02Motion in One Dimension
Analyze position, displacement, velocity, and speed for straight-line motion using kinematics equations and motion graphs.
One-dimensional motion is the simplest case of kinematics. The equations and graphs you learn here form the basis for analyzing all motion — including projectile motion and circular motion.
How can we describe and predict the motion of an object moving in a straight line using mathematical relationships between position, velocity, and acceleration?
Lesson Overview
Kinematics is the branch of physics that describes how objects move — without asking why. In this lesson we restrict motion to a single straight line (one dimension). We begin by distinguishing position from displacement and distance, then build up to average velocity, instantaneous velocity, and average speed. Finally we derive and apply the three kinematic equations for constant acceleration, including the special case of free fall near Earth's surface (a = −9.8 m/s²). Mastering these tools lets you solve a wide range of real-world problems — from a car braking on a highway to a ball thrown straight up into the air.
Key Equations
Δx = x_f − x_iv_avg = Δx / Δtv = v₀ + atx = x₀ + v₀t + ½at²v² = v₀² + 2aΔxg = 9.8 m/s² (a = −g downward)Worked Examples
A car starts at position x = 10 m and drives to x = 85 m in 5 s. Find (a) the displacement and (b) the average velocity.
Given: x_i = 10 m, x_f = 85 m, Δt = 5 s
(a) Δx = x_f − x_i = 85 − 10 = 75 m
(b) v_avg = Δx / Δt = 75 m ÷ 5 s = 15 m/s
A ball is dropped from rest. Find its velocity after 3 s of free fall. (Take downward as positive.)
Given: v₀ = 0 m/s, a = +9.8 m/s² (downward), t = 3 s
Use: v = v₀ + at
v = 0 + (9.8)(3) = 29.4 m/s
A car accelerates from rest to 30 m/s in 6 s. Find (a) the acceleration and (b) the distance traveled.
Given: v₀ = 0 m/s, v = 30 m/s, t = 6 s
(a) a = (v − v₀) / t = (30 − 0) / 6 = 5 m/s²
(b) x = x₀ + v₀t + ½at² (x₀ = 0)
x = 0 + 0(6) + ½(5)(6²) = ½(5)(36) = 90 m
A ball is thrown straight upward at 20 m/s. Find the maximum height reached. (Take upward as positive, a = −9.8 m/s².)
Given: v₀ = +20 m/s, v = 0 m/s at peak, a = −9.8 m/s²
Use: v² = v₀² + 2aΔx
0 = (20)² + 2(−9.8)Δx
0 = 400 − 19.6 Δx
Δx = 400 / 19.6 ≈ 20.4 m
A stone is dropped from rest off an 80 m cliff. How long does it take to reach the ground? (Take downward as positive.)
Given: x₀ = 0, Δx = 80 m, v₀ = 0, a = +9.8 m/s²
Use: Δx = v₀t + ½at²
80 = 0 + ½(9.8)t²
80 = 4.9 t²
t² = 80 / 4.9 ≈ 16.33
t = √16.33 ≈ 4.04 s
Guided Practice
A cyclist rides from x = −5 m to x = 55 m in 10 s. Find the displacement and average velocity.
Hint: Displacement is always x_f − x_i, regardless of the sign of the starting position. Then divide by time.
A skydiver in free fall reaches a velocity of 44.1 m/s downward from rest. How long did she fall? (a = 9.8 m/s² downward)
Hint: Use v = v₀ + at and solve for t. Remember v₀ = 0 since she starts from rest.
A train decelerates uniformly from 25 m/s to rest over 200 m. Find the acceleration.
Hint: Use v² = v₀² + 2aΔx with v = 0 (final), v₀ = 25 m/s, and Δx = 200 m. Solve for a — expect a negative value.
A ball is thrown upward at 15 m/s from the top of a building. How high above the launch point is the ball at t = 1.5 s? (a = −9.8 m/s²)
Hint: Use x = x₀ + v₀t + ½at² with x₀ = 0, v₀ = +15 m/s, a = −9.8 m/s², t = 1.5 s.
A rocket sled starts from rest and travels 360 m in 6 s with constant acceleration. Find the acceleration and the final velocity.
Hint: First find a from Δx = ½at², then use v = v₀ + at (or v² = v₀² + 2aΔx) to find the final velocity.
Key Vocabulary
Position (x)
The location of an object along a number line relative to a chosen reference point (origin). Measured in metres (m).
Example: x = +12 m means 12 m to the right of the origin.
Displacement (Δx)
The change in position of an object: Δx = x_f − x_i. It is a vector — it has both magnitude and direction (sign).
Example: Moving from x = 3 m to x = −2 m gives Δx = −5 m (leftward).
Distance
The total path length traveled, regardless of direction. Distance is always ≥ 0 and is a scalar quantity.
Example: Walking 4 m right then 4 m left: distance = 8 m, but displacement = 0 m.
Velocity (v)
The rate of change of displacement with respect to time. A vector quantity with units m/s. Average velocity = Δx/Δt; instantaneous velocity is the limit as Δt → 0.
Example: v_avg = 20 m / 4 s = 5 m/s to the right.
Speed
The magnitude of velocity, or the rate at which distance is covered. Speed is always ≥ 0 and is a scalar. Average speed = total distance / total time.
Example: A car going 60 km/h north has speed = 60 km/h and velocity = 60 km/h north.
Acceleration (a)
The rate of change of velocity with respect to time: a = Δv/Δt. A vector with units m/s². Positive acceleration in the direction of motion speeds an object up; opposite direction slows it down.
Example: a = (30 − 0) / 6 = 5 m/s² (speeding up).
Free Fall
Motion under the influence of gravity alone, with no air resistance. Near Earth's surface the acceleration is g = 9.8 m/s² directed downward.
Example: A dropped ball accelerates at 9.8 m/s² downward regardless of its mass.
Kinematics
The branch of mechanics that describes motion using position, velocity, acceleration, and time — without reference to the forces causing the motion.
Example: The kinematic equations relate v, v₀, a, t, and Δx for constant acceleration.
Check Your Understanding
Interactive Practice — 5 Questions
A runner goes from x = 2 m to x = 14 m in 4 s. What is the average velocity?
Which quantity is a vector?
A car starts from rest and accelerates at 4 m/s² for 5 s. What is its final velocity?
A ball is dropped from rest. Using a = 9.8 m/s² downward, how far does it fall in 2 s?
An object thrown upward at 14 m/s reaches its peak when v = 0. Using v² = v₀² + 2aΔx with a = −9.8 m/s², what is the maximum height?
Independent Practice
A car starts from rest and reaches 28 m/s in 7.0 s. Calculate its average acceleration.
A ball is dropped from a height of 45 m. How long does it take to reach the ground? (Use g = 9.8 m/s²)
A train traveling at 30 m/s applies brakes and decelerates at 2.5 m/s². How far does it travel before stopping?
An object moves with initial velocity 12 m/s and constant acceleration 3.0 m/s². Find its velocity and displacement after 4.0 s.
★ A stone is thrown vertically upward at 20 m/s from the edge of a cliff 50 m above the ground. Find (a) the maximum height above the ground, (b) the time to reach maximum height, and (c) the total time before the stone hits the ground. (g = 9.8 m/s²)
ChallengeCommon Mistakes
Using total distance instead of displacement in kinematic equations
Displacement is the straight-line change in position (Δx = x_f − x_i); it can be negative
Mixing up average speed and average velocity
Average speed = total distance / time (always ≥ 0); average velocity = displacement / time (can be negative)
Applying v² = v₀² + 2aΔx when the object changes direction mid-trip
Split the motion at the turning point into two separate segments, then apply kinematics to each
Forgetting that at maximum height of a projectile, v_y = 0 but the object is still moving horizontally
At the peak, only the vertical component is zero; horizontal velocity is unchanged throughout flight
Math Tips
Choose a positive direction at the start and stick with it. If upward is positive, then downward displacement, velocity, and acceleration are all negative
The 'Big 4' kinematic equations each omit one variable (a, Δx, v, t). Identify the unknown and the given quantities, then pick the equation that omits the variable you don't need
On a position–time graph, slope = velocity. On a velocity–time graph, slope = acceleration and area under the curve = displacement
Free-fall shortcut: near Earth's surface use g = 9.8 m/s² ≈ 10 m/s² for quick estimates. The sign of g depends on your chosen positive direction