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Motion - Speed and Velocity

Grade 9CBSE

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

🔑Concepts

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Speed is defined as the distance traveled by an object per unit time. It is a scalar quantity, meaning it has only magnitude and no direction. The SI unit of speed is m/sm/s or ms−1ms^{-1}.

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Velocity is the displacement of an object per unit time in a specified direction. It is a vector quantity, possessing both magnitude and direction. The SI unit of velocity is also m/sm/s or ms−1ms^{-1}.

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Average Speed is the total distance traveled divided by the total time taken. It is useful for describing non-uniform motion: Average Speed=Total DistanceTotal Time\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}.

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Average Velocity is the total displacement divided by the total time taken. If the velocity of an object is changing at a uniform rate, the average velocity is the arithmetic mean of the initial velocity uu and the final velocity vv.

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The magnitude of speed and velocity are equal only when an object moves in a straight line without changing its direction. If the path is curved, the magnitude of average velocity is usually less than the average speed.

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Conversion Factor: To convert speed from km/hkm/h to m/sm/s, multiply by 518\frac{5}{18}.

📐Formulae

Speed=DistanceTime=st\text{Speed} = \frac{\text{Distance}}{\text{Time}} = \frac{s}{t}

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

v⃗=s⃗t (where s⃗ is displacement)\vec{v} = \frac{\vec{s}}{t} \text{ (where } \vec{s} \text{ is displacement)}

vavg=u+v2v_{avg} = \frac{u + v}{2}

1 km/h=1000 m3600 s=518 m/s1 \text{ km/h} = \frac{1000 \text{ m}}{3600 \text{ s}} = \frac{5}{18} \text{ m/s}

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

Explanation:

To find average speed, we sum the total distances covered and divide by the sum of the total time intervals, regardless of the change in individual speeds.

Problem 2:

Usha swims in a 90 m90 \text{ m} long pool. She covers 180 m180 \text{ m} in one minute by swimming from one end to the other and back along the same straight path. Find the average speed and average velocity of Usha.

Solution:

Total distance=180 m\text{Total distance} = 180 \text{ m}. \ Total displacement=0 m\text{Total displacement} = 0 \text{ m} (since she returns to the starting point). \ Time taken=1 min=60 s\text{Time taken} = 1 \text{ min} = 60 \text{ s}. \ Average speed=180 m60 s=3 m/s\text{Average speed} = \frac{180 \text{ m}}{60 \text{ s}} = 3 \text{ m/s}. \ Average velocity=0 m60 s=0 m/s\text{Average velocity} = \frac{0 \text{ m}}{60 \text{ s}} = 0 \text{ m/s}.

Explanation:

Average speed depends on the path length (distance), while average velocity depends on the shortest distance between the start and end points (displacement). Since she returned to the start, her displacement is zero.