Comprehensive Study Notes on Kinematics: Graphical Analysis of Motion

Uniform Acceleration and Speed-Time Graphs: Analysis of Car Motion

This section details the analysis of a car moving on a straight road with uniform acceleration, based on a provided data table of speed at specific time intervals.

Data Representation

The relationship between time and speed is captured in the following dataset:

• Time ($t$) in seconds ($s$): $0, 5, 10, 15, 20, 25, 30$
• Speed ($v$) in meters per second ($m/s$): $5, 10, 15, 20, 25, 30, 35$

Graphical Analysis

• Scale Selection: For a convenient scale, the x-axis (Time) should be plotted with $1\,\text{cm} = 5\,s$ and the y-axis (Speed) with $1\,\text{cm} = 5\,m/s$.
• Graph Characteristic: The resulting speed-time graph is a straight line, which confirms that the car is moving with uniform acceleration.

(i) Determination of Acceleration

Acceleration ($a$) is defined as the rate of change of velocity (or speed in a straight line) and is represented by the slope of the speed-time graph.

Using the formula: $a = \frac{v_2 - v_1}{t_2 - t_1}$

Substituting the values from the first two data points ($t_1 = 0, v_1 = 5$ and $t_2 = 5, v_2 = 10$): $a = \frac{10\,m/s - 5\,m/s}{5\,s - 0\,s}$$a = \frac{5\,m/s}{5\,s}$$a = 1\,m/s^2$

(ii) Distance Travelled in 50 Seconds

To determine the distance travelled ($s$) in $50\,s$, we use the kinematic equation for distance under uniform acceleration: $s = ut + \frac{1}{2}at^2$

Given values:

• Initial speed ($u$) = $5\,m/s$
• Acceleration ($a$) = $1\,m/s^2$
• Time ($t$) = $50\,s$

Calculation: $s = (5\,m/s \times 50\,s) + \frac{1}{2}(1\,m/s^2 \times (50\,s)^2)$$s = 250\,m + \frac{1}{2}(1 \times 2500)\,m$$s = 250\,m + 1250\,m$$s = 1500\,m$

Alternatively, utilizing the area under the speed-time graph (trapezium area): Distance = Area of Trapezium with parallel sides $u$ and $v$$v = u + at = 5 + (1 \times 50) = 55\,m/s$$\text{Area} = \frac{1}{2} \times (u + v) \times t$$\text{Area} = \frac{1}{2} \times (5 + 55) \times 50$$\text{Area} = 30 \times 50 = 1500\,m$

Kinematic Interpretation of Position-Time Graphs

This section analyzes the movement of a body through different segments (A to B, B to C, and C to D) specifically identifying speed variations at different time intervals.

Segment (i): Motion from A to B

• Initial Position (A): Time $t = 0\,s$, Distance $d = 0\,km$
• Final Position (B): Time $t = 2\,s$, Distance $d = 3\,km$
• Speed Calculation: $\text{Speed} = \frac{\text{Total Distance}}{\text{Total Time}}$$\text{Speed}_{AB} = \frac{3\,km - 0\,km}{2\,s - 0\,s}$$\text{Speed}_{AB} = 1.5\,km/s$

Segment (ii): Motion from B to C

• Position (B): Time $t = 2\,s$, Distance $d = 3\,km$
• Position (C): Time $t = 5\,s$, Distance $d = 3\,km$
• Observation: The distance remains constant at $3\,km$. This indicates the body is stationary.
• Speed Calculation: $\text{Speed}_{BC} = \frac{3\,km - 3\,km}{5\,s - 2\,s}$$\text{Speed}_{BC} = 0\,km/s$

Segment (iii): Motion from C to D

• Position (C): Time $t = 5\,s$, Distance $d = 3\,km$
• Position (D): Time $t = 7\,s$, Distance $d = 7\,km$
• Speed Calculation: $\text{Speed}_{CD} = \frac{7\,km - 3\,km}{7\,s - 5\,s}$$\text{Speed}_{CD} = \frac{4\,km}{2\,s}$$\text{Speed}_{CD} = 2\,km/s$

Advanced Accelerated Motion: Case Study 2

Analysis of a car moving with uniform acceleration where speed increments occur at regular time intervals.

Data Table

• Time ($t$) in seconds ($s$): $0, 2, 4, 6, 8, 10, 12$
• Speed ($v$) in meters per second ($m/s$): $4, 8, 12, 16, 20, 24, 28$

(i) Determination of Acceleration

Acceleration is the slope of the speed-time plot. $a = \frac{\Delta v}{\Delta t}$$a = \frac{8\,m/s - 4\,m/s}{2\,s - 0\,s}$$a = \frac{4\,m/s}{2\,s}$$a = 2\,m/s^2$

(ii) Distance Travelled in 10 Seconds

Knowns:

• $u = 4\,m/s$
• $a = 2\,m/s^2$
• $t = 10\,s$

Calculation using kinematic formula: $s = ut + \frac{1}{2}at^2$$s = (4 \times 10) + \frac{1}{2}(2 \times (10)^2)$$s = 40 + 100$$s = 140\,m$

Analysis of Velocity-Time Graphs and Differential Acceleration

This section focuses on a velocity-time graph for a body with varying rates of acceleration across segments A, B, and C.

(i) Velocity of the Body at Point C

By directly reading the y-axis value corresponding to point C on the graph:

• Time at C: $5\,hr$
• Velocity at C: $40\,km/hr$

(ii) Acceleration Acting on the Body Between A and B

• Coordinates of A: $(0, 0)$
• Coordinates of B: $(3, 10)$ (Based on graphical data points marked at $t=3$ and $v=10$\,km/hr)
• Acceleration Calculation: $a_{AB} = \frac{v_B - v_A}{t_B - t_A}$$a_{AB} = \frac{10\,km/hr - 0\,km/hr}{3\,hr - 0\,hr}$$a_{AB} = \frac{10}{3}\,km/hr^2 \approx 3.33\,km/hr^2$

(iii) Acceleration Acting on the Body Between B and C

• Coordinates of B: $(3, 10)$
• Coordinates of C: $(5, 40)$
• Acceleration Calculation: $a_{BC} = \frac{v_C - v_B}{t_C - t_B}$$a_{BC} = \frac{40\,km/hr - 10\,km/hr}{5\,hr - 3\,hr}$$a_{BC} = \frac{30\,km/hr}{2\,hr}$$a_{BC} = 15\,km/hr^2$