Chapter Notes

Motion

In physics, motion describes a change in an object's position over time. We see it everywhere: birds flying, cars moving, and even planets orbiting the sun. An object that is not changing its position is said to be at rest.

However, motion is relative. It depends on the observer's point of view.

Motion can be simple, like moving in a straight line, or complex, like a combination of rotating and vibrating. In this chapter, we will learn how to describe motion, especially along a straight line and in a circle, using simple equations and graphs.

Describing Motion

To describe where an object is, we need a point of reference. This fixed point is called the origin. We describe the object's position relative to this origin.

Motion along a straight line

The simplest type of motion is movement along a straight line. To understand this, we use two key physical quantities: distance and displacement.

Distance is the total path length an object covers. It is a scalar quantity, meaning it only has a numerical value, or magnitude, and does not require a direction.

Displacement is the shortest distance measured from the initial position to the final position of an object. It is a vector quantity, meaning it includes both magnitude and direction.

Let's consider an object moving along a straight path. It starts at the origin O, travels to point A, and then returns to point C.

• The distance covered is the total path length: (Path OA) + (Path AC).
• The displacement is the shortest distance between the start (O) and end (C) points, which is simply the length OC.

Uniform motion and non-UNIFORM MOTION

We can classify motion based on how an object's speed changes over time.

• Uniform Motion: An object is in uniform motion if it travels equal distances in equal intervals of time. For example, a car moving at a constant 5 meters every second.
• Non-uniform Motion: An object is in non-uniform motion if it travels unequal distances in equal intervals of time. For instance, a car moving in city traffic or a person jogging in a park will speed up and slow down.

Measuring the Rate of Motion

To measure how fast or slow an object is moving, we use the concept of speed.

Speed is the distance an object travels per unit of time.

• The SI unit for speed is metre per second ($\text{m s}^{-1}$ or m/s).
• Other common units include centimetre per second ($\text{cm s}^{-1}$) and kilometre per hour ($\text{km h}^{-1}$).

The formula for speed is: $v = \frac{s}{t}$ where $v$ is speed, $s$ is distance, and $t$ is time.

In most real-world situations, objects are in non-uniform motion. For these cases, we describe the rate of motion using average speed.

Average speed is the total distance travelled divided by the total time taken. $\text{average speed} = \frac{\text{Total distance travelled}}{\text{Total time taken}}$

Given

• Distance 1, $s_1 = 16$ m
• Time 1, $t_1 = 4$ s
• Distance 2, $s_2 = 16$ m
• Time 2, $t_2 = 2$ s

To Find

The average speed of the object.

Formula

$\text{Average speed} = \frac{\text{Total distance travelled}}{\text{Total time taken}}$

Solution

Total distance travelled = $16 \text{ m} + 16 \text{ m} = 32 \text{ m}$ Total time taken = $4 \text{ s} + 2 \text{ s} = 6 \text{ s}$

Substitute the values into the formula: $\text{Average speed} = \frac{32 \text{ m}}{6 \text{ s}} = 5.33 \text{ m s}^{-1}$

Final Answer The average speed of the object is $5.33 \text{ m s}^{-1}$.

Speed with direction

To provide a more complete description of motion, we need to specify its direction along with its speed. This quantity is called velocity.

Velocity is the speed of an object moving in a definite direction.

• The velocity of an object can be uniform or variable. It can be changed by changing the object's speed, its direction of motion, or both.
• The units for speed and velocity are the same ($\text{m s}^{-1}$).

When an object's velocity is changing at a uniform rate, its average velocity is the arithmetic mean of its initial and final velocities. $\text{average velocity} = \frac{\text{initial velocity} + \text{final velocity}}{2}$ Mathematically, this is written as: $v_{av} = \frac{u+v}{2}$ where $v_{av}$ is the average velocity, $u$ is the initial velocity, and $v$ is the final velocity.

Given

• Initial odometer reading = 2000 km
• Final odometer reading = 2400 km
• Time elapsed, $t = 8$ h

To Find

The average speed in $\text{km h}^{-1}$ and $\text{m s}^{-1}$.

Formula

$v_{av} = \frac{s}{t}$

Solution

First, calculate the total distance covered: Distance, $s = 2400 \text{ km} - 2000 \text{ km} = 400 \text{ km}$

Now, calculate the average speed in $\text{km h}^{-1}$: $v_{av} = \frac{400 \text{ km}}{8 \text{ h}} = 50 \text{ km h}^{-1}$

To convert this to $\text{m s}^{-1}$, we use the conversion factors $1 \text{ km} = 1000 \text{ m}$ and $1 \text{ h} = 3600 \text{ s}$: $50 \frac{\text{km}}{\text{h}} \times \frac{1000 \text{ m}}{1 \text{ km}} \times \frac{1 \text{ h}}{3600 \text{ s}} = 13.9 \text{ m s}^{-1}$

Final Answer The average speed of the car is $50 \text{ km h}^{-1}$ or $13.9 \text{ m s}^{-1}$.

Given

• Total distance covered = 180 m
• Total time taken = 1 minute = 60 s
• Length of pool = 90 m

To Find

Average speed and average velocity.

Formula

$\text{Average speed} = \frac{\text{Total distance covered}}{\text{Total time taken}}$ $\text{Average velocity} = \frac{\text{Displacement}}{\text{Total time taken}}$

Solution

Average Speed: $\text{Average speed} = \frac{180 \text{ m}}{60 \text{ s}} = 3 \text{ m s}^{-1}$

Average Velocity: Usha starts at one end and swims to the other and back. Her final position is the same as her initial position. Therefore, her displacement is 0 m. $\text{Average velocity} = \frac{0 \text{ m}}{60 \text{ s}} = 0 \text{ m s}^{-1}$

Final Answer The average speed of Usha is $3 \text{ m s}^{-1}$ and her average velocity is $0 \text{ m s}^{-1}$.

Rate of Change of Velocity

In non-uniform motion, velocity changes with time. The rate at which velocity changes is described by another physical quantity called acceleration.

Acceleration is the change in the velocity of an object per unit time. $\text{acceleration} = \frac{\text{change in velocity}}{\text{time taken}}$ If an object's velocity changes from an initial value $u$ to a final value $v$ in time $t$, the acceleration $a$ is: $a = \frac{v-u}{t}$

• The SI unit of acceleration is metre per second squared ($\text{m s}^{-2}$).
• Acceleration is positive if it is in the direction of velocity (speeding up).
• Acceleration is negative if it is opposite to the direction of velocity (slowing down).

Uniform Acceleration: The acceleration is uniform if an object's velocity changes by equal amounts in equal intervals of time. A freely falling body is an example of uniformly accelerated motion.

Non-uniform Acceleration: The acceleration is non-uniform if the velocity changes at a non-uniform rate.

Given

Case 1:

• Initial velocity, $u = 0 \text{ m s}^{-1}$ (stationary)
• Final velocity, $v = 6 \text{ m s}^{-1}$
• Time, $t = 30$ s

Case 2:

• Initial velocity, $u = 6 \text{ m s}^{-1}$
• Final velocity, $v = 4 \text{ m s}^{-1}$
• Time, $t = 5$ s

To Find

The acceleration, $a$, in both cases.

Formula

$a = \frac{v-u}{t}$

Solution

(i) First Case (Speeding up) $a = \frac{6 \text{ m s}^{-1} - 0 \text{ m s}^{-1}}{30 \text{ s}} = \frac{6}{30} \text{ m s}^{-2} = 0.2 \text{ m s}^{-2}$ Answer for part (i) = $0.2 \text{ m s}^{-2}$

(ii) Second Case (Slowing down) $a = \frac{4 \text{ m s}^{-1} - 6 \text{ m s}^{-1}}{5 \text{ s}} = \frac{-2}{5} \text{ m s}^{-2} = -0.4 \text{ m s}^{-2}$ Answer for part (ii) = $-0.4 \text{ m s}^{-2}$

Final Answer The acceleration of the bicycle in the first case is $0.2 \text{ m s}^{-2}$ and in the second case, it is $-0.4 \text{ m s}^{-2}$.

Graphical Representation of Motion

Graphs are a powerful tool for visualizing and understanding motion.

Distance-Time Graphs

In a distance-time graph, time is plotted on the x-axis and distance on the y-axis.

• Uniform Speed: The graph is a straight line. The steepness (slope) of the line indicates the speed. A steeper line means a higher speed. The speed can be calculated from the slope: $v = \frac{s_2 - s_1}{t_2 - t_1}$
• Non-uniform Speed: The graph is a curved line. This indicates that the object is accelerating.
• Object at Rest: The graph is a straight line parallel to the time axis, as the distance does not change.

Velocity-Time Graphs

In a velocity-time graph, time is plotted on the x-axis and velocity on the y-axis.

• Uniform Velocity (Constant Velocity): The graph is a straight line parallel to the time axis.
• Uniform Acceleration: The graph is a straight line with a constant slope.
• Non-uniform Acceleration: The graph can be any curved shape.

Equations of Motion

For an object moving along a straight line with uniform acceleration, we can use a set of three equations to relate its velocity, acceleration, distance, and time. These are known as the equations of motion.

1. Velocity-Time Relation: $v = u + at$
2. Position-Time Relation: $s = ut + \frac{1}{2}at^2$
3. Position-Velocity Relation: $2as = v^2 - u^2$

Where:

• $u$ = initial velocity
• $v$ = final velocity
• $a$ = uniform acceleration
• $t$ = time
• $s$ = distance travelled

Given

• Initial velocity, $u = 0$ (starts from rest)
• Final velocity, $v = 72 \text{ km h}^{-1} = 20 \text{ m s}^{-1}$
• Time, $t = 5 \text{ minutes} = 300 \text{ s}$

To Find

(i) The acceleration, $a$ (ii) The distance travelled, $s$

Formula

$a = \frac{v-u}{t}$ $2as = v^2 - u^2$

Solution

(i) Calculate the acceleration $a = \frac{20 \text{ m s}^{-1} - 0 \text{ m s}^{-1}}{300 \text{ s}} = \frac{20}{300} \text{ m s}^{-2} = \frac{1}{15} \text{ m s}^{-2}$ Answer for part (i) = $\frac{1}{15} \text{ m s}^{-2}$

(ii) Calculate the distance travelled Using the third equation of motion: $2as = v^2 - u^2$ $s = \frac{v^2 - u^2}{2a} = \frac{(20 \text{ m s}^{-1})^2 - 0^2}{2 \times (\frac{1}{15}) \text{ m s}^{-2}}$ $s = \frac{400}{2/15} \text{ m} = 400 \times \frac{15}{2} \text{ m} = 3000 \text{ m} = 3 \text{ km}$ Answer for part (ii) = $3$ km

Final Answer The acceleration of the train is $\frac{1}{15} \text{ m s}^{-2}$ and the distance travelled is 3 km.

Given

• Initial velocity, $u = 18 \text{ km h}^{-1} = 5 \text{ m s}^{-1}$
• Final velocity, $v = 36 \text{ km h}^{-1} = 10 \text{ m s}^{-1}$
• Time, $t = 5$ s

To Find

(i) The acceleration, $a$ (ii) The distance covered, $s$

Formula

$a = \frac{v-u}{t}$ $s = ut + \frac{1}{2}at^2$

Solution

(i) Calculate the acceleration $a = \frac{10 \text{ m s}^{-1} - 5 \text{ m s}^{-1}}{5 \text{ s}} = \frac{5}{5} \text{ m s}^{-2} = 1 \text{ m s}^{-2}$ Answer for part (i) = $1 \text{ m s}^{-2}$

(ii) Calculate the distance covered $s = (5 \text{ m s}^{-1})(5 \text{ s}) + \frac{1}{2}(1 \text{ m s}^{-2})(5 \text{ s})^2$ $s = 25 \text{ m} + \frac{1}{2}(1)(25) \text{ m}$ $s = 25 \text{ m} + 12.5 \text{ m} = 37.5 \text{ m}$ Answer for part (ii) = $37.5$ m

Final Answer The acceleration of the car is $1 \text{ m s}^{-2}$ and the distance covered is 37.5 m.

Given

• Acceleration, $a = -6 \text{ m s}^{-2}$ (opposite direction)
• Time, $t = 2$ s
• Final velocity, $v = 0 \text{ m s}^{-1}$ (car stops)

To Find

The distance the car travels, $s$.

Formula

$v = u + at$ $s = ut + \frac{1}{2}at^2$

Solution

First, we need to find the initial velocity ($u$) before the brakes were applied. Using the first equation of motion: $0 = u + (-6 \text{ m s}^{-2})(2 \text{ s})$ $0 = u - 12 \text{ m s}^{-1}$ $u = 12 \text{ m s}^{-1}$

Now, we can calculate the distance travelled using the second equation of motion: $s = (12 \text{ m s}^{-1})(2 \text{ s}) + \frac{1}{2}(-6 \text{ m s}^{-2})(2 \text{ s})^2$ $s = 24 \text{ m} + \frac{1}{2}(-6)(4) \text{ m}$ $s = 24 \text{ m} - 12 \text{ m} = 12 \text{ m}$

Final Answer The car travels 12 m before it stops after the application of brakes.

Uniform Circular Motion

When an object moves in a circular path with uniform speed, its motion is called uniform circular motion.

An object is accelerating if its velocity is changing. Velocity includes both speed and direction. In uniform circular motion, the speed is constant, but the direction of motion is continuously changing at every point on the circle. Because the velocity is changing, uniform circular motion is an example of accelerated motion.

If an athlete takes $t$ seconds to go once around a circular path of radius $r$, the speed $v$ is given by the circumference divided by the time: $v = \frac{2 \pi r}{t}$

Examples of uniform circular motion include:

• The motion of the Moon and the Earth.
• An artificial satellite orbiting the Earth.
• A cyclist on a circular track moving at a constant speed.