Chapter 3: Motion in a Straight Line
This chapter studies the simplest type of motion — motion along a straight line. This branch of physics is called kinematics: it describes · how · objects move without asking · why · they move.
Key Definitions
- Position (x): Location of an object relative to a chosen reference point (origin). Has both magnitude and direction (positive or negative along the line).
- Distance: Total path length covered. Always positive. Scalar quantity.
- Displacement (s): Change in position = final position - initial position. Can be positive, negative, or zero. Vector quantity.
- Speed: Distance / time. Always positive. Scalar.
- Velocity (v): Displacement / time. Can be positive or negative. Vector.
- Average velocity = total displacement / total time
- Instantaneous velocity: velocity at a particular instant = dx/dt (slope of x-t graph)
- Acceleration (a): rate of change of velocity = dv/dt = slope of v-t graph
Uniform and Non-Uniform Motion
Uniform motion: equal displacements in equal time intervals → constant velocity, zero acceleration.
Non-uniform motion: velocity changes with time → non-zero acceleration.
Equations of Motion (Uniform Acceleration)
For constant acceleration a, initial velocity u, final velocity v, displacement s in time t:
Key formulas
These are valid only when acceleration is constant.
Motion Under Gravity (Free Fall)
Key formulas
For an object thrown upward with speed u: it reaches maximum height H = u2/(2g) in time t = u/g, then falls back.
Graphical Analysis
Key formulas
---
A car accelerates from rest to 20 m/s in 10 s. Find acceleration and distance covered.
a = (v - u)/t = (20 - 0)/10 = 2 m/s2
s = ut + (1/2)at2 = 0 + (1/2)(2)(100) = 100 m
A ball is thrown upward at 30 m/s. Find maximum height (g = 10 m/s2).
At max height, v = 0. Using v2 = u2 - 2gh:
0 = 900 - 20h → h = 45 m
A train moving at 72 km/h brakes and stops in 200 m. Find deceleration.
Convert: 72 km/h = 20 m/s. v2 = u2 + 2as → 0 = 400 + 2a(200) → a = -1 m/s2
A stone is dropped from a 80 m cliff. When does it hit the ground? (g = 10 m/s2)
h = (1/2)gt2 → 80 = 5t2 → t2 = 16 → t = 4 s
Find displacement in 3rd second for initial velocity 5 m/s, acceleration 2 m/s2.
sn = u + a(2n - 1)/2 = 5 + 2(2×3 - 1)/2 = 5 + 2(5)/2 = 5 + 5 = 10 m
The v-t graph of a particle is a straight line from (0, 2) to (4, 10). Find acceleration and displacement.
a = (10 - 2)/(4 - 0) = 2 m/s2; Displacement = area under graph = (1/2)(2 + 10)(4) = 24 m
Two vehicles start from rest from the same point. A accelerates at 2 m/s2, B at 4 m/s2. After 5 s, how far apart are they?
sA = (1/2)(2)(25) = 25 m; sB = (1/2)(4)(25) = 50 m; Distance apart = 25 m
---
Common mistakes
- Confusing distance (scalar) with displacement (vector) — a person walking 10 m east then 10 m west has distance 20 m but displacement 0.
- Using kinematic equations when acceleration is NOT constant — these equations only hold for uniform acceleration.
- Sign errors: always define a positive direction first and stick to it throughout.
Summary
Kinematics describes motion using position, velocity, and acceleration. For constant acceleration, the three equations of motion (and their variants) fully describe straight-line motion. Graphical methods using x-t and v-t plots provide additional insight. Free fall is a special case with constant downward acceleration g.