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Motion - Scalar and Vector Quantities

Grade 7ICSE

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

🔑Concepts

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A Physical Quantity is a property of a material or system that can be quantified by measurement.

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Scalar Quantities: These are physical quantities that have only magnitude (numerical value) and no direction. Examples include mass (mm), time (tt), distance (dd), speed (vv), and temperature (TT).

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Vector Quantities: These are physical quantities that have both magnitude and a specific direction. Examples include displacement (s⃗\vec{s}), velocity (v⃗\vec{v}), acceleration (a⃗\vec{a}), and force (F⃗\vec{F}).

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Distance: It is the actual length of the path covered by a moving body irrespective of the direction. It is a scalar quantity and its S.I. unit is the meter (mm).

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Displacement: It is the shortest straight-line distance between the initial and final positions of a body in a particular direction. It is a vector quantity and its S.I. unit is the meter (mm).

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The magnitude of displacement is always less than or equal to the distance traveled: ∣s⃗∣≤d|\vec{s}| \le d.

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If a body returns to its starting point, its displacement is 0 m0\text{ m}, even if the distance traveled is non-zero.

📐Formulae

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

v⃗=s⃗t\vec{v} = \frac{\vec{s}}{t}

Net Displacement=x⃗f−x⃗i\text{Net Displacement} = \vec{x}_f - \vec{x}_i

Average Speed=dtotalttotal\text{Average Speed} = \frac{d_{total}}{t_{total}}

💡Examples

Problem 1:

An athlete runs along a circular track of radius r=7 mr = 7\text{ m} and completes one full revolution. Calculate the distance and displacement.

Solution:

Distance =2πr=2×227×7 m=44 m= 2\pi r = 2 \times \frac{22}{7} \times 7\text{ m} = 44\text{ m}. Displacement =0 m= 0\text{ m}.

Explanation:

Since the athlete returns to the starting point, the shortest distance between the start and end is zero, making displacement 0 m0\text{ m}. The distance is the circumference of the circle.

Problem 2:

A car travels 30 km30\text{ km} towards the East and then 40 km40\text{ km} towards the North. Find the total distance and the magnitude of the displacement.

Solution:

Distance =30 km+40 km=70 km= 30\text{ km} + 40\text{ km} = 70\text{ km}. Displacement =(30)2+(40)2=900+1600=2500=50 km= \sqrt{(30)^2 + (40)^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50\text{ km}.

Explanation:

Distance is the simple sum of paths (30+4030 + 40). For displacement, we use the Pythagoras theorem because East and North are perpendicular to each other.

Problem 3:

A person walks 5 m5\text{ m} North, then turns around and walks 3 m3\text{ m} South. What is the distance and displacement?

Solution:

Distance =5 m+3 m=8 m= 5\text{ m} + 3\text{ m} = 8\text{ m}. Displacement =5 m−3 m=2 m= 5\text{ m} - 3\text{ m} = 2\text{ m} North.

Explanation:

Distance is scalar, so we add both movements. Displacement is a vector; since the directions are opposite, we subtract the magnitudes to find the resultant vector pointing North.

Scalar and Vector Quantities - Revision Notes & Key Formulas | ICSE Class 7 Science