

$\to$ The time taken by swimmer to reach opposite bank with shortest path is: $t = \dfrac{d}{\sqrt{v_s^2 - v_r^2}}$

Crossing of River with least time:

$\to$ To reach the opposite side of the river with the least time, the swimmer should approach with angle $90^{\circ}$

$\to$ The magnitude of velocity of swimmer across river is: $\sqrt{v_s^2 + v_r^2}$

$\to$ The least time taken by swimmer to reach opposite bank is: $t = \dfrac{d}{\sqrt{v_s}} = \dfrac{AC}{\sqrt{v_s^2 + v_r^2}} = \dfrac{BC}{v_r}$

Graphs of Motion

Displacement time graph:

$\to$ The slope at a point in the displacement time graph gives velocity.

$\to$ The y-intercept gives the initial displacement.

$\to$ The ordinate gives the instantaneous displacement.

$\to$ displacement time graph can never be parallel to displacement axis.

Velocity time graph:

$\to$ The slope at a point in the velocity-time graph gives acceleration.

$\to$ The y-intercept gives the initial velocity.

$\to$ The ordinate gives the instantaneous velocity.

$\to$ velocity-time graph can never be parallel to the velocity axis.

$\to$ The area under the curve on a velocity-time graph gives the displacement covered in a given interval of time.

Acceleration time graph:

$\to$ The slope at a point in the velocity-time graph gives jerk.

$\to$ The y-intercept gives the initial acceleration.

$\to$ The ordinate gives the instantaneous acceleration.

$\to$ velocity-time graph can never be parallel to the acceleration axis.

$\to$ The area under the curve on a velocity-time graph gives the change in velocity in a given interval of time.

Better You Know

When body moving with uniform velocity (uniform motion) then its acceleration is zero.

When a particle returns to its initial position, its displacement is zero, but the distance covered is not zero. It means its average velocity is zero, but average speed is not zero.

Displacement of the body in nth second is given by

$s_t = u + \dfrac{a}{2} ( 2 t - 1)$

If $s \propto t^n$

$\to$ n = 0 , then v = 0.

$\to$ n = 1 , then a = 0.

$\to$ n = 2, then ’a’ = constant.

$\to$ n > 2, then ’a’ increases with time.

$\to$ n < 0, then a decreases with time.

When a body is moving with a uniform acceleration, then its average velocity, u is initial velocity and v is final velocity. The average velocity is the arithmetic mean of two velocities.

$V_{av} =\dfrac{(u + v)}{2}$

If a body is moving with a uniform acceleration $a_1$ for time $t_1$ and with uniform acceleration $a_2$ for a time $t_2$, then its average acceleration is

$a= \dfrac{a_1t_1+ a_2t_2}{ ( t_1 +t_2 )}$

If a body is thrown upwards, it will go vertically until its vertical velocity becomes zero.

If a body A moving with velocity u and body B moving with velocity v, the velocity of body A relative to ground = u + v, which is in the direction of motion of bodies and is equal to (u - v) in the direction of motion of body A, if they are moving in opposite directions.

When a body is in equilibrium, its acceleration is zero.

The velocity and acceleration of a body may not necessarily be in the same direction. Similarly, the velocity and acceleration of a body may not be zero simultaneously.

When the location of a body has changed it must have covered some distance and undergo some displacement.

When a body is thrown upwards. At highest point, the velocity of the body becomes zero but acceleration is not zero.

For motion in semi-circle, distance = πr and displacement = 2r and For complete revolution in a circle, distance = 2πr and displacement = 0.

If ’$s_0$' is the distance at which a car can be stopped when initially it was moving with speed ’u’ then on making the speed of the car $‘nu’$, the distance at which the car can be stopped $s = n^2s_0$

Speedometer measures instantaneous velocity.

A person is throwing balls into the other. He throws the second ball when the first ball is at the highest point. If he throws ‘n’ balls every second, the height to which each ball will rise is $\dfrac{g}{2n^2}$

Two ends of a train moving with a constant acceleration pass a certain point with velocities ’u’ and ‘v’. The velocity a the middle point of train is $\sqrt{\dfrac{u^2+v^2}{2}}$

A person standing on an escalator takes time $t_1$ to reach the top of a tower when the escalator is moving. He takes time $t_2$ to reach the top of the tower when the escalator is standing. If he walks up an escalator , then the time taken by him to reach the top is given by

$t = \dfrac{t_1t_2}{( t_1+t_2 )}$

A body falls freely under gravity from the top of a tower. It is stopped in the middle after time $t_1$ and then allowed to fall again. It reaches the ground again after time $t_2$. If it was not stopped in the middle, it would reach the ground in time t. Then $t^2 = t_1^2 + t_2^2$

A body starts from rest and moves with uniform acceleration. The ratio of distance covered in the nth sec to distance covered in n sec is $\dfrac{2}{n} – \dfrac{1}{n^2}$

A train is moving at constant speed v when its driver sights another train in front of him on the same track and moving in the same direction with constant speed u. If the distance between the two trains is x, then the minimum retardation of the train so as to avoid collision is $\dfrac{(v-u)^2}{2x}$

Similarly, the minimum time required to avoid collision is $\dfrac{x}{v-u}$

If man and rain are in relative motion the man can protect himself from the rain , if he holds his umbrella in the direction of relative velocity of rain with respect to man .

A car is moving on a road when rain is falling vertically downward. Rain will strike the front screen.

If displacement covered by a body in certain time is zero then distance may be zero but if the displacement covered by a moving body is zero then distance is never zero.

A simple pendulum hangs from the roof of a train. If the string is inclined towards the rear of the train. The nature of the motion of the train is accelerated.

If the distance time graph of a moving particle is parabola, then it has constant acceleration.

If the location of a particle has changed then displacement and the distance covered by the particle neither can be zero.

A ball is thrown up it reaches a maximum height and then comes back. If $t_1$ and $t_2$ are times that the ball takes to be at a particular height, the time taken by the ball to reach height point is $\dfrac{(t_2+t_1)}{2}$

When a body is thrown vertically upward through the air with initial velocity u and fall down then taking air resistance into account

$\to$ Acceleration of the body while moving up

$a= \dfrac{(F_g+F_v)}{m}$

($F_g$ = wt. of body, $F_v$ = viscous force)

So, $a = g + (\dfrac{6πηr}{m}) v > g$

Hence acceleration of the body moving upward through the air or any viscous medium is always more than g and is gradually decreasing. At the highest point, it is equal to g.

$\to$ Acceleration of the body while moving down is given by

$a= \dfrac{(F_g- F_v)}{m}$

So, $a = g - (\dfrac{6πηr}{m}) v < g$

Hence acceleration of a body moving downward through the air (medium) always less than g and is gradually decreasing.

$\to t_{\text{ascent}} < \dfrac{u}{ g} < t_{\text{descent}}$

$\to$ $H_{max} < \dfrac{u^2}{2g}$