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Motion - Uniform and Non-uniform Motion

Grade 9CBSE

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

🔑Concepts

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Uniform Motion: An object is said to be in uniform motion if it covers equal distances in equal intervals of time, however small these time intervals may be. In this case, the velocity remains constant (v=constantv = \text{constant}) and acceleration is zero (a=0a = 0).

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Non-uniform Motion: An object is said to be in non-uniform motion if it covers unequal distances in equal intervals of time. Most motions in our day-to-day life are non-uniform, such as a car moving in traffic. In this case, acceleration is non-zero (a≠0a \neq 0).

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Distance-Time Graph: For uniform motion, the distance-time graph is a straight line inclined to the time axis. For non-uniform motion, the distance-time graph is a curve (usually a parabola for constant acceleration).

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Speed: It is the distance covered per unit time. It is a scalar quantity with the SI unit m/sm/s or m⋅s−1m \cdot s^{-1}.

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Velocity: It is the displacement per unit time in a given direction. It is a vector quantity with the SI unit m/sm/s.

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Acceleration: The rate of change of velocity of an object is called acceleration. If uu is initial velocity and vv is final velocity over time tt, then a=v−uta = \frac{v - u}{t}.

📐Formulae

v=stv = \frac{s}{t}

Average Speed=Total DistanceTotal Time\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}

Average Velocity=Total DisplacementTotal Time\text{Average Velocity} = \frac{\text{Total Displacement}}{\text{Total Time}}

a=v−uta = \frac{v - u}{t}

v=u+atv = u + at

s=ut+12at2s = ut + \frac{1}{2}at^2

v2=u2+2asv^2 = u^2 + 2as

💡Examples

Problem 1:

An object travels 16 m16\text{ m} in 4 s4\text{ s} and then another 16 m16\text{ m} in 2 s2\text{ s}. What is the average speed of the object?

Solution:

Total distance traveled by the object =16 m+16 m=32 m= 16\text{ m} + 16\text{ m} = 32\text{ m}. Total time taken =4 s+2 s=6 s= 4\text{ s} + 2\text{ s} = 6\text{ s}. Average speed =Total distanceTotal time=32 m6 s=5.33 m/s= \frac{\text{Total distance}}{\text{Total time}} = \frac{32\text{ m}}{6\text{ s}} = 5.33\text{ m/s}.

Explanation:

Since the object covers the same distance in different time intervals, this is an example of non-uniform motion. Average speed is used to describe the overall rate of motion.

Problem 2:

A bus decreases its speed from 80 km/h80\text{ km/h} to 60 km/h60\text{ km/h} in 5 s5\text{ s}. Find the acceleration of the bus.

Solution:

Initial velocity u=80 km/h=80×518 m/s=22.22 m/su = 80\text{ km/h} = 80 \times \frac{5}{18}\text{ m/s} = 22.22\text{ m/s}. Final velocity v=60 km/h=60×518 m/s=16.67 m/sv = 60\text{ km/h} = 60 \times \frac{5}{18}\text{ m/s} = 16.67\text{ m/s}. Time t=5 st = 5\text{ s}. Acceleration a=v−ut=16.67−22.225=−1.11 m/s2a = \frac{v - u}{t} = \frac{16.67 - 22.22}{5} = -1.11\text{ m/s}^2.

Explanation:

The negative sign indicates retardation or deceleration, meaning the object is slowing down. This change in velocity indicates non-uniform motion.