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Motion - Graphical Representation of Motion

Grade 9CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Distance-Time Graphs (s−ts-t): These graphs represent the change in position of an object over time. A straight line with a constant slope represents uniform speed, while a curved line represents non-uniform motion.

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Slope of s−ts-t Graph: The slope of a distance-time graph at any point gives the speed of the object. Mathematically, v=ΔsΔtv = \frac{\Delta s}{\Delta t}.

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Velocity-Time Graphs (v−tv-t): These graphs show how velocity changes with time. A horizontal line parallel to the time axis indicates zero acceleration (uniform velocity).

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Slope of v−tv-t Graph: The slope of a velocity-time graph represents the acceleration of the object. A positive slope indicates acceleration, while a negative slope indicates retardation (a=v−uta = \frac{v - u}{t}).

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Area under v−tv-t Graph: The total area enclosed by the velocity-time graph and the time axis represents the displacement or distance traveled by the object during that time interval.

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Uniform Acceleration: In a v−tv-t graph, uniform acceleration is represented by a straight line inclined to the time axis.

📐Formulae

v=s2−s1t2−t1v = \frac{s_2 - s_1}{t_2 - t_1}

a=v2−v1t2−t1a = \frac{v_2 - v_1}{t_2 - t_1}

Distance (Area of Rectangle)=v×t\text{Distance (Area of Rectangle)} = v \times t

Distance (Area of Triangle)=12×base×height\text{Distance (Area of Triangle)} = \frac{1}{2} \times \text{base} \times \text{height}

Distance (Area of Trapezium)=12×(u+v)×t\text{Distance (Area of Trapezium)} = \frac{1}{2} \times (u + v) \times t

💡Examples

Problem 1:

An object moves along a straight line. Its velocity increases from 10 m/s10\text{ m/s} to 30 m/s30\text{ m/s} in 5 s5\text{ s} at a constant rate. Using a v−tv-t graph approach, find the acceleration and the total distance covered.

Solution:

  1. Acceleration (aa): a=v−ut=30−105=205=4 m/s2a = \frac{v - u}{t} = \frac{30 - 10}{5} = \frac{20}{5} = 4\text{ m/s}^2. 2. Distance (ss): The area under the v−tv-t graph (trapezium) is given by s=12×(u+v)×ts = \frac{1}{2} \times (u + v) \times t. s=12×(10+30)×5=12×40×5=20×5=100 ms = \frac{1}{2} \times (10 + 30) \times 5 = \frac{1}{2} \times 40 \times 5 = 20 \times 5 = 100\text{ m}.

Explanation:

The acceleration is calculated by finding the slope of the velocity-time graph. The distance is calculated by finding the area of the trapezium formed between the graph line and the time axis from t=0t = 0 to t=5 st = 5\text{ s}.

Problem 2:

A car travels at a constant speed of 20 m/s20\text{ m/s} for 10 seconds10\text{ seconds}. Calculate the distance using the area under the v−tv-t graph.

Solution:

For constant speed, the v−tv-t graph is a horizontal line. The distance is the area of the rectangle: s=velocity×times = \text{velocity} \times \text{time}. s=20 m/s×10 s=200 ms = 20\text{ m/s} \times 10\text{ s} = 200\text{ m}.

Explanation:

Since the velocity is uniform, the acceleration is 0 m/s20\text{ m/s}^2. The area under the graph is a simple rectangle with height vv and width tt.

Graphical Representation of Motion Class 9 Notes & Examples