Motion in One Dimension
Introduction
This chapter explains how to describe motion constrained to a straight line, i.e., motion in one dimension. One-dimensional motion models situations such as cars on a straight road or trains on straight tracks. To describe motion precisely we use three related quantities:
- Position - where the object is located (or displacement - change of location).
- Speed and velocity - how fast the object is moving; velocity also indicates direction.
- Acceleration - how fast the object's velocity is changing.
There is also a derived concept called jerk, which describes how fast acceleration itself changes, but this is beyond the basic scope of one-dimensional motion problems in this chapter.
Reference frame and one-dimensional coordinate systems
When discussing position we must specify a reference point and a direction. A reference point combined with a choice of positive/negative directions is called a frame of reference. In one-dimensional problems the coordinate system has only one axis (x). Choose an origin and a positive direction; positions to the left of the origin are then negative if the right was chosen as positive, and vice versa.
Defining position
Position is a measurement of location with respect to the origin of a chosen frame of reference.
Quantity: Position (x) Unit name: metre Unit symbol: m
Position values can be positive or negative depending on the choice of origin and positive direction.
One-dimensional motion
A body undergoes one-dimensional motion when it is constrained to move only along a single straight line (the x-axis). To show direction, we use sign: positive x indicates motion in the chosen positive direction; negative x indicates the opposite.
Distance and displacement
It is important to distinguish between distance and displacement.
Distance
Distance is the total length of the path travelled from the initial to the final point. Distance is a scalar (it has magnitude only).
Quantity: Distance (D) Unit name: metre Unit symbol: m
Displacement
Displacement is the change in position; it is a vector that points from the initial position to the final position. Using Δ to denote change,
Δx⃗ = x⃗f - x⃗i
Quantity: Displacement (Δx⃗) Unit name: metre Unit symbol: m
Displacement depends only on initial and final positions, not on the path travelled. In contrast, distance depends on the actual path taken.
Key differences
- Distance depends on the path; displacement does not.
- Distance is always positive; displacement can be positive, negative, or zero.
- Distance is a scalar; displacement is a vector.
Speed and velocity
Speed and velocity are related but distinct.
Average speed
Average speed is the total distance travelled divided by the time interval:
vav = D / Δt
Unit: m·s-1
Average velocity
Average velocity is the displacement divided by the time interval:
v⃗av = Δx⃗ / Δt
Velocity is a vector and can be positive or negative according to the chosen sign convention.
Instantaneous velocity and speed
Instantaneous velocity is the velocity at a specific instant in time and is defined as the limit of Δx⃗/Δt as Δt → 0. Its magnitude is the instantaneous speed. Instantaneous speed is a scalar; instantaneous velocity is a vector.
Examples: average speed and average velocity
Example 1: Average speed and average velocity
QUESTIONJames walks 2 km away from home in 30 minutes. He then turns around and walks back home along the same path, also in 30 minutes. Calculate James' average speed and average velocity.
SOLUTIONConvert units to SI.
2 km = 2000 m. 30 minutes = 30 × 60 = 1 800 s.
James walked 2000 m out and 2000 m back, so total distance D = 4000 m.
Total time Δt = 1800 s + 1800 s = 3600 s.
Average speed vav = D / Δt = 4000 m / 3600 s ≈ 1.11 m·s-1.
Displacement Δx⃗ = 0 (he returned to his starting point), so average velocity v⃗av = Δx⃗ / Δt = 0 / 3600 s = 0 m·s-1.
Acceleration
Acceleration measures how fast velocity changes.
Average acceleration is defined as
a⃗av = Δv⃗ / Δt
Unit: m·s-2
In this chapter we treat only uniform (constant) acceleration, so average acceleration equals instantaneous acceleration and we denote it simply by a⃗.
Acceleration is a vector. The sign of acceleration together with the sign of velocity determines whether the object is speeding up or slowing down. If velocity and acceleration have the same sign, the speed increases; if they have opposite signs, the speed decreases. Avoid using "deceleration" as a synonym for negative acceleration - deceleration means "slowing down", which can occur for positive or negative acceleration depending on velocity sign.
Example: Uniform acceleration then deceleration
Example 2: Acceleration
QUESTIONA car accelerates uniformly from an initial velocity of 2 m·s-1 to a final velocity of 10 m·s-1 in 8 seconds. It then slows down uniformly to a final velocity of 4 m·s-1 in 6 seconds. Calculate the acceleration of the car during the first 8 seconds and during the last 6 seconds.
SOLUTIONFor the first 8 s: initial velocity v⃗i = 2 m·s-1, final v⃗f = 10 m·s-1, Δt = 8 s.
a⃗ = (v⃗f - v⃗i) / Δt = (10 - 2) / 8 = 8 / 8 = 1 m·s-2 (positive).
For the next 6 s: initial velocity v⃗i = 10 m·s-1, final v⃗f = 4 m·s-1, Δt = 6 s.
a⃗ = (4 - 10) / 6 = -6 / 6 = -1 m·s-2 (negative).
Interpretation: during the first interval the car speeds up; during the second it slows down (velocity positive but acceleration negative).
Describing motion: words, diagrams and graphs
We describe motion in three common ways:
- in words;
- by diagrams showing positions along a line;
- by graphs: position vs. time (x vs. t), velocity vs. time (v vs. t), and acceleration vs. time (a vs. t).
Stationary object
For a stationary object, position is constant, velocity is zero, and acceleration is zero. The x vs. t graph is a horizontal line. The v vs. t and a vs. t graphs are both lines on the horizontal axis (zero).
Motion at constant velocity (uniform motion)
Constant velocity means position changes linearly with time. The x vs. t graph is a straight line with slope equal to the velocity. The v vs. t graph is a horizontal line (constant value). The a vs. t graph is a horizontal line at zero.
Example calculation: if an object moves from x = 0 m to x = 100 m in 100 s, v = (100 - 0)/(100 - 0) = 1 m·s-1. If the direction towards the origin is taken negative, the velocity sign can be negative accordingly.
Motion at constant acceleration (uniform acceleration)
With constant acceleration, the velocity changes linearly with time and the x vs. t graph is a curve (parabolic). The v vs. t graph is a straight line whose slope is a (the acceleration). The a vs. t graph is a horizontal line at the acceleration value.
Using graphs
- The gradient (slope) of an x vs. t graph gives the velocity.
- The gradient of a v vs. t graph gives the acceleration.
- The area under a v vs. t graph (between the curve and the time axis) gives displacement.
- The area under an a vs. t graph gives change in velocity.
Worked examples with graphs
Example 3: From x vs. t to v vs. t and a vs. t
QUESTIONA position vs. time graph for a car shows: from t = 0 to 2 s the position is constant; from 2 s to 4 s the position increases with increasing slope; from 4 s to 6 s the position increases linearly with constant slope. Draw corresponding v vs. t and a vs. t graphs and describe the motion.
SOLUTIONFrom 0 to 2 s: x constant → v = 0. Then a = 0 because v is constant (zero).
From 2 to 4 s: x increases and slope of x vs. t increases → v increases with time. Because v increases at a constant rate here (the slope of x vs. t increases steadily), acceleration is positive and constant.
From 4 to 6 s: x increases with constant slope → v is constant and positive; hence acceleration is zero.
Description: object is at rest from 0-2 s, accelerates positively from 2-4 s, then moves with constant positive velocity from 4-6 s.
Example 4: Areas under v vs. t
QUESTIONA truck has the v vs. t graph shown (positive region 0-14 s with plateaus and triangles, and negative region 14-15 s). Calculate distance and displacement after 15 s given the graph shapes described.
SOLUTIONCompute areas between the curve and the time axis for each time interval; sum positive areas to get forward distance and add magnitudes of negative areas for total distance. To find displacement, sum positive areas and subtract the magnitude of negative area(s).
Example numeric breakdown (illustrative): area 0-5 s is a triangle = (1/2) × 5 s × 4 m·s-1 = 10 m. Area 5-12 s rectangle = 7 s × 4 m·s-1 = 28 m. Area 12-14 s triangle above axis = (1/2) × 2 s × 4 = 4 m. Area 14-15 s triangle below axis = (1/2) × 1 s × 2 = 1 m (negative).
Total distance D = 10 + 28 + 4 + 1 = 43 m. Total displacement Δx = 10 + 28 + 4 - 1 = 41 m (positive).
Equations of motion for constant acceleration
For motion with constant acceleration, the following equations (kinematic equations) relate initial velocity v⃗i, final velocity v⃗f, acceleration a⃗ (constant), time t, and displacement Δx⃗. These are valid when acceleration is uniform.
v⃗f = v⃗i + a⃗ t
Δx⃗ = (v⃗i + v⃗f) / 2 × t
Δx⃗ = v⃗i t + (1/2) a⃗ t2
v⃗f2 = v⃗i2 + 2 a⃗ Δx⃗
To solve problems: identify the three known quantities, pick the equation that contains them and the unknown, convert to SI units, substitute and solve, then check units and sign conventions.
Example 7: Using equations of motion
QUESTIONA racing car is travelling north. It accelerates uniformly, covering a distance of 725 m in 10 s. If it has an initial velocity of 10 m·s-1, find its acceleration.
SOLUTIONGiven: v⃗i = 10 m·s-1, Δx⃗ = 725 m, t = 10 s. Find a⃗.
Use Δx⃗ = v⃗i t + (1/2) a⃗ t2.
725 = 10 × 10 + (1/2) a × (10)2.
725 - 100 = 50 a → 625 = 50 a → a = 625 / 50 = 12.5 m·s-2 North.
Example 8: Multiple quantities from kinematics
QUESTIONA motorcycle, travelling East, starts from rest and moves with constant acceleration, covering 64 m in 4 s. Calculate its acceleration, final velocity, at what time it had covered half the total distance, and the distance covered at half the total time.
SOLUTIONGiven: v⃗i = 0 m·s-1, Δx⃗ = 64 m, t = 4 s.
Use Δx⃗ = v⃗i t + (1/2) a t2 to find a:
64 = 0 + (1/2) a × 42 = 8 a → a = 64 / 8 = 8 m·s-2 East.
Final velocity: v⃗f = v⃗i + a t = 0 + 8 × 4 = 32 m·s-1 East.
Time to reach half the distance (Δx = 32 m): use Δx = vi t + (1/2) a t2 → 32 = 0 + 4 t2 → t2 = 8 → t ≈ 2.83 s.
Distance at half the time (t = 2 s): Δx = 0 + (1/2) × 8 × 22 = 16 m East.
Applications: stopping distance
Example 9: Stopping distance and reaction time
QUESTIONA truck travels at 10 m·s-1 when the driver sees a child 50 m ahead. Reaction time is 0.5 s (during which the truck continues at constant speed). After braking, acceleration is -1.25 m·s-2. Will the truck stop before hitting the child?
SOLUTIONDuring reaction time: distance AB = v × Δt = 10 × 0.5 = 5 m.
After brakes applied: initial velocity for deceleration v⃗i = 10 m·s-1, final v⃗f = 0, a = -1.25 m·s-2.
Time to stop: v⃗f = v⃗i + a t → 0 = 10 + (-1.25) t → t = 10 / 1.25 = 8 s.
Distance during braking (BC): Δx = (v⃗i + v⃗f)/2 × t = (10 + 0)/2 × 8 = 5 × 8 = 40 m.
Total stopping distance = AB + BC = 5 + 40 = 45 m, which is less than 50 m. The truck will not hit the child.
Summary of graph relations and tips
- The gradient of an x vs. t graph gives the velocity.
- The gradient of a v vs. t graph gives the acceleration.
- The area under a v vs. t graph gives displacement.
- The area under an a vs. t graph gives change in velocity.
- When solving kinematics problems with constant acceleration, use the four standard kinematic equations and pick the one containing the known and unknown quantities.
Exercises
Exercise 21-1
1. Write down the positions for objects at A, B, D and E on the number line (include units).
2. Write down the positions for objects at F, G, H and J on the given number line (include units).
3. There are five houses on a street A, B, C, D and E separated by 20 m each (A-B-C-D-E). Assume positions to the right are positive.
- Draw a frame of reference with house A as the origin and write positions of B, C, D and E.
- If you live in house C, what is your position relative to house E?
- What are the positions of houses A, B and D if house B is taken as the origin?
Exercise 21-2
1. Use the figure with houses and distances to answer: (a) Kogis walks to Kosma's house then to school - find distance and displacement. (b) Kholo walks to Kosma's house then to school - find distance and displacement. (c) Komal walks to the shop then to school - find distance and displacement. (d) State the reference point used for each calculation.
2. You stand at your front door (Δx = 0). The street is 10 m away. You walk to the street and back again.
- What is the distance you walked?
- What is your final displacement?
- Is displacement a vector or scalar? Give a reason.
Exercise 21-3
1. Bongani walks 100 m to a shop and goes back home; the round trip took 2 minutes. Find:
- Time out of the house in seconds.
- Total distance walked.
- Displacement.
- Average velocity (m·s-1).
- Average speed (m·s-1).
2. Two poles are 50 m apart. A car takes 3 s to travel from one pole to the other. (a) Calculate the car's velocity. (b) If a measured velocity is -16.67 m·s-1, which direction is the car travelling? (c) Given times when taxi and bus pass the poles, calculate time intervals and average velocities and speeds in m·s-1 and km·h-1.
3. A rabbit runs across a freeway; a car is 100 m away travelling towards the rabbit:
- Convert car speed 120 km·h-1 to m·s-1.
- How long will it take the car to travel 100 m?
- Convert rabbit speed 10 km·h-1 to m·s-1.
- If each lane is 3 m and the rabbit must cross 3 lanes, how long to cross all lanes?
- Will the rabbit safely cross if the car is in the farthest lane?
Exercise 21-4
- An athlete accelerates uniformly from 0 to 4 m·s-1 in 2 s. Calculate acceleration.
- A bus accelerates uniformly from 15 to 7 m·s-1 in 4 s. Calculate acceleration (positive direction is direction of motion).
- An aeroplane accelerates uniformly from 100 to 200 m·s-1 in 10 s, then to 240 m·s-1 in 20 s. Calculate accelerations for each phase.
Exercise 21-5 (graphs)
- Using given x vs. t and v vs. t figures, calculate: (a) Vivian's velocity between 50 s and 100 s (use gradient); (b) Vivian's acceleration for whole motion (from v vs. t); (c) Vivian's displacement (area under v vs. t).
- Thandi walks 100 m in 200 s to the bus stop each morning and back in the evening. (a) Draw x vs. t, v vs. t, a vs. t for morning. (b) Draw graphs for evening. (c) Compare the two sets of graphs and explain differences.
Exercise 21-6
- A car is parked 10 m from home for 10 minutes. Draw displacement-time, velocity-time and acceleration-time graphs and label axes.
- A bus travels at constant velocity of 12 m·s-1 for 6 s. Draw x vs. t, v vs. t, and a vs. t graphs and label axes.
- An athlete runs with constant acceleration 1 m·s-2 for 4 s. Draw a vs. t, v vs. t and x vs. t graphs. Accurate values needed for a vs. t and v vs. t.
- Given a v vs. t graph for a car, draw x vs. t and a vs. t and explain the motion.
- Given another v vs. t graph for a truck, draw x vs. t and a vs. t and explain the motion.
Exercise 21-7 (kinematics problems)
- A car starts at 10 m·s-1 and accelerates at 1 m·s-2 for 10 s. What is its final velocity?
- A train starts from rest and accelerates at 1 m·s-2 for 10 s. How far does it move?
- A bus travels at 30 m·s-1 and stops in 5 s. What is its stopping distance?
- A racing car at 20 m·s-1 stops in 20 m. What is its acceleration?
- A ball has uniform acceleration 4 m·s-2, starts from rest. Find velocity and displacement after 10 s.
- A motorcycle has uniform acceleration 4 m·s-2 and initial velocity 20 m·s-1. Find velocity and displacement after 12 s.
- An aeroplane accelerates from rest to 144 km·h-1 in 8 s. Find acceleration and distance travelled.
Derivations (extension)
The following derivations are extension material and are not required for standard problem solving but help to understand the origin of the kinematic equations.
Derivation of v⃗f = v⃗i + a⃗ t
Acceleration a⃗ = Δv⃗ / t. So Δv⃗ = a⃗ t and v⃗f = v⃗i + Δv⃗ = v⃗i + a⃗ t.
Derivation of Δx⃗ = (v⃗i + v⃗f) / 2 × t
For constant acceleration, velocity changes linearly, so the average velocity over the interval is (v⃗i + v⃗f)/2. Displacement is average velocity times time:
Δx⃗ = [(v⃗i + v⃗f) / 2] t.
Derivation of Δx⃗ = v⃗i t + (1/2) a⃗ t2
Substitute v⃗f = v⃗i + a⃗ t into the previous result:
Δx⃗ = [(v⃗i + v⃗i + a⃗ t) / 2] t = v⃗i t + (1/2) a⃗ t2.
Derivation of v⃗f2 = v⃗i2 + 2 a⃗ Δx⃗
Eliminate time using t = (v⃗f - v⃗i) / a⃗ and substitute into Δx⃗ = v⃗i t + (1/2) a⃗ t2, then simplify to obtain v⃗f2 = v⃗i2 + 2 a⃗ Δx⃗.
Applications and final remarks
The kinematic ideas in one dimension are widely used in analysing vehicle motion and road safety (stopping distances), sports motion, and many basic physics problems. Key points to remember:
- Always choose a clear frame of reference and positive direction before starting a calculation.
- Keep units consistent; use SI units (metre, second).
- Check signs: a negative result means the quantity is in the negative direction chosen by your coordinate system.
Summary of important formulas and quantities
Key formulas (constant acceleration):
- v⃗f = v⃗i + a⃗ t
- Δx⃗ = (v⃗i + v⃗f) / 2 × t
- Δx⃗ = v⃗i t + (1/2) a⃗ t2
- v⃗f2 = v⃗i2 + 2 a⃗ Δx⃗
| Quantity | Vector? | Unit name | Unit symbol |
|---|---|---|---|
| Position (x) | No | metre | m |
| Distance (D) | No | metre | m |
| Displacement (Δx⃗) | Yes | metre | m |
| Speed (vav) | No | metre per second | m·s-1 |
| Average velocity (v⃗av) | Yes | metre per second | m·s-1 |
| Instantaneous velocity (v⃗) | Yes | metre per second | m·s-1 |
| Instantaneous speed (v) | No | metre per second | m·s-1 |
| Acceleration (a⃗) | Yes | metre per second per second | m·s-2 |
