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Understanding Motion through Experience - Scalars and Vectors-advanced

Grade 9CBSE

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

🔑Concepts

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Scalars are physical quantities described by a magnitude (numerical value) only, such as mass (mm), distance (dd), and speed (vv).

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Vectors are physical quantities that require both magnitude and direction for a complete description, such as displacement (s⃗\vec{s}), velocity (v⃗\vec{v}), and acceleration (a⃗\vec{a}).

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Displacement is the shortest straight-line distance between the initial and final positions of an object. It is a vector quantity and its magnitude can be zero even if the distance covered is non-zero.

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The magnitude of the resultant displacement for two perpendicular movements xx and yy is calculated using the Pythagorean theorem: R=x2+y2R = \sqrt{x^2 + y^2}.

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In vector notation, direction is often represented by signs (++ or −-). For example, if East is ++, then West is −-.

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Speed is the rate of change of distance (v=dtv = \frac{d}{t}), while Velocity is the rate of change of displacement (v⃗=Δs⃗Δt\vec{v} = \frac{\Delta \vec{s}}{\Delta t}).

📐Formulae

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}}

Resultant Displacement (s)=Δx2+Δy2\text{Resultant Displacement } (s) = \sqrt{\Delta x^2 + \Delta y^2}

a⃗=v⃗−u⃗t\vec{a} = \frac{\vec{v} - \vec{u}}{t}

Δv⃗=v⃗final−v⃗initial\Delta \vec{v} = \vec{v}_{final} - \vec{v}_{initial}

💡Examples

Problem 1:

A person walks 8 m8\text{ m} East and then 6 m6\text{ m} North. Find the total distance and the magnitude of the displacement.

Solution:

Distance=8 m+6 m=14 m\text{Distance} = 8\text{ m} + 6\text{ m} = 14\text{ m} Displacement=(8)2+(6)2\text{Displacement} = \sqrt{(8)^2 + (6)^2} Displacement=64+36=100=10 m\text{Displacement} = \sqrt{64 + 36} = \sqrt{100} = 10\text{ m}

Explanation:

Distance is a scalar sum of the paths taken. Displacement is a vector quantity representing the shortest path from start to finish, found here using the Pythagorean theorem because East and North are perpendicular.

Problem 2:

A ball is thrown vertically upwards with a velocity of 20 m/s20\text{ m/s} and returns to the thrower's hand. What is the total distance and total displacement?

Solution:

Let the maximum height reached be HH. Using v2−u2=2asv^2 - u^2 = 2as where v=0,u=20,a=−10v=0, u=20, a=-10: 02−(20)2=2(−10)H0^2 - (20)^2 = 2(-10)H −400=−20H  ⟹  H=20 m-400 = -20H \implies H = 20\text{ m} Total Distance=Hup+Hdown=20+20=40 m\text{Total Distance} = H_{up} + H_{down} = 20 + 20 = 40\text{ m} Total Displacement=0 m\text{Total Displacement} = 0\text{ m}

Explanation:

Distance is the total path length (20 m20\text{ m} up and 20 m20\text{ m} down). Displacement is zero because the initial and final positions are the same.

Problem 3:

A car moving at 15 m/s15\text{ m/s} East changes its direction and moves at 15 m/s15\text{ m/s} West. Calculate the change in velocity.

Solution:

Let East be positive (+)\text{Let East be positive (+)} u⃗=+15 m/s\vec{u} = +15\text{ m/s} v⃗=−15 m/s\vec{v} = -15\text{ m/s} Δv⃗=v⃗−u⃗\Delta \vec{v} = \vec{v} - \vec{u} Δv⃗=−15−(+15)=−30 m/s\Delta \vec{v} = -15 - (+15) = -30\text{ m/s} Magnitude of change=30 m/s towards West\text{Magnitude of change} = 30\text{ m/s towards West}

Explanation:

Even though the speed remains constant, the velocity changes because the direction changes. The vector subtraction yields a total change of 30 m/s30\text{ m/s}.