Position, reference point, motion and rest
You say your friend sits 'near the school gate'. Near compared with what? A place can only be described when we say where we measure from.
Position tells where an object is: its distance and direction from a chosen reference point (the origin). On a straight line we mark one side as positive (+) and the other as negative (−). An object is in motion if its position changes with time and at rest if it does not. Both depend on the reference point chosen. An instant is a single moment, while an interval is the gap between two moments.
In the figure : Position needs a reference point and a sign for direction.
Rule 1: Always fix a reference point first — Motion and rest are relative. A passenger is at rest compared with the bus seat but in motion compared with the road.
- Easy : A book on a moving train is at rest compared with you, but moving compared with the platform.
- Medium : A tree seems to move backwards when you look at it from a moving car.
- Exam-type : Choosing the school gate as origin, a shop 50 m to the east is at +50 m.
Rule 2: Direction is given by the sign — Positions on one side of the origin are positive and on the other side negative, so one number can show both distance and direction.
- Easy : A well 30 m west of the origin has position −30 m if east is positive.
- Medium : Moving from +20 m to +50 m means position increased by 30 m.
- Exam-type : Position −10 m and +10 m are the same distance from the origin but on opposite sides.
Name the reference point, choose the positive direction, then read the position with its sign.
Thinking an object is 'in motion' or 'at rest' in an absolute sense. It always depends on the reference point.
Memory clue: R-S-P — Reference, Sign, Position. First pick R, then the sign, and only then say the position.
Spot the difference
| Term | Meaning | Example |
|---|---|---|
| Position | Distance and direction from the origin | +40 m from the gate |
| Motion | Position changes with time | A cycle going along the road |
| Rest | Position does not change | A parked cycle |
Practice questions
- 1What is a reference point?
- 2When is an object said to be in motion?
- 3What does a negative position mean?
- 4Why can the same passenger be at rest and in motion at the same time?
- 5Why is a sign needed along with the distance when giving position?
- 6How is an instant different from an interval?
- 7A shop is 60 m east of the gate and a temple is 80 m west of it. If east is positive, write both positions.
- 8A boy moves from +15 m to −5 m. By how much did his position change?
Show answers (try them yourself first)
- 1The fixed point from which position is measured.
- 2When its position changes with time compared with a reference point.
- 3The object is on the side of the origin chosen as negative.
- 4Rest and motion depend on the reference point, so the passenger can be at rest compared with the seat and moving compared with the road.
- 5Distance alone does not tell which side of the origin the object is on.
- 6An instant is one moment of time; an interval is the time gap between two instants.
- 7Shop +60 m, temple −80 m.
- 8−5 − 15 = −20 m, so the position decreased by 20 m.
Distance travelled and displacement
A cyclist rides 400 m around a park and stops exactly where she started. Her legs are tired, yet she says she has not 'gone anywhere'. Both statements are true, and two different quantities explain them.
Distance is the total length of the path travelled; it has only size and no direction. Displacement is the change in position, that is, final position minus initial position; it has size and direction. Both are measured in metres (m). The size of displacement is never more than the distance, and the two are equal only when the object moves along a straight line without turning back.
In the figure : Distance counts the path; displacement counts only the change in position.
Rule 1: Displacement = final position − initial position — It depends only on where you start and where you end, not on the route taken.
- Easy : A walk from +10 m to +70 m gives displacement +60 m.
- Medium : A walk from +70 m back to +10 m gives displacement −60 m.
- Exam-type : Going round a track and returning to the start gives displacement 0.
Rule 2: Distance is never less than the size of displacement — Distance adds up every bit of the path, so turning back increases distance but may reduce displacement.
- Easy : Walk 30 m east and 10 m back west: distance 40 m, displacement 20 m east.
- Medium : A taxi meter shows distance, not displacement.
- Exam-type : Walk 5 m east and 5 m west: distance 10 m but displacement 0.
Distance: add the whole path. Displacement: final position minus initial position, with sign.
Believing displacement can never be zero if the object has moved. A round trip has zero displacement but a large distance.
Memory clue: D-P, D-F-I — Distance is the Path; Displacement is Final minus Initial.
Spot the difference
| Point | Distance | Displacement |
|---|---|---|
| Meaning | Length of the path | Change in position |
| Direction | None | Has direction (sign) |
| Can it be zero after a round trip? | No | Yes |
Practice questions
- 1Define distance and displacement.
- 2What is the SI unit of displacement?
- 3When are distance and the size of displacement equal?
- 4A girl walks 12 m east and then 5 m west. Find her distance and displacement.
- 5Why can displacement be zero while distance is not?
- 6Why does a taxi meter record distance and not displacement?
- 7A runner completes one full 200 m track and ends at the start. State her distance and displacement.
- 8A boy moves from −8 m to +12 m on a line. Find his displacement and the distance he travelled.
Show answers (try them yourself first)
- 1Distance is the total path length; displacement is the change in position with direction.
- 2Metre (m).
- 3When the object moves in a straight line without turning back.
- 4Distance 17 m; displacement 7 m east.
- 5Because the starting and final positions can be the same even after a long path.
- 6Because the fare depends on how long the route is, not on where the trip ended.
- 7Distance 200 m; displacement 0.
- 8Displacement +20 m; distance 20 m.
Average speed and average velocity
A bus covers 120 km in 3 hours, but the driver kept stopping at signals and sometimes drove very fast. The bus did not travel at one speed all the time, yet we can still give one number to describe the trip.
Average speed = total distance ÷ total time, and average velocity = displacement ÷ time interval. Speed has no direction, while velocity does. Both are measured in m/s or km/h, and 1 km/h = 5/18 m/s. If the speed stays the same in equal intervals of time, the motion is uniform; otherwise it is non-uniform.
In the figure : Speed uses distance; velocity uses displacement.
Average speed and average velocity
In the app you can move this figure with your finger.
Open in the appRule 1: Divide by time to get the average — Average speed = distance ÷ time and average velocity = displacement ÷ time.
- Easy : 50 m covered in 10 s gives average speed 5 m/s.
- Medium : A displacement of 30 m east in 6 s gives velocity 5 m/s east.
- Exam-type : A round trip of 100 m in 20 s has average speed 5 m/s but average velocity 0.
Rule 2: Convert units before using the formula — To change km/h to m/s multiply by 5/18; to change m/s to km/h multiply by 18/5.
- Easy : 36 km/h = 36 × 5/18 = 10 m/s.
- Medium : 20 m/s = 20 × 18/5 = 72 km/h.
- Exam-type : A speedometer in a vehicle shows speed, not velocity.
Same time in the denominator; put distance on top for speed and displacement on top for velocity.
Using distance in the velocity formula. Velocity needs displacement, so a round trip has zero average velocity.
Memory clue: S-D-V-D — Speed uses Distance, Velocity uses Displacement; both divide by time.
Spot the difference
| Point | Average speed | Average velocity |
|---|---|---|
| Formula | Distance ÷ time | Displacement ÷ time |
| Direction | No | Yes |
| Round trip | Not zero | Zero |
Practice questions
- 1Write the formula for average speed.
- 2Convert 54 km/h into m/s.
- 3What kind of motion is it when speed changes in equal time intervals?
- 4Why is average velocity zero for a complete round trip but average speed is not?
- 5Why does a speedometer give speed and not velocity?
- 6A runner covers 100 m in 20 s. What is her average speed and what does it tell?
- 7A train covers 90 km in 1.5 hours. Find its average speed in km/h and in m/s.
- 8A boy walks 40 m east and 40 m back west in 20 s. Find his average speed and average velocity.
Show answers (try them yourself first)
- 1Average speed = total distance ÷ total time.
- 254 × 5/18 = 15 m/s.
- 3Non-uniform motion.
- 4Displacement is zero in a round trip, but the total distance covered is not zero.
- 5It shows only how fast the vehicle moves, not the direction.
- 65 m/s; she covers 5 m on average every second.
- 760 km/h, which is about 16.7 m/s.
- 8Average speed 4 m/s; average velocity 0.
Average acceleration
Press the accelerator and a scooter feels as if it is pulled forward; press the brake and it slows down. In both cases the velocity is changing, and the push you feel depends on how quickly it changes.
Average acceleration is the change in velocity divided by the time taken: a = (v − u) ÷ t, where u is the initial velocity and v the final velocity. Its SI unit is m/s². The sign shows the direction: if velocity increases in the positive direction, a is positive; if the object slows down, a is negative (retardation). If velocity is constant, acceleration is zero. Objects falling freely near the Earth have acceleration g = 9.8 m/s² downward.
In the figure : Acceleration is the rate of change of velocity, with a sign.
Average acceleration
In the app you can move this figure with your finger.
Open in the appRule 1: a = (v − u) ÷ t — Subtract the initial velocity from the final velocity, then divide by the time taken.
- Easy : Velocity 0 to 12 m/s in 4 s gives a = 3 m/s².
- Medium : Velocity 20 m/s to 5 m/s in 3 s gives a = −5 m/s².
- Exam-type : A car moving at a steady 15 m/s has a = 0.
Rule 2: Direction of acceleration vs velocity — If acceleration points along the velocity, speed increases; if it points opposite, speed decreases.
- Easy : A ball thrown up slows down because g acts opposite to its velocity.
- Medium : A ball falling down speeds up because g acts along its velocity.
- Exam-type : Braking gives acceleration opposite to the motion.
Find v − u first, keep its sign, then divide by t.
Thinking negative acceleration always means slowing down. It depends on the direction of velocity as well.
Memory clue: V-U-T — (Velocity final minus Velocity initial) over Time. Say 'v minus u over t'.
Spot the difference
| Situation | Velocity | Acceleration |
|---|---|---|
| Steady speed in a line | Constant | Zero |
| Speeding up | Increasing | Along velocity |
| Slowing down | Decreasing | Opposite to velocity |
Practice questions
- 1Write the formula for average acceleration.
- 2What is the SI unit of acceleration?
- 3What is the value of g near the Earth?
- 4Why can a body moving at high speed still have zero acceleration?
- 5Why is the acceleration of a braking vehicle negative?
- 6Explain why g acts opposite to velocity for a ball thrown upward.
- 7A scooter goes from 4 m/s to 16 m/s in 6 s. Find its acceleration.
- 8A cyclist slows from 12 m/s to 2 m/s in 5 s. Find the acceleration and say what its sign shows.
Show answers (try them yourself first)
- 1a = (v − u) ÷ t.
- 2m/s².
- 39.8 m/s² downward.
- 4Acceleration is the change in velocity; if velocity stays constant, acceleration is zero at any speed.
- 5The velocity decreases, so the change in velocity is opposite to the motion.
- 6Because g points down while the ball moves up, so the ball slows down.
- 7a = (16 − 4) ÷ 6 = 2 m/s².
- 8a = −2 m/s²; the negative sign shows velocity is decreasing.
Position-time graphs
A GPS tracker draws a line showing how far a runner is from the starting point at every minute. Reading that line carefully tells you whether she is running, resting or speeding up, without seeing her at all.
A position–time graph plots position on the vertical axis and time on the horizontal axis. Its slope gives the velocity. A straight sloping line means constant velocity, a horizontal line means the object is at rest, and a curved line means the velocity is changing, that is, the motion is accelerated. A steeper line means a higher speed, and a line sloping downward means motion in the negative direction.
In the figure : The slope of a position–time graph is the velocity.
Position-time graphs
In the app you can move this figure with your finger.
Open in the appRule 1: Slope = velocity — Velocity = change in position ÷ change in time between two points of the graph.
- Easy : Position rising from 0 to 20 m in 4 s gives velocity 5 m/s.
- Medium : A horizontal line has zero slope, so zero velocity.
- Exam-type : A downward sloping line gives negative velocity.
Rule 2: Shape tells the type of motion — Straight line is uniform motion; curve is non-uniform motion.
- Easy : A straight steep line shows fast uniform motion.
- Medium : A line bending upward shows increasing speed.
- Exam-type : A line that rises and then goes flat shows motion followed by rest.
Look at the slope of the line: steep, flat, curved or downward.
Reading a flat line as 'moving at constant speed'. A flat position–time line means the object is at rest.
Memory clue: X-T slope = V. Flat = rest, straight = steady, curve = changing.
Spot the difference
| Shape of x–t graph | Meaning | Velocity |
|---|---|---|
| Horizontal line | At rest | Zero |
| Straight sloping line | Uniform motion | Constant |
| Curved line | Non-uniform motion | Changing |
Practice questions
- 1What does the slope of a position–time graph give?
- 2What does a horizontal line on a position–time graph show?
- 3What does a curved position–time graph indicate?
- 4Why is a steeper line on a position–time graph a faster motion?
- 5Why does a downward sloping position–time line show negative velocity?
- 6How can you tell from the graph that an object stopped?
- 7On a position–time graph the position changes from 10 m to 40 m in 6 s along a straight line. Find the velocity.
- 8A graph rises from 0 to 30 m in 10 s, stays flat for 5 s, and then rises to 60 m in 5 s. Find the velocity in each part.
Show answers (try them yourself first)
- 1Velocity.
- 2The object is at rest.
- 3Non-uniform motion, with velocity changing.
- 4A steeper line covers more position change in the same time.
- 5Position decreases with time, so displacement is in the negative direction.
- 6The line becomes horizontal.
- 7(40 − 10) ÷ 6 = 5 m/s.
- 8First part 3 m/s, flat part 0, last part 6 m/s.
Velocity-time graphs
The speedometer needle of a bike climbs as the rider speeds up. If you record that needle reading every second and draw it, you get a picture that tells you the acceleration and the distance covered at once.
A velocity–time graph plots velocity against time. Its slope gives the acceleration, and the area under the graph gives the displacement. A horizontal line shows constant velocity (zero acceleration), a rising straight line shows constant positive acceleration, and a falling straight line shows constant negative acceleration. For irregular shapes the area is split into rectangles and triangles.
In the figure : Slope of a v–t graph is acceleration; area under it is displacement.
Velocity-time graphs
In the app you can move this figure with your finger.
Open in the appRule 1: Slope = acceleration — Acceleration = change in velocity ÷ time between two points of the graph.
- Easy : Velocity from 2 to 14 m/s in 6 s gives a = 2 m/s².
- Medium : A horizontal line has zero slope, so a = 0.
- Exam-type : A falling line gives negative acceleration.
Rule 2: Area = displacement — Split the area into a rectangle and a triangle and add them.
- Easy : Constant 6 m/s for 5 s gives area 6 × 5 = 30 m.
- Medium : Velocity rising from 0 to 8 m/s in 4 s gives area ½ × 4 × 8 = 16 m.
- Exam-type : Starting at 3 m/s and rising to 9 m/s in 4 s gives 3 × 4 + ½ × 4 × 6 = 24 m.
Slope gives acceleration; area gives displacement. Split the area into simple shapes.
Confusing the v–t graph with the path of the object. The graph does not show the shape of the path.
Memory clue: S-A, A-D — Slope gives Acceleration, Area gives Displacement.
Spot the difference
| Part of v–t graph | Meaning | Value |
|---|---|---|
| Horizontal line | Constant velocity | a = 0 |
| Rising straight line | Constant positive acceleration | a > 0 |
| Area under the graph | Displacement | Rectangle + triangle |
Practice questions
- 1What does the slope of a velocity–time graph give?
- 2What does the area under a velocity–time graph give?
- 3What does a horizontal line on a velocity–time graph show?
- 4Why is the area under a v–t graph a displacement?
- 5How do you find displacement when the graph is a rectangle plus a triangle?
- 6How does a v–t graph show a body is slowing down?
- 7A body's velocity increases uniformly from 0 to 20 m/s in 5 s. Find its acceleration and displacement.
- 8A car moves at a steady 10 m/s for 4 s and then slows down uniformly to rest in 5 s. Find the total displacement.
Show answers (try them yourself first)
- 1Acceleration.
- 2Displacement.
- 3Constant velocity, so zero acceleration.
- 4Each thin strip of velocity × time is a small displacement, and the area adds them all.
- 5Find the area of the rectangle and the area of the triangle separately and add them.
- 6The line slopes downward.
- 7a = 4 m/s²; displacement = ½ × 5 × 20 = 50 m.
- 840 + ½ × 5 × 10 = 40 + 25 = 65 m.
Kinematic equations of motion
A car driver must know whether the car will stop before the pedestrian crossing. Instead of testing on the road, we can calculate it, as long as the acceleration stays the same during braking.
For motion along a straight line with constant acceleration there are three equations connecting u, v, a, t and s: v = u + at, s = ut + ½at² and v² = u² + 2as. Here s is the displacement. These equations are valid only when the acceleration is constant. They also give the stopping distance of a vehicle: put v = 0 and use a negative a, so s = −u² ÷ 2a.
In the figure : Three equations work only when acceleration is constant.
Rule 1: Pick the equation that has the unknown and avoids the one you don't need — List what is given and what is asked, then choose the equation that contains exactly those.
- Easy : If s is not given and not asked, use v = u + at.
- Medium : If t is not given and not asked, use v² = u² + 2as.
- Exam-type : If v is not given and not asked, use s = ut + ½at².
Rule 2: Keep signs consistent — Take the direction of initial motion as positive; braking then gives negative a.
- Easy : A car stopping from 20 m/s with a = −5 m/s² has v = 0.
- Medium : Its stopping time is t = 4 s and its stopping distance is 40 m.
- Exam-type : Doubling the speed makes the stopping distance four times, since s depends on u².
Write u, v, a, t, s; mark the one asked; choose the equation that skips the missing one.
Using these equations when acceleration is changing. They hold only for constant acceleration.
Memory clue: V-U-A-T, S-U-T-A, V²-U²-A-S — match the missing letter: no s, no v, no t.
Spot the difference
| Missing quantity | Use this equation | Form |
|---|---|---|
| s not needed | First equation | v = u + at |
| v not needed | Second equation | s = ut + ½at² |
| t not needed | Third equation | v² = u² + 2as |
Practice questions
- 1Write the three equations of motion.
- 2For which kind of acceleration are they valid?
- 3What is the final velocity when a vehicle stops?
- 4Why does the stopping distance become four times when the speed is doubled?
- 5Why should the sign of acceleration be chosen carefully?
- 6Which equation is the best when time is not given?
- 7A car at 10 m/s accelerates at 2 m/s² for 5 s. Find its final velocity and the distance covered.
- 8A bike moving at 20 m/s brakes with a = −4 m/s². Find the time to stop and the stopping distance.
Show answers (try them yourself first)
- 1v = u + at; s = ut + ½at²; v² = u² + 2as.
- 2Constant (uniform) acceleration.
- 3Zero.
- 4Because s = u² ÷ 2a, and u² becomes four times when u is doubled.
- 5A wrong sign gives wrong answers such as a car that speeds up while braking.
- 6v² = u² + 2as.
- 7v = 20 m/s; s = 10 × 5 + ½ × 2 × 25 = 75 m.
- 8t = 5 s; s = 400 ÷ 8 = 50 m.
Motion in a plane and uniform circular motion
A child on a merry-go-round moves at a steady speed all the time, yet it is wrong to say the child's velocity is constant. The direction keeps changing every moment.
Motion on a straight line is one-dimensional, but motion on a flat surface is motion in a plane (two dimensions) and in space it is three-dimensional. In uniform circular motion an object moves on a circle of radius R with constant speed. In one revolution the distance is 2πR and the displacement is zero. The speed is v = 2πR ÷ T, where T is the time for one revolution. The velocity is always along the tangent, and because its direction keeps changing, the motion is accelerated even though the speed is constant.
In the figure : In circular motion speed is constant but velocity keeps changing.
Rule 1: One round gives distance 2πR and displacement 0 — The path length is the circumference, but you end at the starting point.
- Easy : A runner doing one lap of a circular track of radius 50 m runs about 314 m.
- Medium : After one full lap her displacement is 0.
- Exam-type : After half a lap the displacement is 2R, the diameter.
Rule 2: Velocity is along the tangent and changes direction — Since velocity changes direction, there is acceleration even when speed is steady.
- Easy : A stone whirled on a string flies off along the tangent when the string breaks.
- Medium : The Earth's orbit around the Sun is treated as nearly circular motion.
- Exam-type : The tip of a clock's minute hand moves with constant speed but changing velocity.
Distance round = 2πR, speed = 2πR ÷ T, velocity = tangent, acceleration present.
Saying uniform circular motion is non-accelerated because the speed is constant. The velocity changes direction, so it is accelerated.
Memory clue: 2πR ÷ T — 'two pie R over T' gives the speed; tangent gives the direction.
Spot the difference
| Quantity | In one revolution | Remark |
|---|---|---|
| Distance | 2πR | Circumference |
| Displacement | 0 | Back to the start |
| Velocity | Along the tangent | Direction keeps changing |
Practice questions
- 1What is the distance covered in one revolution of a circle of radius R?
- 2What is the displacement after one complete revolution?
- 3In which direction does the velocity act in circular motion?
- 4Why is uniform circular motion accelerated although the speed is constant?
- 5Why is the circular path considered an idealised model for planets?
- 6How are motion in a plane and motion in space different?
- 7A toy car goes round a circular track of radius 7 m in 11 s (take π = 22/7). Find its speed.
- 8A girl runs one full lap of a circular track and then a half lap more. Find her total distance and displacement in terms of R.
Show answers (try them yourself first)
- 12πR.
- 2Zero.
- 3Along the tangent to the circle at that point.
- 4The direction of velocity keeps changing, so velocity changes and there is acceleration.
- 5Real orbits are slightly elongated; treating them as circles keeps the calculation simple.
- 6In a plane the position needs two coordinates, in space three.
- 7v = 2 × 22/7 × 7 ÷ 11 = 4 m/s.
- 8Distance = 3πR; displacement = 2R (the diameter).