Describing Motion Around Us — Notes

Motion in a Straight Line

Describing Position

To describe the motion of an object, we first need to specify its position. This is done by choosing a reference point (or origin) and stating the object's distance and direction from this point. The position of an object is defined by its distance and direction relative to a fixed reference point.

• An object is in motion if its position changes with respect to the reference point over time.
• An object is at rest if its position does not change with respect to the reference point over time.

For motion in a straight line, the direction can be indicated by plus (+) and minus (-) signs. Positions in one direction (e.g., to the right of the origin) are typically taken as positive, and in the opposite direction (e.g., to the left of the origin) as negative.

Distance Travelled and Displacement

• Distance travelled is the total length of the path covered by an object during its motion. It is a scalar quantity, meaning it only has magnitude and no direction.
• Displacement is the net change in the position of an object between two instants of time. It is a vector quantity, meaning it has both magnitude and direction. The magnitude of displacement is the straight-line distance between the initial and final positions, and its direction is from the initial position to the final position.

Key Points:

• The SI unit for both distance travelled and displacement is the metre (m).
• If an object moves without changing its direction, the magnitude of its displacement is equal to the distance travelled.
• If an object changes its direction of motion, the magnitude of its displacement is always less than or equal to the distance travelled.
• Displacement can be zero even if the distance travelled is not zero (e.g., if an object returns to its starting point).

Example: An athlete starts at point O (0 m), runs to point A (100 m), and then runs back to point B (40 m).

• Distance travelled: OA + AB = 100 m + 60 m = 160 m.
• Displacement: The net change in position from O (0 m) to B (40 m) is 40 m in the positive direction.

Average Speed and Average Velocity

• Average speed is defined as the total distance travelled by an object divided by the total time interval during which the distance was covered.

Average speed is a scalar quantity.

• Average velocity is defined as the displacement of an object divided by the time interval in which the displacement occurred.

Average velocity is a vector quantity, having both magnitude and direction. The direction of average velocity is the same as the direction of the displacement.

Comparison:

• Uniform motion in a straight line occurs when an object travels equal distances in equal time intervals (constant speed).
• Non-uniform motion occurs when an object travels unequal distances in equal time intervals (speed changes).

Example: A swimmer swims 50 m across a pool and then swims back to the starting point in 50 seconds.

• Total distance travelled: 50 m + 50 m = 100 m
• Displacement: 0 m (since the swimmer returns to the start)
• Average speed: $\frac{100\text{ m}}{50\text{ s}}=2\text{ m/s}$
• Average velocity: $\frac{0\text{ m}}{50\text{ s}}=0\text{ m/s}$

Average Acceleration

• Acceleration is the rate at which the velocity of an object changes.
• Average acceleration is defined as the change in velocity divided by the time interval in which the change occurred.

where $u$ is the initial velocity, $v$ is the final velocity, and $t$ is the time interval.

Key Points:

• Acceleration is a vector quantity. Its direction is the same as the direction of the change in velocity.
• If the velocity is increasing, acceleration is in the same direction as the velocity.
• If the velocity is decreasing (deceleration or retardation), acceleration is in the opposite direction to the velocity.
• The SI unit for acceleration is metres per second squared (m/s² or m s⁻²).
• Acceleration can be caused by a change in the magnitude (speed) of velocity, the direction of velocity, or both.

Example: A car moving at 10 m/s applies brakes and comes to a stop (0 m/s) in 5 seconds.

• Initial velocity ($u$) = 10 m/s
• Final velocity ($v$) = 0 m/s
• Time interval ($t$) = 5 s
• Average acceleration ($a$) = $\frac{0\text{ m/s}-10\text{ m/s}}{5\text{ s}}=\frac{-10\text{ m/s}}{5\text{ s}}=-2{\text{ m/s}}^2$ The negative sign indicates that the acceleration is opposite to the direction of motion, meaning the car is slowing down.

Remember

For motion in a straight line without changing direction, distance travelled is a scalar, and displacement is a vector. Speed is the magnitude of velocity. 🎯 Exam-smart: Questions often involve calculating distance, displacement, average speed, and average velocity for scenarios where an object moves back and forth. Pay close attention to whether the question asks for distance or displacement, as this significantly impacts the answer. Also, be careful with units (km/h vs. m/s) and conversions.

Graphical Representation of Motion

Graphs offer a powerful visual method to depict motion, showing how physical quantities change over time. This allows for comparison of motions, calculation of physical quantities, and identification of motion types (uniform or non-uniform).

Plotting a Graph

Plotting a graph involves representing data from a table on graph paper.

Steps:

1. Obtain graph paper: Use pre-divided squares for accuracy.
2. Draw axes: Draw two perpendicular lines, the X-axis (horizontal) and the Y-axis (vertical). Their intersection is the origin (O).
3. Assign quantities to axes: Decide which quantity goes on the X-axis (usually independent variable, like time) and which on the Y-axis (usually dependent variable, like position).
4. Choose scales: Select appropriate scales for each axis to conveniently represent the data range and utilize graph paper space. For example, on the X-axis, 5 divisions might represent 1 second; on the Y-axis, 5 divisions might represent 20 meters.
5. Mark values: Use the chosen scales to mark numerical values along each axis starting from the origin.
6. Plot points: For each data pair (e.g., time and position), locate the value on the X-axis, draw a line parallel to the Y-axis, and simultaneously locate the corresponding value on the Y-axis and draw a line parallel to the X-axis. The intersection of these two lines is the plotted point.
7. Join points: Connect the plotted points. This could form a straight line (for uniform motion) or a curve (for non-uniform motion).

Example: For data with time (0s, 20m), (1s, 40m), (2s, 60m)...

• X-axis: Time (e.g., 5 divisions = 1s)
• Y-axis: Position (e.g., 5 divisions = 20m)
• Plot (0,0), (1,20), (2,40) and connect them.

Remember

The chosen scale should be consistent throughout the graph and allow for clear representation of all data points.

Exam-smart

You might be given a table of values and asked to plot a specific graph, or interpret a given graph to extract values or determine the type of motion.

Position-Time Graphs

A position-time graph shows how an object's position changes with respect to time.

• Straight Line: If the position-time graph is a straight line, it indicates that the object is moving with constant velocity. This is because for equal time intervals, the object covers equal displacements.
• Curved Line: A curved position-time graph signifies that the object's velocity is not constant. This means the object is undergoing accelerated motion.

Calculating Velocity from a Position-Time Graph:

The velocity of an object can be determined from the slope of its position-time graph. For any two points A $(t_1,s_1)$ and B $(t_2,s_2)$ on the graph:

This calculated value is the average velocity over the time interval $(t_2-t_1)$. Geometrically, this is the slope of the line segment connecting A and B.

• Stationary Object: A horizontal line (parallel to the time axis) on a position-time graph represents an object that is at rest at a particular position. Its position does not change with time.

Interpreting Graphs:

• A steeper slope on a position-time graph indicates a higher velocity.
• If two objects' position-time graphs are plotted, the one with the steeper slope has a higher magnitude of average velocity.

Remember

The slope of a position-time graph represents velocity. A positive slope means moving away from the origin, a negative slope means moving towards the origin, and zero slope means being stationary.

Exam-smart

Questions may provide a position-time graph and ask to find the velocity of an object at a specific time (by calculating the slope of a tangent if it's a curve, though for Class 9 it's usually a straight line) or compare the velocities of multiple objects.

Velocity-Time Graphs

A velocity-time graph illustrates how an object's velocity changes over time.

Types of Velocity-Time Graphs and Interpretation:

Calculating Acceleration from a Velocity-Time Graph:

The acceleration of an object can be determined from the slope of its velocity-time graph. For any two points A $(t_1,u)$ and B $(t_2,v)$ on the graph, where $u$ is initial velocity and $v$ is final velocity:

This value represents the average acceleration over the time interval $(t_2-t_1)$.

• A positive slope indicates acceleration in the direction of motion.
• A negative slope indicates acceleration opposite to the direction of motion (deceleration).

Calculating Displacement from a Velocity-Time Graph:

The displacement of an object can be found by calculating the area enclosed by the velocity-time graph and the time axis over a specific time interval.

• Constant Velocity: Displacement = Area of rectangle = velocity $\times$ time interval.

• Constant Acceleration: Displacement = Area of a trapezoid (or a rectangle plus a triangle).

This is equivalent to the kinematic equation $s=ut+\frac{1}{2}a{t}^2$, where $(v-u)\mathrm{/}t=a$.

Remember

The slope of a velocity-time graph gives acceleration, and the area under the velocity-time graph gives displacement.

Exam-smart

You might be given a velocity-time graph and asked to find the acceleration (by calculating the slope) or the total displacement (by calculating the area under the curve). For example, a question could ask for the distance covered by a car accelerating from rest to a certain speed in a given time.

Kinematic Equations for Motion in a Straight Line with Constant Acceleration

When an object moves with constant acceleration, its velocity changes uniformly over time. We can use a set of three kinematic equations to describe and predict its motion. These equations relate five key quantities:

• Initial velocity ($u$): The velocity of the object at the beginning of the time interval.
• Final velocity ($v$): The velocity of the object at the end of the time interval.
• Acceleration ($a$): The constant rate at which the velocity changes.
• Time interval ($t$): The duration over which the motion occurs.
• Displacement ($s$): The change in position of the object.

These equations are derived from the definitions of velocity and acceleration and are valid only when the acceleration is constant.

The Kinematic Equations

The three primary kinematic equations are:

1. Velocity-Time relation: This equation relates final velocity, initial velocity, acceleration, and time.

• How it works: It states that the final velocity ($v$) is equal to the initial velocity ($u$) plus the change in velocity due to constant acceleration ($a$) over time ($t$). The term $at$ represents the total change in velocity.
• Why it matters: This equation is useful when you know the initial velocity, acceleration, and time, and you want to find the final velocity. It also helps visualize motion: a velocity-time graph for constant acceleration is a straight line with a slope equal to the acceleration. If $u=0$, the graph starts from the origin. If $u\mathrm{\ne }0$, the graph starts at a height $u$ on the y-axis.

2. Displacement-Time relation: This equation relates displacement, initial velocity, acceleration, and time.

• How it works: It describes the displacement ($s$) as the sum of two parts: the distance the object would have traveled if its velocity remained constant at $u$ (i.e., $ut$), plus an additional term $\frac{1}{2}a{t}^2$ that accounts for the increase in distance due to acceleration.
• Why it matters: This equation is useful when you know the initial velocity, acceleration, and time, and you want to find the displacement. It also means that for constant acceleration, the displacement is not a linear function of time but rather a quadratic one.

3. Velocity-Displacement relation: This equation relates final velocity, initial velocity, acceleration, and displacement.

• How it works: This equation is derived by eliminating time ($t$) from the first two equations. It directly connects the velocities and displacement without needing to know the time taken.
• Why it matters: This equation is particularly useful when the time interval is unknown or not relevant to the problem, but you need to relate initial velocity, final velocity, acceleration, and displacement.

Using the Kinematic Equations

To effectively use these equations:

• Identify knowns and unknowns: Determine which of the five quantities ($u,v,a,t,s$) are given and which one you need to find.
• Choose the appropriate equation: Select the equation that contains the knowns and the unknown variable.
• Ensure consistent units: All quantities must be in the same system of units (e.g., meters for displacement, seconds for time, meters per second for velocity, and meters per second squared for acceleration).
• Pay attention to direction: For motion in one direction (straight line), if you choose a positive direction, quantities in that direction are positive, and those in the opposite direction are negative. This is crucial for acceleration (e.g., braking causes negative acceleration) and displacement. If the object reverses direction, displacement and distance travelled might differ.

Example Application: Imagine a car starting from rest ($u=0$ m/s) and accelerating at a constant rate of $a=2$ m/s². After 10 seconds ($t=10$ s), what is its final velocity and how far has it traveled?

• To find final velocity ($v$): Use $v=u+at$

• To find displacement ($s$): Use $s=ut+\frac{1}{2}a{t}^2$

We could also use $v^2=u^2+2as$ to find displacement once we know $v$:

Remember

These equations are a powerful tool for analyzing motion, but they are strictly applicable only for constant acceleration. If acceleration changes, you need to break the motion into segments where acceleration is constant.

Exam-smart

A case study question might describe a scenario (e.g., a vehicle braking, a falling object) and provide some of the initial velocity, final velocity, acceleration, or time. You would need to identify the correct kinematic equation to calculate the missing quantity, often involving unit conversions (like km/h to m/s). Pay close attention to the wording for signs of velocity and acceleration.

Motion in a Plane

Motion that occurs in two dimensions, occupying both length and width (like on a flat surface), is called motion in a plane. This is also known as two-dimensional motion.

• Examples:
• A car moving on a road.
• A projectile, like a kicked football or a thrown ball, moving through the air.
• A satellite orbiting the Earth in a circular path.

In motion in a plane, the object’s position needs two coordinates (e.g., x and y) to be described at any given time. This is a departure from one-dimensional motion where only one coordinate was sufficient.

Uniform Circular Motion

A specific type of motion in a plane is uniform circular motion. This occurs when an object moves along a circular path at a constant speed.

• Key Characteristics:
• Constant Speed: The magnitude of the velocity (speed) remains the same.
• Changing Velocity: Even though the speed is constant, the direction of velocity is continuously changing because the object is turning along the circular path. This means there is an acceleration present, directed towards the center of the circle.
• Distance vs. Displacement: In one complete revolution, the object travels a distance equal to the circumference of the circle ($2\pi R$, where $R$ is the radius). However, its displacement is zero because it returns to its starting position.
• Average Speed vs. Average Velocity: For one revolution, the average speed is the total distance divided by the time taken ($v=\frac{2\pi R}{T}$), while the average velocity is zero because the displacement is zero.

Remember

Uniform circular motion is an example of accelerated motion because the velocity is changing (due to change in direction), even if the speed is constant.

Exam-smart

Questions might ask to differentiate between distance and displacement for an object in uniform circular motion, or to calculate average speed versus average velocity over one or more revolutions. You might also be asked to identify that although speed is constant, velocity is not.

Uniform Circular Motion

Uniform circular motion is a specific type of motion where an object travels along a circular path at a constant speed. While the speed remains the same, the direction of velocity at any given instant is continuously changing.

Consider an athlete running along different tracks.

• On a rectangular track, the athlete changes direction four times per round.
• On a hexagonal track, the athlete changes direction six times per round.
• As the number of sides of a polygon increases, the number of turns increases, and the path increasingly resembles a circle. In the limit, as the number of sides becomes infinite, the path becomes a circle, and the direction of motion changes continuously at every point.

Velocity in Uniform Circular Motion

The velocity of an object is a vector quantity that has both magnitude (speed) and direction. In uniform circular motion:

• The magnitude of velocity (i.e., the speed) is constant.
• The direction of velocity is continuously changing.

At any point on the circular path, the velocity vector is tangent to the circle at that point and points in the direction of motion.

• A tangent is a line that touches a circle at exactly one point.

Acceleration in Uniform Circular Motion

Since the direction of velocity is continuously changing in uniform circular motion, the object is accelerating. This acceleration is not due to a change in speed, but solely due to the change in direction. This acceleration is often referred to as centripetal acceleration.

Example: Marble in a Ring

Imagine rolling a marble inside a ring on a smooth surface.

1. While the marble is moving along the inside of the ring, its motion is approximately circular.
2. If you lift the ring, the marble will continue to move in a straight line along the direction it was moving at the instant the ring was removed. This demonstrates that when the constraint of the circular path is removed, the object moves in the direction of its instantaneous velocity.

Remember

In uniform circular motion, speed is constant, but velocity is not, because its direction is always changing.

Exam-smart

A question might describe an object moving in a circle with constant speed (like a stone whirled by a string) and ask about its acceleration. You need to identify that since the direction is changing, acceleration is present, even though speed is constant. You might also be asked to infer the direction of motion if a constraint is removed (like the string breaking).