Describing Motion Around Us Class 9 PPT & Notes (Ch 4)

1. Motion in a straight line: reference point and position, distance and displacement, speed and velocity, acceleration, free fall (4.1)
2. Graphical representation of motion: position-time and velocity-time graphs (4.2)
3. Kinematic equations of motion (4.3)
4. Motion in a plane and uniform circular motion (4.4)

Think it over: “keep a safe distance” on the road. Does the safe distance depend on the speed at which we are moving? If a truck suddenly applies brakes, how much distance should we keep to avoid a collision? This chapter helps you answer these questions using physics.

• A reference point (the origin) is needed to describe where an object is. It can be anything: a start line, a tree or your home
• Position = the distance from the reference point + the direction towards it. Without a reference point, position is meaningless
• An object is at rest if its position does not change with time with respect to the reference point, and in motion if its position changes

Rest and motion are relative: a passenger in a moving train is at rest with respect to a co-passenger, but in motion with respect to a person on the platform or the trees outside. A book on a table is at rest with respect to the table, but moves with the Earth around the Sun at about 30 km/s. There is no absolute rest or absolute motion.

Distance ≥ |displacement| always. They are equal only when the object moves in one direction without turning back.

An athlete runs from O to A (100 m), turns back and stops at B (40 m from O). Total distance = OA + AB = 100 + 60 = 160 m. Displacement = 40 m (in the positive direction).

As the ball returns, the distance keeps increasing but the displacement decreases back to zero. The magnitude of displacement is less than or equal to the total distance; the displacement is never greater than the distance.

• If the athlete runs from O to A and back to O, the displacement is zero (start and end at the same point), but the distance is 100 + 100 = 200 m
• Father’s shop round trip: 4 trips of 250 m each: distance = 250 + 250 + 250 + 250 = 1000 m = 1 km, but the net change in position is 0 m, so the displacement from home is zero
• A car that goes 10 km from home to market and back: distance = 10 km, displacement = 0 km. Fuel consumption depends on the distance (the total path), the odometer records the distance, and the net change in position is the displacement

A lift goes from the ground floor (0 m) up to the 4th floor (12 m) and then down to the 2nd floor (6 m). Step 1: total distance = 12 m + 6 m = 18 m. Step 2: displacement = final position − initial position = 6 m − 0 m = 6 m upward.

Distance is a scalar and displacement is a vector. Speed is a scalar and velocity is a vector.

An instant of time is a single reading of the clock at one moment (like a photograph), for example t = 3 s. A time interval is the duration between two instants, (t₂ − t₁) (like a video), for example from t = 0 s to t = 10 s the interval is 10 s. Position and displacement are measured at an instant, whereas average speed, velocity and acceleration are calculated over a time interval.

Uniform motion is equal distances in equal time intervals. Non-uniform motion is unequal distances in equal time intervals, so the speed is changing. In uniform motion the position-time graph is a straight line.

km/h ↔ m/s: 1 km/h = 1000 m ÷ 3600 s = 5/18 m/s. Multiply by 5 and divide by 18 to go from km/h to m/s, and multiply by 18 and divide by 5 to go from m/s to km/h. For example, 18 km/h = 5 m/s, 36 km/h = 10 m/s, 54 km/h = 15 m/s, 72 km/h = 20 m/s, 90 km/h = 25 m/s and 108 km/h = 30 m/s.

The Aryabhatiya (5th century CE) gives speed = distance ÷ time. The Ganitakaumudi by Narayana Pandita (14th century CE) has a classic problem: two postmen 210 yojanas apart walk towards each other, one at 9 yojanas a day and the other at 5. Combined speed = 9 + 5 = 14 yojanas/day, and time = 210 ÷ 14 = 15 days. In 15 days the first covers 135 yojanas and the second 75 yojanas, so they meet.

• Velocity has direction. If the displacement is zero, the velocity is zero even if the speed is not
• Swimming pool: Sarang swims one length (25 m) and back in 50 s. Total distance = 50 m and displacement = 0 m, so average speed = 50 ÷ 50 = 1 m/s but average velocity = 0 ÷ 50 = 0 m/s
• Road trip: 200 km North in 3 h, then 200 km South in 2 h. Total distance = 400 km, total time = 5 h, so average speed = 400 ÷ 5 = 80 km/h, but the displacement is 0, so the average velocity = 0 km/h
• |velocity| = speed only when the object moves in one direction without turning back (for example 100 m East in 10 s gives 10 m/s)

The speedometer reading is the magnitude of the velocity at that instant. Taking shrinking time intervals (Δt → 0), the average velocity approaches the velocity at one instant. A long trip can have an average velocity of 40 km/h, but the velocity at one instant can be very different.

• Acceleration is the rate of change of velocity: a = (v − u) ÷ t, where u is the initial velocity, v is the final velocity and t is the time interval
• SI unit = m/s². It is a vector
• Positive acceleration: velocity increasing (acceleration in the same direction as the velocity). Negative acceleration (retardation or deceleration): velocity decreasing (acceleration opposite to the velocity)
• When a bus brakes, you feel pushed forward (acceleration opposite to the motion); when it starts, you feel pushed back

Taking 100 km/h = 27.78 m/s, the average acceleration from 0 to 100 km/h is: an economy car (about 15 s) 1.85 m/s², a family sedan (about 10 s) 2.78 m/s², a sports car (about 3.5 s) 7.94 m/s² and a Formula 1 race car (about 2.6 s) about 10.7 m/s². The Formula 1 car has the highest acceleration, which is why fast cars feel more “pushy” when they accelerate.

With constant acceleration, the velocity increases by an equal amount in every equal time interval. In non-constant acceleration the change is different in each interval (for example +2, +5, +1, +8 m/s). In this chapter we study only constant acceleration cases.

Dropped from rest (u = 0), the velocity increases by 9.8 m/s every second: (9.8 − 0) ÷ 1 = (19.6 − 9.8) ÷ 1 = (29.4 − 19.6) ÷ 1 = (39.2 − 29.4) ÷ 1 = 9.8 m/s². This constant acceleration, directed downward, is the acceleration due to gravity, g = 9.8 m/s².

A scooter on a curved road with the speedometer locked at 30 km/h has constant speed, but its direction keeps changing, so its velocity keeps changing and the rate of change of velocity is not zero. Acceleration exists even though the speed is constant. On a straight road with constant speed, acceleration = 0; on a curved road with constant speed, acceleration ≠ 0 because the direction of the velocity changes.

For a ball thrown upward, you may take the origin at the ground with upward as positive (and downward as negative), or choose another origin. The choice is yours, but once chosen, do not change it during the same problem.

The graphs show how position or velocity changes with time, give a visual picture of the entire journey, help to identify uniform versus non-uniform motion, and make it easy to compare the motion of two objects.

Draw the X-axis (time) and Y-axis (position), mark the origin, choose a scale (for example 5 divisions = 1 s and 5 divisions = 20 m), mark the values on both axes, plot each point, and join the points. For the data Time (s): 0, 1, 2, 3, 4, 5, 6 and Position (m): 0, 20, 40, 60, 80, 100, 120, the points lie on a straight line: uniform motion at 20 m/s.

Table 4.4: for time 0, 2, 4, 6, 8, 10, 12 s the position is 0, 1, 4, 9, 16, 25, 36 m. The points do not fall on a straight line, so join them with a smooth curve, not straight segments. The curve gets steeper with time: early on (0 to 4 s) the position changes slowly, and later (10 to 12 s) it changes a lot, so the speed is increasing: accelerated motion.

The slope of a line is its steepness: slope = (change in position) ÷ (change in time) = velocity. A steeper line means a higher velocity. For a curve (non-uniform motion) the slope at any point gives the velocity at that instant. Two objects have the same velocity when their lines have the same slope.

Take A (t₁ = 10 s, u = 5 m/s) and B (t₂ = 20 s, v = 10 m/s) on the graph. Slope = BC ÷ CA = (10 − 5) ÷ (20 − 10) = 5 ÷ 10 = 0.5 m/s². The slope of a p-t graph is velocity, and the slope of a v-t graph is acceleration: one level higher each time.

For a constant velocity of 20 m/s for 6 s, the area of the rectangle = 20 × 6 = 120 m. For a velocity of 5 m/s that increases to 10 m/s from t = 10 s to t = 20 s, the area = rectangle (5 × 10 = 50 m) + triangle (½ × 10 × 5 = 25 m) = 75 m. Displacement = the area under any velocity-time graph, whatever its shape. Displacement = average velocity × time.

Here u = initial velocity, v = final velocity, a = acceleration, s = displacement and t = time. These equations are valid only when the acceleration is constant. (Two more useful forms: s = vt − ½at² when the initial velocity is not given, and s = ½(u + v)t from the trapezium area.)

• First equation: from the definition a = (v − u) ÷ t, multiply both sides by t: a × t = v − u, so v = u + at. (In the v-t graph the slope is the acceleration.)
• Second equation: displacement = area under the v-t graph = a rectangle (u × t = ut) + a triangle (½ × t × (v − u) = ½ × t × at = ½at²), so s = ut + ½at²
• Third equation: from the graph the displacement is s = ½(u + v)t. Put t = (v − u) ÷ a to get 2as = v² − u², so v² = u² + 2as

A car brakes with acceleration −4 m/s². With v = 0, v² = u² + 2as gives 0 = u² − 8s, so s = u² ÷ 8. (i) At 54 km/h (15 m/s): s = 225 ÷ 8 ≈ 28.1 m. (ii) At 108 km/h (30 m/s): s = 900 ÷ 8 = 112.5 m. Double the speed gives four times the stopping distance, because s ∝ u².

A bus at 36 km/h (10 m/s) sees an obstacle 30 m ahead. Reaction phase (0.5 s at 10 m/s): distance = 10 × 0.5 = 5 m. Braking phase (a = −2.5 m/s²): 0 = 100 − 5s, so s = 20 m. Total = 5 + 20 = 25 m, which is less than 30 m, so the bus stops safely with a 5 m gap. Reaction time also adds to the stopping distance, so always stay alert.

• Motion in a plane is motion in two dimensions (2D), so we need two coordinates (x and y) to describe the position. Examples: a kicked ball (a projectile) and a satellite in a circular orbit
• Uniform circular motion (UCM) is motion in a circle with constant speed but changing direction
• Speed in UCM: v = 2πR ÷ T (R = radius, T = time for one revolution)
• The velocity in UCM is always along the tangent to the circle
• UCM is accelerated motion, because the direction of the velocity changes continuously

For one full revolution the distance is the circumference 2πR, and the displacement is 0 (back to the start). Average speed = 2πR ÷ T, and average velocity = 0 ÷ T = 0 m/s.

In UCM the speed is constant (the same magnitude of velocity at every point) but the velocity changes continuously because its direction keeps changing. Therefore acceleration ≠ 0 even though the speed is constant. Examples: the Earth orbiting the Sun, a fan blade tip, and a car at constant speed on a circular roundabout. UCM is an idealised model; real circular motion often has varying speed. A polygon with more and more sides becomes a circle, and then the change of direction becomes continuous.

While the ring is present, it pushes the marble inward and the marble moves in a circular path. When the ring is removed, there is no inward push and the marble continues in the direction it had at that instant: a straight line along the tangent. Objects continue in the direction of their velocity at the instant the force is removed (you will learn why in Newton’s first law).

Rohan studies from 6 PM to 7:30 PM (90 min). The minute hand has radius R = 7 cm. One revolution: distance = 2πR = 2 × 3.14 × 7 ≈ 44 cm. 1.5 revolutions: distance = 1.5 × 44 = 66 cm. Displacement: the tip is at 12 at 6:00 PM and at 6 at 7:30 PM, so the displacement = the diameter = 2R = 14 cm (downward). Time = 90 × 60 = 5400 s. Speed = 66 ÷ 5400 ≈ 0.0122 cm/s; velocity magnitude = 14 ÷ 5400 ≈ 0.0026 cm/s.

• Spinning disc: the numbers and letters complete the same number of revolutions per second (same time period T), but the outer letters travel farther per revolution (2π × 7 ≈ 44 cm) than the inner ones (2π × 4 ≈ 25 cm), so the outer speed is higher
• Phyphox app: a smartphone accelerometer can measure tiny accelerations, even when you try to hold still, and the same sensor rotates the screen
• 1D, 2D and 3D motion: a train on a straight track (1D), a ball thrown horizontally (2D) and a bird in the sky (3D)

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