What this chapter is about
This chapter introduces kinematics — the study of motion without considering the forces that cause it. You focus entirely on objects moving along a straight path, which physicists call one-dimensional or rectilinear motion. Understanding this restricted case builds the vocabulary and mathematical tools you will need before tackling motion in two and three dimensions.
You will learn to describe where an object is (position), how its position changes (displacement), how fast it moves (speed and velocity), and how its velocity changes (acceleration). The chapter develops equations that connect these quantities when acceleration stays constant, a situation common enough to be practically useful: a ball rolling on a level floor, a car speeding up steadily, or an object falling freely under gravity near the Earth's surface.
After working through this material you should be able to draw and interpret position–time and velocity–time graphs, convert word problems into equations, and predict where or when a moving object will be. These skills underpin almost every topic you will meet later in mechanics.
Key ideas
- Frame of reference: Motion is always described relative to a chosen origin and coordinate axis. An object at rest in one frame may be moving in another.
- Distance vs displacement: Distance is the total path length (always positive); displacement is the change in position with a direction (can be positive, negative or zero).
- Speed vs velocity: Speed is distance travelled per unit time (scalar); velocity is displacement per unit time (vector). Average speed can differ from the magnitude of average velocity.
- Acceleration: The rate of change of velocity. In straight-line motion, acceleration along the direction of velocity speeds the object up; acceleration opposite to velocity slows it down.
- Uniform motion: Velocity is constant, so acceleration is zero. The position–time graph is a straight line with slope equal to velocity.
- Uniformly accelerated motion: Acceleration is constant. Three kinematic equations relate initial velocity, final velocity, acceleration, displacement and time.
- Free fall: Near the Earth's surface an object falling under gravity alone has constant downward acceleration g ≈ 9.8 m/s². Air resistance is ignored at this level.
- Graphical interpretation: The slope of a position–time (x–t) graph gives velocity; the slope of a velocity–time (v–t) graph gives acceleration; the area under a v–t curve gives displacement.
Formulas and facts to remember
- Displacement: Δx = x₂ − x₁ (final position minus initial position, with sign).
- Average velocity: v_avg = Δx / Δt (displacement divided by time interval).
- Instantaneous velocity: v = dx/dt (the limit of Δx/Δt as Δt → 0).
- Average acceleration: a_avg = Δv / Δt.
- Instantaneous acceleration: a = dv/dt = d²x/dt².
- First kinematic equation (constant a): v = u + at, where u is initial velocity.
- Second kinematic equation: x = ut + (1/2)at², giving displacement in time t.
- Third kinematic equation: v² = u² + 2ax, linking velocities and displacement without time.
- Free-fall acceleration: g ≈ 9.8 m/s² downward near the Earth's surface.
Worked examples
Example 1: Finding displacement and distance
A cyclist rides 400 m east, then turns around and rides 150 m west. Find (a) total distance, (b) displacement.
Solution
(a) Distance is the full path length: 400 m + 150 m = 550 m.
(b) Choose east as the positive direction. Displacement = +400 m + (−150 m) = +250 m, i.e., 250 m east.
Example 2: Uniform acceleration on a straight road
An auto-rickshaw starts from rest and accelerates uniformly at 1.5 m/s² for 8 s on a straight road. Find (a) final velocity, (b) distance covered.
Solution
Given: u = 0, a = 1.5 m/s², t = 8 s.
(a) Using v = u + at: v = 0 + 1.5 × 8 = 12 m/s.
(b) Using x = ut + (1/2)at²: x = 0 + 0.5 × 1.5 × 8² = 0.5 × 1.5 × 64 = 48 m.
The rickshaw reaches 12 m/s and covers 48 m.
Example 3: Free fall from a water tank
A coconut falls from rest from a tank 19.6 m above the ground. Taking g = 9.8 m/s² and ignoring air resistance, find the time to hit the ground and the speed just before impact.
Solution
Take downward as positive; u = 0, a = 9.8 m/s², x = 19.6 m.
Time: Using x = (1/2)gt², 19.6 = 0.5 × 9.8 × t² t² = 19.6 / 4.9 = 4 t = 2 s.
Speed: Using v = u + gt, v = 0 + 9.8 × 2 = 19.6 m/s downward.
Common mistakes
- Treating distance and displacement as the same → Remember displacement has direction; if you return to your starting point, displacement is zero even though distance is not.
- Forgetting signs when objects reverse direction → Choose a positive direction and keep acceleration negative if it opposes velocity.
- Using kinematic equations when acceleration is not constant → These formulas require constant acceleration; otherwise use calculus or graphs.
- Confusing the slope of an x–t graph with acceleration → Slope of x–t gives velocity; for acceleration, read the slope of the v–t graph.
- Dropping the factor 1/2 in x = ut + (1/2)at² → Derive it once by integration or area method and the factor will stick.
Quick revision
- Displacement = change in position (vector); distance = path length (scalar).
- Velocity = dx/dt; acceleration = dv/dt.
- Three constant-acceleration equations: v = u + at, x = ut + (1/2)at², v² = u² + 2ax.
- Slope of x–t graph → velocity; slope of v–t graph → acceleration; area under v–t graph → displacement.
- Free fall near Earth: a = g ≈ 9.8 m/s² downward, with air resistance ignored.