Chapter 2: Motion in a Straight Line
Unit II – Kinematics
2.1 Frame of Reference
A frame of reference is a coordinate system relative to which we describe the position and motion of objects.
- Origin: The fixed reference point.
- Position: Described by coordinates \((x, y, z)\).
- In 1D motion, we use a single axis (usually the x-axis).
2.2 Types of Motion
| Type | Description |
|---|---|
| Uniform motion | Equal distances in equal time intervals |
| Non-uniform motion | Unequal distances in equal time intervals |
| Uniformly accelerated | Constant acceleration (e.g., free fall) |
2.3 Distance, Displacement, Speed, Velocity
- Distance: Total path length (scalar)
- Displacement: Change in position — shortest path from start to end (vector)
\[ \vec{s} = \vec{r_f} - \vec{r_i} \]
- Average speed: \(\bar{v} = \dfrac{\text{total distance}}{\text{total time}}\)
- Average velocity: \(\bar{v} = \dfrac{\Delta x}{\Delta t} = \dfrac{x_2 - x_1}{t_2 - t_1}\)
- Instantaneous velocity: \(v = \displaystyle\lim_{\Delta t \to 0} \dfrac{\Delta x}{\Delta t} = \dfrac{dx}{dt}\)
- Instantaneous speed = magnitude of instantaneous velocity
2.4 Acceleration
\[ a = \frac{dv}{dt} = \frac{d^2x}{dt^2} \]
- Average acceleration: \(\bar{a} = \dfrac{v_2 - v_1}{t_2 - t_1}\)
- Positive acceleration: velocity increasing in +x direction
- Deceleration (retardation): acceleration opposite to velocity
2.5 Equations of Uniformly Accelerated Motion
These three kinematic equations are fundamental:
\[ \boxed{v = u + at} \]
\[ \boxed{s = ut + \frac{1}{2}at^2} \]
\[ \boxed{v^2 = u^2 + 2as} \]
Also:
\[ s_n = u + a\left(n - \frac{1}{2}\right) \quad \text{(displacement in } n\text{th second)} \]
Where:
- \(u\) = initial velocity
- \(v\) = final velocity
- \(a\) = uniform acceleration
- \(s\) = displacement
- \(t\) = time
2.6 Free Fall
An object falling under gravity alone (no air resistance) is in free fall.
\[ g = 9.8 \text{ m s}^{-2} \approx 10 \text{ m s}^{-2} \]
For a body dropped from rest: \[ v = gt, \quad h = \frac{1}{2}gt^2, \quad v^2 = 2gh \]
Time to fall height \(h\): \[ t = \sqrt{\frac{2h}{g}} \]
2.7 Position-Time (x-t) Graph
Key Points:
- Slope of x-t graph = (instantaneous) velocity
- A straight line: uniform motion (constant velocity)
- Curved line: non-uniform motion
- Horizontal line: body at rest
- Negative slope: motion in negative direction
2.8 Velocity-Time (v-t) Graph
Key Points:
- Slope of v-t graph = acceleration
- Area under v-t graph = displacement
- \(a\)-\(t\) graph: area = change in velocity
2.9 Relative Velocity
If two objects A and B move with velocities \(v_A\) and \(v_B\) in the same direction:
\[ v_{AB} = v_A - v_B \quad \text{(velocity of A relative to B)} \]
If moving in opposite directions:
\[ v_{AB} = v_A + v_B \]
Key Formulas Summary
| Quantity | Formula |
|---|---|
| Instantaneous velocity | \(v = \dfrac{dx}{dt}\) |
| Instantaneous acceleration | \(a = \dfrac{dv}{dt}\) |
| First equation of motion | \(v = u + at\) |
| Second equation of motion | \(s = ut + \frac{1}{2}at^2\) |
| Third equation of motion | \(v^2 = u^2 + 2as\) |
| Displacement in nth second | \(s_n = u + a(n - \frac{1}{2})\) |
| Free fall time | \(t = \sqrt{\frac{2h}{g}}\) |
Practice Questions
Section A – Multiple Choice Questions (1 mark each)
Q1. The slope of a position-time graph gives:
(a) Acceleration (b) Displacement (c) Velocity (d) Force
Answer
(c) Velocity — \(v = \dfrac{dx}{dt}\) = slope of x-t graph.
Q2. A body starts from rest and moves with uniform acceleration. The ratio of distances covered in the 1st, 2nd, and 3rd seconds is:
(a) 1:2:3 (b) 1:3:5 (c) 1:4:9 (d) 1:2:4
Answer
(b) 1:3:5
Using \(s_n = u + a(n - \frac{1}{2})\) with \(u = 0\):
\(s_1 = \frac{a}{2},; s_2 = \frac{3a}{2},; s_3 = \frac{5a}{2}\) → ratio 1:3:5.
Q3. A ball is thrown vertically upward with speed \(20 \text{ m/s}\). The maximum height reached is (\(g = 10 \text{ m/s}^2\)):
(a) 10 m (b) 20 m (c) 30 m (d) 40 m
Answer
(b) 20 m
\(v^2 = u^2 - 2gh\), at max height \(v = 0\):
\(h = \dfrac{u^2}{2g} = \dfrac{400}{20} = 20 \text{ m}\)
Section B – Short Answer Questions (2–3 marks)
Q4. A train starts from rest and accelerates uniformly at \(2 \text{ m/s}^2\) for \(10\text{ s}\). It then moves at constant velocity for \(20\text{ s}\) and finally decelerates to rest at \(4 \text{ m/s}^2\). Find the total distance.
Answer
Phase 1 (Acceleration):
\(v = u + at = 0 + 2 \times 10 = 20 \text{ m/s}\)
\(s_1 = \frac{1}{2} \times 2 \times 100 = 100 \text{ m}\)
Phase 2 (Constant velocity):
\(s_2 = 20 \times 20 = 400 \text{ m}\)
Phase 3 (Deceleration):
\(t_3 = \dfrac{20}{4} = 5 \text{ s}\), \(s_3 = \dfrac{v^2}{2 \times 4} = \dfrac{400}{8} = 50 \text{ m}\)
Total = \(100 + 400 + 50 = \mathbf{550 \text{ m}}\)
Q5. From the top of a tower 40 m high, a stone is thrown vertically upward with a speed of \(10 \text{ m/s}\). When does the stone hit the ground? (\(g = 10 \text{ m/s}^2\))
Answer
Taking upward as positive, initial position at top (+40 m from ground):
\[ y = y_0 + ut - \frac{1}{2}gt^2 \Rightarrow 0 = 40 + 10t - 5t^2 \]
\[ 5t^2 - 10t - 40 = 0 \Rightarrow t^2 - 2t - 8 = 0 \Rightarrow (t-4)(t+2) = 0 \]
\(t = 4 \text{ s}\) (taking positive root)
Section C – Long Answer Questions (5 marks)
Q6. Derive the three equations of uniformly accelerated motion using the graphical (v-t graph) method.
Answer
Consider a v-t graph where a body accelerates uniformly from velocity \(u\) at \(t=0\) to velocity \(v\) at time \(t\).
Equation 1: Slope of v-t graph = acceleration
\[ a = \frac{v - u}{t} \Rightarrow \boxed{v = u + at} \]
Equation 2: Area under v-t graph = displacement
Area = trapezium = \(\dfrac{1}{2}(u + v)t = \dfrac{1}{2}(u + u + at)t\)
\[ \boxed{s = ut + \frac{1}{2}at^2} \]
Equation 3: From equations 1 and 2:
\(s = \dfrac{(v+u)}{2} \times t\) and \(t = \dfrac{v-u}{a}\)
\[ s = \dfrac{(v+u)(v-u)}{2a} \Rightarrow \boxed{v^2 = u^2 + 2as} \]
Section D – Competency-Based Questions
Q7. (Graph-based) Study the v-t graph below and answer:
A car starts from rest, accelerates to \(20 \text{ m/s}\) in \(10 \text{ s}\), maintains this speed for \(20 \text{ s}\), then decelerates to rest in \(5 \text{ s}\).
(i) What is the acceleration in the first phase?
(ii) What is the retardation in the last phase?
(iii) Calculate total displacement using the area method.
(iv) What does the area of the v-t graph represent physically?
Answer
(i) \(a = \dfrac{20 - 0}{10} = \mathbf{2 \text{ m/s}^2}\)
(ii) Retardation = \(\dfrac{20 - 0}{5} = \mathbf{4 \text{ m/s}^2}\)
(iii)
- Triangle 1 (0–10 s): \(\frac{1}{2} \times 10 \times 20 = 100\) m
- Rectangle (10–30 s): \(20 \times 20 = 400\) m
- Triangle 2 (30–35 s): \(\frac{1}{2} \times 5 \times 20 = 50\) m
- Total = 550 m
(iv) Area under v-t graph = displacement (not distance, unless motion is one-directional)
Q8. (Assertion-Reason) Assertion (A): A body with zero velocity may have non-zero acceleration.
Reason (R): Velocity and acceleration are independent quantities.
(a) Both A and R are true, and R is the correct explanation of A
(b) Both A and R are true, but R is not the correct explanation of A
(c) A is true, but R is false
(d) A is false, but R is true
Answer
(a) Both are true, and R explains A.
Example: At the highest point of a vertically thrown ball, velocity = 0 but \(a = g \neq 0\).