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Motion - Distance and Displacement

Grade 9CBSE

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

🔑Concepts

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Distance is the total path length traveled by an object during its motion. It is a scalar quantity, meaning it has only magnitude and no direction.

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Displacement is the shortest straight-line distance measured from the initial position to the final position of an object. It is a vector quantity, possessing both magnitude and direction.

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The SI unit for both distance and displacement is the meter (mm).

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Distance is always greater than or equal to the magnitude of displacement: Distance≥∣Displacement∣\text{Distance} \geq |\text{Displacement}|.

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If an object returns to its starting point, its displacement is 00, even if it has traveled a significant distance.

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Motion in a straight line in a single direction results in distance being equal to the magnitude of displacement.

📐Formulae

Distance=Total path length traveled\text{Distance} = \text{Total path length traveled}

Δx=xf−xi\Delta x = x_{f} - x_{i}

Circumference of a circle=2πr\text{Circumference of a circle} = 2\pi r

Magnitude of displacement in a rectangle (diagonal)=l2+b2\text{Magnitude of displacement in a rectangle (diagonal)} = \sqrt{l^2 + b^2}

💡Examples

Problem 1:

An object moves through a distance of 3 m3\text{ m} towards the East and then 4 m4\text{ m} towards the North. Calculate the total distance and the magnitude of the displacement.

Solution:

Distance =3 m+4 m=7 m= 3\text{ m} + 4\text{ m} = 7\text{ m}. Displacement magnitude =32+42=9+16=25=5 m= \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\text{ m}.

Explanation:

Distance is the simple sum of the paths. Displacement is the hypotenuse of the right-angled triangle formed by the two paths, calculated using the Pythagorean theorem.

Problem 2:

An athlete completes one round of a circular track of diameter 200 m200\text{ m} in 40 s40\text{ s}. What will be the distance covered and the displacement at the end of 2 min 20 s2\text{ min } 20\text{ s}?

Solution:

Total time =140 s= 140\text{ s}. Number of rounds =14040=3.5= \frac{140}{40} = 3.5. Radius r=100 mr = 100\text{ m}. Distance =3.5×2×π×100=700π≈2198 m= 3.5 \times 2 \times \pi \times 100 = 700\pi \approx 2198\text{ m}. Displacement =diameter=200 m= \text{diameter} = 200\text{ m}.

Explanation:

After 3.53.5 rounds, the athlete is at the opposite end of the diameter from the starting point. Thus, displacement is the diameter (2r2r), while distance is 3.53.5 times the circumference.

Problem 3:

A body travels from point AA to BB and then returns back to AA. If the distance between AA and BB is ss, find the total distance and net displacement.

Solution:

Total distance =s+s=2s= s + s = 2s. Net displacement =s−s=0= s - s = 0.

Explanation:

Since the initial and final positions are identical (point AA), the shortest distance between them is zero, resulting in zero displacement.

Distance and Displacement - Revision Notes & Key Formulas | CBSE Class 9 Science