Motion in a Straight Line

CBSE Class 11 Physics â€¢ Chapter 02

Chapter Overview

This chapter covers the fundamental concepts of Motion in a Straight Line. Ensure you understand the definitions and derivations thoroughly.

2.1 Introduction to Motion

Motion: Change in position of an object with respect to its surroundings with time.

Frame of Reference: Coordinate system with respect to which motion is described.

Types of Motion:

1D Motion: Motion along a straight line (e.g., train on straight track)
2D Motion: Motion in a plane (covered in Chapter 3)
3D Motion: Motion in space

2.2 Position, Path Length & Displacement

Position: Location of particle with respect to origin. Represented by position vector $\vec{r}$ or coordinate x.

Path Length (Distance): Actual length of path traveled. Scalar quantity, always positive.

Displacement: Change in position. Vector from initial to final position.

$$ \Delta x = x_f - x_i $$

Key Differences:

Path length â‰¥ |Displacement|
For straight line motion in one direction: Path length = |Displacement|
For round trip: Displacement = 0, but path length â‰ 0

Worked Example 1

Q: A person walks 40 m East, then 30 m West. Find (a) Distance (b) Displacement.

Solution:
(a) Distance = Total path = 40 + 30 = 70 m
(b) Taking East as positive: Displacement = 40 - 30 = +10 m (10 m East)

2.3 Average & Instantaneous Velocity

Average Velocity: Displacement per unit time.

$$ v_{avg} = \frac{\Delta x}{\Delta t} = \frac{x_f - x_i}{t_f - t_i} $$

Average Speed: Total distance per unit time.

$$ speed_{avg} = \frac{Total\ distance}{Total\ time} $$

Instantaneous Velocity: Velocity at a particular instant.

$$ v = \lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t} = \frac{dx}{dt} $$

Instantaneous speed = |v| (magnitude of instantaneous velocity)

2.4 Acceleration

Average Acceleration: Change in velocity per unit time.

$$ a_{avg} = \frac{\Delta v}{\Delta t} = \frac{v_f - v_i}{t_f - t_i} $$

Instantaneous Acceleration:

$$ a = \lim_{\Delta t \to 0} \frac{\Delta v}{\Delta t} = \frac{dv}{dt} = \frac{d^2x}{dt^2} $$

Types:

Uniform Acceleration: Constant acceleration (a = constant)
Non-uniform Acceleration: Variable acceleration (a â‰ constant)

2.5 Kinematic Equations (Equations of Motion)

Applicable only for uniformly accelerated motion

Standard Notation:

u = initial velocity
v = final velocity
a = acceleration (constant)
t = time
s = displacement

Derivation of First Equation: v = u + at

Given: Uniform acceleration a

By definition: $a = \frac{v - u}{t}$

$at = v - u$

$$ v = u + at $$

Physical Meaning: Final velocity = Initial velocity + Gain in velocity

Derivation of Second Equation: $s = ut + \frac{1}{2}at^2$

Method 1: Using average velocity

For uniform acceleration, average velocity = $\frac{u + v}{2}$

Displacement: $s = v_{avg} \times t = \frac{u + v}{2} \times t$

From first equation: $v = u + at$

$s = \frac{u + (u + at)}{2} \times t = \frac{2u + at}{2} \times t$

$s = \frac{2ut + at^2}{2} = ut + \frac{1}{2}at^2$

$$ s = ut + \frac{1}{2}at^2 $$

Method 2: Using calculus

$v = \frac{ds}{dt} = u + at$

$ds = (u + at)dt$

Integrating both sides from 0 to t:

$\int_0^s ds = \int_0^t (u + at)dt$

$s = [ut]_0^t + [a\frac{t^2}{2}]_0^t = ut + \frac{1}{2}at^2$ âœ“

Derivation of Third Equation: $v^2 = u^2 + 2as$

Method 1: Eliminating time

From first equation: $t = \frac{v - u}{a}$ ... (i)

From second equation: $s = ut + \frac{1}{2}at^2$ ... (ii)

Substitute (i) in (ii):

$s = u \cdot \frac{v - u}{a} + \frac{1}{2}a\left(\frac{v - u}{a}\right)^2$

$s = \frac{uv - u^2}{a} + \frac{(v - u)^2}{2a}$

$s = \frac{2(uv - u^2) + (v - u)^2}{2a} = \frac{2uv - 2u^2 + v^2 - 2uv + u^2}{2a}$

$s = \frac{v^2 - u^2}{2a}$

$2as = v^2 - u^2$

$$ v^2 = u^2 + 2as $$

Method 2: Using calculus

$a = \frac{dv}{dt} = \frac{dv}{ds} \cdot \frac{ds}{dt} = v\frac{dv}{ds}$

$a \, ds = v \, dv$

Integrating: $\int_0^s a \, ds = \int_u^v v \, dv$

$as = \left[\frac{v^2}{2}\right]_u^v = \frac{v^2}{2} - \frac{u^2}{2}$

$2as = v^2 - u^2$ â†’ $v^2 = u^2 + 2as$ âœ“

Summary of Three Equations:

Equation	Missing Quantity	Use When
$v = u + at$	s (displacement)	Finding v when s is not given/needed
$s = ut + \frac{1}{2}at^2$	v (final velocity)	Finding s when v is not given/needed
$v^2 = u^2 + 2as$	t (time)	Finding v or s when time is not given/needed

Worked Example 2 - Comprehensive Kinematics

Q: A car accelerates uniformly from 18 km/h to 54 km/h in 5 s. Find (a) Acceleration (b) Distance covered.

Solution:
Convert to m/s: $u = 18 \times \frac{5}{18} = 5$ m/s, $v = 54 \times \frac{5}{18} = 15$ m/s
Time: $t = 5$ s

(a) Using $v = u + at$:
$15 = 5 + a(5)$
$5a = 10$ â†’ $a = \mathbf{2 \text{ m/s}^2}$

(b) Using $s = ut + \frac{1}{2}at^2$:
$s = 5(5) + \frac{1}{2}(2)(5)^2 = 25 + 25 = \mathbf{50 \text{ m}}$

Verification using $v^2 = u^2 + 2as$:
$15^2 = 5^2 + 2(2)s$ â†’ $225 = 25 + 4s$ â†’ $s = 50$ m âœ“

2.6 Graphical Analysis of Motion

Position-Time (s-t) Graph

Interpretation:

Slope = Velocity ($v = \frac{ds}{dt}$)
Steeper slope â†’ Higher velocity
Horizontal line â†’ At rest (v = 0)
Positive slope â†’ Moving forward
Negative slope â†’ Moving backward

Velocity-Time (v-t) Graph

Interpretation:

Slope = Acceleration ($a = \frac{dv}{dt}$)
Area under curve = Displacement
Horizontal line â†’ Uniform motion (a = 0)
Straight line with slope â†’ Uniform acceleration
Curved line â†’ Non-uniform acceleration

Deriving $s = ut + \frac{1}{2}at^2$ from v-t graph:

For uniform acceleration starting from velocity u:

v-t graph is a straight line from (0, u) to (t, v)

Area = Area of rectangle + Area of triangle

$s = ut + \frac{1}{2}(v - u)t$

Since $v - u = at$:

$s = ut + \frac{1}{2}at^2$ âœ“

Acceleration-Time (a-t) Graph

Interpretation:

Area under curve = Change in velocity
Horizontal line â†’ Uniform acceleration
Zero line â†’ Uniform velocity

Worked Example 3 - Graph Interpretation

Q: A body starts from rest and moves with uniform acceleration 2 m/sÂ² for 5 s, then moves with constant velocity for 3 s. Find total displacement using v-t graph.

Solution:
Phase 1 (0-5s): $u = 0$, $a = 2$ m/sÂ², $t = 5$ s
Final velocity: $v = 0 + 2(5) = 10$ m/s
Displacement = Area of triangle = $\frac{1}{2} \times 5 \times 10 = 25$ m

Phase 2 (5-8s): Constant velocity = 10 m/s, time = 3 s
Displacement = Area of rectangle = $10 \times 3 = 30$ m

Total displacement = 25 + 30 = 55 m

2.7 Relative Velocity in One Dimension

Relative Velocity: Velocity of one object with respect to another.

If A moves with velocity $v_A$ and B moves with velocity $v_B$ (both w.r.t. ground):

$$ v_{AB} = v_A - v_B $$

$v_{AB}$ = Velocity of A relative to B

Note: $v_{AB} = -v_{BA}$

Worked Example 4 - Relative Velocity

Q: Train A moves east at 60 km/h. Train B moves west at 40 km/h. Find velocity of A relative to B.

Solution:
Taking east as positive:
$v_A = +60$ km/h, $v_B = -40$ km/h
$v_{AB} = v_A - v_B = 60 - (-40) = \mathbf{100 \text{ km/h (east)}}$
(A appears to move at 100 km/h eastward to observer in B)

2.8 Motion Under Gravity (Free Fall)

Free Fall: Motion under gravity alone (air resistance neglected).

Acceleration due to gravity: $g = 9.8$ m/sÂ² â‰ˆ 10 m/sÂ² (downward)

Sign Convention:

Taking upward as positive: $a = -g$
Taking downward as positive: $a = +g$

Equations for Vertical Motion (taking upward as +ve):

$v = u - gt$ (velocity decreases when going up)
$h = ut - \frac{1}{2}gt^2$ (h = height)
$v^2 = u^2 - 2gh$

For object dropped (u = 0):

$v = gt$ (downward)
$h = \frac{1}{2}gt^2$
$v^2 = 2gh$

For object thrown upward:

At maximum height: $v = 0$
Max height: $H = \frac{u^2}{2g}$
Time to reach max height: $t_up = \frac{u}{g}$
Total time of flight (up and down): $T = \frac{2u}{g}$
Velocity on return = -u (same magnitude, opposite direction)

Worked Example 5 - Free Fall

Q: A stone is dropped from a height of 80 m. Find (a) Time to reach ground (b) Velocity just before hitting ground. (g = 10 m/sÂ²)

Solution:
Given: $u = 0$, $h = 80$ m, $g = 10$ m/sÂ²

(a) Using $h = \frac{1}{2}gt^2$:
$80 = \frac{1}{2}(10)t^2$
$t^2 = 16$ â†’ $t = \mathbf{4 \text{ s}}$

(b) Using $v^2 = 2gh$:
$v^2 = 2(10)(80) = 1600$
$v = \mathbf{40 \text{ m/s (downward)}}$

OR using $v = gt = 10 \times 4 = 40$ m/s âœ“

Worked Example 6 - Upward Throw

Q: A ball is thrown vertically upward with velocity 20 m/s. Find (a) Maximum height (b) Total time in air (c) Velocity after 1.5 s. (g = 10 m/sÂ²)

Solution:
Given: $u = 20$ m/s (upward), $g = 10$ m/sÂ²

(a) Max height: $H = \frac{u^2}{2g} = \frac{(20)^2}{2(10)} = \frac{400}{20} = \mathbf{20 \text{ m}}$

(b) Total time: $T = \frac{2u}{g} = \frac{2(20)}{10} = \mathbf{4 \text{ s}}$

(c) At t = 1.5 s (taking upward as positive):
$v = u - gt = 20 - 10(1.5) = 20 - 15 = \mathbf{+5 \text{ m/s}}$ (still going up)

Study Tip: Master the derivations in this chapter as they are frequently asked in exams. Practice numericals based on the formulas.