Distance & Displacement · Speed & Velocity · Acceleration · Graphs · Equations of Motion · Circular Motion
Everything in nature is in motion -- from massive astronomical objects to tiny subatomic particles. And what a variety of motion we have in nature! Flitting butterflies, slithering snakes, hopping hares, galloping horses, tendrils of climbers twining around a support, closing of flytraps, dancing dust particles in a sunbeam, smoke particles moving in air, rising and falling of ocean tides, and gathering clouds!
To study this wide variety of complex motions, scientists first study motion in its idealised simplified forms: linear motion, circular motion, and oscillatory motion. In this chapter, you will learn more about linear motion (motion in a straight line) and uniform circular motion.
Distance, time, and speed from earlier grades.
Displacement, average velocity, average acceleration, graphs, and equations of motion.
Hints:
• When brakes are applied, the vehicle does not stop instantly -- it travels some distance before stopping.
• This stopping distance depends on the speed of the vehicle. The faster you go, the longer it takes to stop!
• By the end of this chapter, you will be able to calculate this distance using equations of motion.
When an object moves in a straight line, its motion is called linear motion. It is the simplest kind of motion. Examples include children in a swimming race, a vertically falling ball, a car moving along a straight highway, or a train on a straight track.
To describe the position of an object, we first need to specify a fixed point as the reference point (also called the origin, marked as 'O'). The distance and direction of the object with respect to the reference point describes its position at any instant of time.
If the position of the object with respect to the reference point changes with time, the object is said to be in motion.
If the position of the object with respect to the reference point does not change with time, it is said to be at rest.
Consider an athlete running on a straight track. We take her starting point as the reference point (origin O). Positions to the right of O are taken as positive (+) and to the left as negative (-).
On a number line: O — B — A where O = 0 m, B = 40 m, A = 100 m
For motion in a straight line, the object can move only in one of two directions -- forward (+) and backward (-).
A ball is thrown straight up inside a moving train. Watch how the same ball appears to follow different paths depending on who is watching!
Suppose an athlete starts running from point O at time t = 0 s, reaches point A (100 m) at t = 10 s, then runs back to point B (40 m) at t = 16 s.
OA + AB = 100 + 60 = 160 m
OB = 40 m (positive direction)
Displacement is the net change in the position of an object between two given instants of time. It requires specifying both a direction and a numerical value (with units).
A ball is thrown upward from point O (0 cm). It passes through A (40 cm), reaches the highest point B (140 cm), then comes back down through C (100 cm) and returns to O (0 cm).
Path: O(0) → A(40) → B(140) → C(100) → O(0). Click each blue cell to reveal the answer!
| Journey | Distance (cm) | Displacement (cm) |
|---|---|---|
| O → A | 👉 Tap | 👉 Tap |
| O → B | 👉 Tap | 👉 Tap |
| O → B → C | 👉 Tap | 👉 Tap |
| O → B → O | 👉 Tap | 👉 Tap |
| A → B | 👉 Tap | 👉 Tap |
| A → B → C | 👉 Tap | 👉 Tap |
💡 Notice: Distance is always ≥ |Displacement|. They are equal only when there is no turning back!
Draw any path on the dark canvas below. The app will automatically calculate your total distance (path length) and displacement (start to end) in real-time!
👉 Click/tap and drag to draw. Try zigzags, curves, loops — see how distance and displacement differ!
How can you describe how fast or slow an object is moving? That is where average speed and average velocity come in!
Object travels equal distances in equal time intervals (for all possible choices of time intervals). Speed is constant.
Object travels unequal distances in equal time intervals. Speed is changing (increasing or decreasing or both).
The SI unit of both average speed and average velocity is metre per second (m s-1 or m/s). It is also commonly measured in km h-1.
Problem: Two postmen start walking towards each other from a distance of 210 yojanas. One travels 9 yojanas per day and the other covers 5 yojanas per day. In how many days will they meet?
Answer: Total distance covered per day = 9 + 5 = 14 yojanas
Time to meet = 210 / 14 = 15 days
(In 15 days: first postman covers 135 yojanas, second covers 75 yojanas)
Problem: Sarang takes 50 seconds to swim from one end to the other end (25 m) and back in a swimming pool. Find his average speed and average velocity.
Answer:
Total distance = 25 + 25 = 50 m | Displacement = 0 m (returns to start)
Average speed = 50 m / 50 s = 1 m s-1
Average velocity = 0 m / 50 s = 0 m s-1
The average speed is non-zero but the average velocity is zero because the displacement is zero!
Enter any two values and leave the third blank. Click calculate to find the missing value!
💡 Fun: Try calculating how long it takes to reach your school!
Watch two objects travel from Start to Finish by different paths. Compare their speed and velocity in real time!
A swimmer goes back and forth in a 25 m pool. Watch how distance keeps growing but displacement returns to zero! (NCERT Example 4.2)
Drag each motion scenario into the correct zone. Uniform motion has constant speed in a straight line. Non-uniform motion has changing speed or direction.
Convert between km/h and m/s as fast as you can! Remember: multiply by 5/18 to go from km/h to m/s, and by 18/5 for m/s to km/h.
When you sit in a vehicle and it suddenly moves from rest, you feel a noticeable jolt. Similarly, when it suddenly stops, you experience a jolt. These occurrences capture the feeling of the change in velocity.
Where u = initial velocity, v = final velocity, t = time interval
SI Unit: m s-2 (metre per second squared)
When magnitude of velocity is increasing, acceleration is in the same direction as velocity.
When magnitude of velocity is decreasing, acceleration is opposite to the direction of velocity.
Problem: A bus moving at 36 km h-1 accelerates for 10 s to reach 54 km h-1. Later, it brakes and stops in 5 s from 54 km h-1. Find the average acceleration in both cases.
(i) When accelerator is pressed:
u = 36 km h-1 = 10 m s-1, v = 54 km h-1 = 15 m s-1, t = 10 s
a = (15 - 10) / 10 = 0.5 m s-2 (in the direction of velocity)
(ii) When brakes are pressed:
u = 15 m s-1, v = 0 m s-1, t = 5 s
a = (0 - 15) / 5 = -3 m s-2 (minus sign means opposite to velocity direction)
Problem: An object is dropped from a height. Its velocity at different instants:
t = 0 s: 0 m/s | t = 1 s: 9.8 m/s | t = 2 s: 19.6 m/s | t = 3 s: 29.4 m/s | t = 4 s: 39.2 m/s
Answer: The average acceleration in every successive 1 s interval = 9.8 m s-2 (constant!)
This constant acceleration is called the acceleration due to gravity and is denoted by g. It acts in the direction of motion (downward).
Enter the initial speed, final speed, and time to calculate the acceleration of a car in m/s².
🚗 Compare: A typical family car does 0 to 100 km/h in ~10s (a ≈ 2.78 m/s²). A sports car does it in ~4s (a ≈ 6.94 m/s²). Gravity pulls at 9.8 m/s²!
A ticker timer prints dots on a tape at equal time intervals. The spacing between dots tells us about the motion. Click on any two dots to measure the distance and calculate speed!
Graphs provide a visual representation of how position, velocity, and acceleration change with time. They help in comparing the motion of two objects, calculating physical quantities, and identifying whether the motion is uniform or non-uniform.
A straight line position-time graph indicates constant velocity. Equal displacements in equal time intervals.
A curved position-time graph indicates changing velocity (accelerated motion). Unequal displacements in equal time intervals.
A horizontal line parallel to time axis means the object is at rest (position not changing).
Velocity is constant (zero acceleration). Object is in uniform motion.
Velocity is increasing with constant acceleration. Acceleration is in the direction of velocity.
Velocity is decreasing with constant acceleration. Acceleration is opposite to the direction of velocity.
Click on the dark canvas below to plot data points. The app will connect them, draw the graph, calculate slope (velocity or acceleration) and area under the graph (displacement) automatically!
Select a velocity-time profile, then click "Show Area" to watch the area fill up strip by strip. The total area equals the distance travelled!
For the special case of motion with constant acceleration, we can derive equations that relate five physical quantities: displacement (s), time interval (t), initial velocity (u), final velocity (v), and acceleration (a).
Problem: A car brakes with acceleration of -4 m s-2. Find the stopping distance if the car was moving at (i) 54 km h-1, and (ii) 108 km h-1.
Using v² = u² + 2as with v = 0:
0 = u² + 2(-4)s → s = u² / 8
(i) u = 54 km/h = 15 m/s → s = 225/8 = 28.1 m
(ii) u = 108 km/h = 30 m/s → s = 900/8 = 112.5 m
Doubling the speed makes the stopping distance 4 times longer! That is why maintaining a safe following distance is so important.
The stopping distance depends on the vehicle's velocity, road surface (wet/dry), braking capacity, and driver's reaction time. Vehicle-to-Vehicle (V2V) communication technology is now being developed in India to warn drivers of possible collisions!
Pick an equation, enter the known values, and solve for the unknown. Step-by-step solution included!
Enter any 3 values. Leave one blank to solve for it.
Set the initial velocity and acceleration, then hit Play to watch the object move! Live graphs of position-time and velocity-time draw in real-time. See all three equations come to life.
Adjust the car's initial speed and hit the brakes! See how doubling the speed quadruples the stopping distance. Uses the 3rd equation: v² = u² + 2as with deceleration a = -5 m/s².
| Speed (km/h) | Speed (m/s) | Stopping Distance (m) | Ratio to 20 km/h |
|---|
When an object moves in a circular path with constant (uniform) speed, its motion is called uniform circular motion. Examples include a child on a merry-go-round, planets revolving around the Sun, and a vehicle making a circular turn.
Where R = radius of circular path, T = time for one revolution
Distance = circumference = 2πR
Displacement = 0 (returns to start)
Average velocity = 0
At any point on the circular path, the velocity is along the tangent to the circle at that point, in the direction of motion.
Even though the speed is constant, the direction of velocity keeps changing continuously at every point on the circle. Since velocity is changing (in direction), the motion is accelerated!
Imagine an athlete running along a rectangular track -- they change direction 4 times per lap. Along a hexagonal path, they change direction 6 times. As the number of sides increases indefinitely, the track approaches a circle, and the direction of velocity changes continuously.
Take a ring (like an adhesive tape ring) and roll a marble along its inner boundary. When you lift the ring while the marble is moving, the marble moves in a straight line (not a circle). This is because once released, the marble continues in the direction it was moving at that instant!
Watch the dot orbit in a circle. Adjust the speed with the slider. Notice: the speed is constant, but the direction changes at every point!
Radius (R): 80 units | Circumference: 2πR = 502.7 units
Time Period (T): 2.00 s | Speed: 2πR/T = 251.3 units/s
🔴 Red arrow shows the velocity direction (tangent to circle) -- it keeps changing!
Described by distance and direction from a reference point. Object is in motion if position changes with time.
Distance: total path length (scalar). Displacement: net change in position (vector). They are equal only when there is no turning back.
Average speed = distance/time. Average velocity = displacement/time. Speed has no direction; velocity has direction.
Rate of change of velocity. a = (v-u)/t. Can be due to change in magnitude, direction, or both.
Position-time graph: slope = velocity. Velocity-time graph: slope = acceleration, area = displacement.
v = u + at | s = ut + ½at² | v² = u² + 2as. Valid only for constant acceleration.
Constant speed in circular path = uniform circular motion. Direction changes continuously, so motion is accelerated. v = 2πR/T.
Stopping distance depends on speed, road conditions, and reaction time. Doubling speed quadruples stopping distance!
First Equation (v = u + at): By the definition of acceleration, a = (v - u)/t. Rearranging: at = v - u, therefore v = u + at.
Second Equation (s = ut + ½at²): The displacement equals the area under the velocity-time graph. For an object with initial velocity u and constant acceleration, the v-t graph is a straight line. The area under it = area of rectangle + area of triangle = u x t + ½ x t x (v - u). Substituting (v - u) = at: s = ut + ½at².
Third Equation (v² = u² + 2as): From the first equation, t = (v - u)/a. Substituting in the second equation: s = u(v-u)/a + ½a[(v-u)/a]². Simplifying: 2as = 2u(v-u) + (v-u)² = v² - u². Therefore: v² = u² + 2as.
Phase 1 (Acceleration): u = 0, v = 20 m/s, t = 5 s. Using s = ut + ½at², first find a = (20-0)/5 = 4 m s-2. s1 = 0 + ½(4)(25) = 50 m.
Phase 2 (Constant velocity): v = 20 m/s, t = 10 s. s2 = 20 x 10 = 200 m.
Phase 3 (Braking): u = 20 m/s, v = 0, t = 6 s. a = (0-20)/6 = -10/3 m s-2. s3 = 20(6) + ½(-10/3)(36) = 120 - 60 = 60 m.
Total distance = 50 + 200 + 60 = 310 m
Position-Time Graphs: These show how the position of an object changes with time. A straight line indicates constant velocity. A curved line indicates changing velocity (acceleration). A horizontal line means the object is at rest. The slope of the position-time graph gives the velocity (slope = change in position / change in time).
Velocity-Time Graphs: These show how velocity changes with time. A horizontal line indicates constant velocity (zero acceleration). A straight line with a positive slope indicates constant positive acceleration (speeding up). A straight line with a negative slope indicates constant negative acceleration (slowing down).
From velocity-time graphs: The slope gives the acceleration (slope = change in velocity / change in time). The area enclosed between the graph line and the time axis gives the displacement. For a rectangle: displacement = velocity x time. For a triangle: displacement = ½ x base x height. For a trapezium: the standard area formula applies.
Given: u = 36 km h-1 = 10 m s-1, obstacle at 30 m, reaction time = 0.5 s, a = -2.5 m s-2.
Step 1 -- Distance during reaction time: The bus travels at 10 m/s for 0.5 s before braking. Distance = 10 x 0.5 = 5 m.
Step 2 -- Braking distance: Using v² = u² + 2as with v = 0: 0 = (10)² + 2(-2.5)s → 5s = 100 → s = 20 m.
Step 3 -- Total distance: 5 + 20 = 25 m.
Since 25 m < 30 m, the bus will stop before reaching the obstacle. It stops with 5 m to spare.