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Motion in a Plane

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Chapter Notes

Overview
This chapter extends the concepts of motion from one dimension to two dimensions (a plane) using vectors. It introduces scalars and vectors, vector addition and subtraction, resolution of vectors, and analytical methods for vector addition. The key takeaway is that motion in a plane can be treated as two independent simultaneous one-dimensional motions, which is applied to projectile motion and uniform circular motion.
Key Concepts (16)
Scalar quantity
A quantity that has magnitude only and is specified completely by a single number along with the proper unit.
e.g. Mass of an object, temperature of a body, distance between two points.
Vector quantity
A quantity that has both magnitude and direction and obeys the triangle law of addition or equivalently the parallelogram law of addition.
e.g. Displacement, velocity, acceleration, force.
Position vector
A vector that gives the position of an object with reference to the origin of a coordinate system.
e.g. For a point P with coordinates (x, y), the position vector is r = x î + y ĵ.
Displacement vector
The vector that gives the change in position of an object, directed from the initial position to the final position.
e.g. If an object moves from P to P', the displacement vector is PP' = r' - r.
Equality of vectors
Two vectors A and B are said to be equal if, and only if, they have the same magnitude and the same direction.
e.g. Two forces of 5 N each acting eastward are equal vectors.
Null vector (zero vector)
A vector with zero magnitude and unspecified direction, obtained by adding two equal and opposite vectors or multiplying a vector by zero.
e.g. The displacement of an object that returns to its starting point is a null vector.
Unit vector
A vector of unit magnitude used to specify a direction only; it has no dimension and unit.
e.g. î, ĵ, and k̂ are unit vectors along the x-, y-, and z-axes respectively.
Resolution of vectors
The process of expressing a vector as the sum of two or more component vectors along specified directions.
e.g. A vector A in the x-y plane can be resolved as A = Ax î + Ay ĵ.
Component of a vector
The effective value of a vector along a particular axis, given by the product of its magnitude and the cosine of the angle it makes with that axis.
e.g. Ax = A cos θ, Ay = A sin θ.
Projectile
An object that is in flight after being thrown or projected, under the action of gravity alone (neglecting air resistance).
e.g. A cricket ball thrown in the air, a football kicked from the ground.
Trajectory
The path followed by a projectile, which is a parabola under ideal conditions.
e.g. The curved path of a stone thrown horizontally from a cliff.
Time of flight
The total time during which a projectile is in flight from launch to the point where it returns to the same vertical level.
e.g. For a ball thrown upwards, the time from throw to catch.
Maximum height
The greatest vertical height reached by a projectile above its launch point.
e.g. The highest point of a javelin's path.
Horizontal range
The horizontal distance travelled by a projectile from its launch point to the point where it returns to the same vertical level.
e.g. The distance a cannonball travels before hitting the ground.
Uniform circular motion
Motion of an object along a circular path with constant speed.
e.g. A stone tied to a string and whirled in a horizontal circle at constant speed.
Centripetal acceleration
The acceleration of an object in uniform circular motion, always directed towards the centre of the circle, with magnitude v²/R.
e.g. The acceleration of a satellite in a circular orbit around Earth.
Formulas & Equations (19)
A = |A| n̂
Reads as: Vector A equals magnitude of A times unit vector n hat
Variables: A = vector; |A| = magnitude of A; n̂ = unit vector along A
Use when: Expressing any vector in terms of its magnitude and direction.
⚠ Forgetting that n̂ is a unit vector, not a scalar.
A = Ax î + Ay ĵ
Reads as: Vector A equals A x i hat plus A y j hat
Variables: Ax = x-component of A; Ay = y-component of A; î, ĵ = unit vectors along x and y axes
Use when: Resolving a vector into its rectangular components in a plane.
⚠ Writing A = Ax + Ay (missing unit vectors).
Ax = A cos θ, Ay = A sin θ
Reads as: A x equals A cos theta, A y equals A sin theta
Variables: A = magnitude of vector; θ = angle with x-axis; Ax, Ay = components
Use when: Finding components when magnitude and direction are given.
⚠ Using sin for x-component and cos for y-component.
A = √(Ax² + Ay²)
Reads as: A equals square root of A x squared plus A y squared
Variables: A = magnitude; Ax, Ay = components
Use when: Finding magnitude from components.
⚠ Forgetting to take the square root.
tan θ = Ay / Ax
Reads as: tan theta equals A y over A x
Variables: θ = angle with x-axis; Ax, Ay = components
Use when: Finding direction from components.
⚠ Not considering the quadrant of the vector.
R = A + B = (Ax + Bx) î + (Ay + By) ĵ
Reads as: R equals A plus B equals (A x plus B x) i hat plus (A y plus B y) j hat
Variables: R = resultant vector; A, B = vectors; Ax, Bx, Ay, By = components
Use when: Adding vectors analytically using components.
⚠ Adding magnitudes directly without considering direction.
R² = A² + B² + 2AB cos θ
Reads as: R squared equals A squared plus B squared plus 2 A B cos theta
Variables: R = magnitude of resultant; A, B = magnitudes; θ = angle between A and B
Use when: Finding magnitude of resultant of two vectors using law of cosines.
⚠ Using sin instead of cos.
R / sin(180° - θ) = A / sin α = B / sin β
Reads as: R over sin(180 minus theta) equals A over sin alpha equals B over sin beta
Variables: R, A, B = magnitudes; θ = angle between A and B; α, β = angles opposite to A and B
Use when: Applying law of sines to vector triangles.
⚠ Using the wrong angle opposite to the side.
v = v0 + a t
Reads as: v equals v zero plus a t
Variables: v = velocity at time t; v0 = initial velocity; a = constant acceleration; t = time
Use when: Finding velocity in motion with constant acceleration.
⚠ Using this for non-constant acceleration.
r = r0 + v0 t + ½ a t²
Reads as: r equals r zero plus v zero t plus half a t squared
Variables: r = position at time t; r0 = initial position; v0 = initial velocity; a = constant acceleration; t = time
Use when: Finding position in motion with constant acceleration.
⚠ Forgetting the factor ½.
x = (v0 cos θ0) t
Reads as: x equals v zero cos theta zero times t
Variables: x = horizontal displacement; v0 = initial speed; θ0 = projection angle; t = time
Use when: Finding horizontal position of a projectile.
⚠ Using sin instead of cos for horizontal component.
y = (v0 sin θ0) t - ½ g t²
Reads as: y equals v zero sin theta zero t minus half g t squared
Variables: y = vertical displacement; v0 = initial speed; θ0 = projection angle; g = acceleration due to gravity; t = time
Use when: Finding vertical position of a projectile.
⚠ Forgetting the negative sign for g.
y = x tan θ0 - (g x²) / (2 v0² cos² θ0)
Reads as: y equals x tan theta zero minus g x squared over 2 v zero squared cos squared theta zero
Variables: y = vertical displacement; x = horizontal displacement; θ0 = projection angle; g = acceleration due to gravity; v0 = initial speed
Use when: Finding the equation of the parabolic trajectory of a projectile.
⚠ Mixing up the terms or forgetting the square on cos.
tm = (v0 sin θ0) / g
Reads as: t m equals v zero sin theta zero over g
Variables: tm = time to reach maximum height; v0 = initial speed; θ0 = projection angle; g = acceleration due to gravity
Use when: Finding the time at which projectile reaches its highest point.
⚠ Using cos instead of sin.
Tf = 2 (v0 sin θ0) / g
Reads as: T f equals 2 v zero sin theta zero over g
Variables: Tf = time of flight; v0 = initial speed; θ0 = projection angle; g = acceleration due to gravity
Use when: Finding total time a projectile is in the air.
⚠ Forgetting the factor of 2.
hm = (v0² sin² θ0) / (2g)
Reads as: h m equals v zero squared sin squared theta zero over 2 g
Variables: hm = maximum height; v0 = initial speed; θ0 = projection angle; g = acceleration due to gravity
Use when: Finding the maximum height reached by a projectile.
⚠ Forgetting the square on sin or the factor 2 in denominator.
R = (v0² sin 2θ0) / g
Reads as: R equals v zero squared sin 2 theta zero over g
Variables: R = horizontal range; v0 = initial speed; θ0 = projection angle; g = acceleration due to gravity
Use when: Finding the horizontal range of a projectile.
⚠ Using sin θ0 instead of sin 2θ0.
Rm = v0² / g (for θ0 = 45°)
Reads as: R m equals v zero squared over g for theta zero equals 45 degrees
Variables: Rm = maximum horizontal range; v0 = initial speed; g = acceleration due to gravity
Use when: Finding the maximum possible range for a given initial speed.
⚠ Forgetting that this is only valid at 45°.
a = v² / R (centripetal acceleration)
Reads as: a equals v squared over R
Variables: a = centripetal acceleration; v = speed; R = radius of circular path
Use when: Finding the acceleration of an object in uniform circular motion.
⚠ Using v instead of v².
How It Works
Vector addition using the triangle (head-to-tail) method
  1. Place vector B such that its tail is at the head of vector A.
  2. Draw a vector from the tail of A to the head of B.
  3. This new vector is the resultant R = A + B.
💡 The order of addition does not matter (commutative law: A + B = B + A).
Vector addition using the parallelogram method
  1. Bring the tails of vectors A and B to a common origin.
  2. Draw a line from the head of A parallel to B.
  3. Draw a line from the head of B parallel to A to complete a parallelogram.
  4. The diagonal from the common origin to the opposite vertex represents the resultant R.
💡 This method is equivalent to the triangle method.
Resolution of a vector into rectangular components
  1. Draw perpendicular lines from the head of the vector to the x- and y-axes.
  2. The component along x-axis is Ax = A cos θ, represented as Ax î.
  3. The component along y-axis is Ay = A sin θ, represented as Ay ĵ.
  4. The vector is then A = Ax î + Ay ĵ.
💡 The component can be positive, negative, or zero depending on the angle θ.
Analytical method of vector addition
  1. Resolve each vector into its x- and y-components.
  2. Add the x-components: Rx = Ax + Bx.
  3. Add the y-components: Ry = Ay + By.
  4. The resultant vector is R = Rx î + Ry ĵ.
  5. Magnitude: R = √(Rx² + Ry²).
  6. Direction: tan θ = Ry / Rx.
💡 This method is more accurate than the graphical method.
Projectile motion analysis
  1. Resolve initial velocity v0 into horizontal (v0 cos θ0) and vertical (v0 sin θ0) components.
  2. Treat horizontal motion as uniform (ax = 0) and vertical motion as uniformly accelerated (ay = -g).
  3. Use equations of motion separately for x and y directions.
  4. Eliminate time t to get the equation of the parabolic path.
  5. Calculate time of flight, maximum height, and range using derived formulas.
💡 The horizontal and vertical motions are independent of each other.
Derivation of centripetal acceleration in uniform circular motion
  1. Consider an object moving with constant speed v in a circle of radius R.
  2. At two nearby points P and P', the position vectors r and r' and velocity vectors v and v' are perpendicular to them.
  3. The triangle formed by r and r' is similar to the triangle formed by v, v', and Δv.
  4. From similarity: |Δv| / v = |Δr| / R.
  5. For small Δt, |Δr| ≈ v Δt, so |Δv| ≈ (v² Δt) / R.
  6. Acceleration a = lim(Δt→0) |Δv|/Δt = v² / R, directed towards the centre.
💡 The acceleration is always directed towards the centre of the circle, hence called centripetal acceleration.
Exam Traps — Don't Make These Mistakes
⚠ Confusing scalar and vector addition
✗ Adding magnitudes of two vectors directly to get the magnitude of the resultant, e.g., R = A + B.
→ Vectors have direction; the magnitude of the resultant depends on the angle between them via the law of cosines.
✓ Use R² = A² + B² + 2AB cos θ to find the magnitude of the resultant.
⚠ Using wrong trigonometric function for components
✗ Writing Ax = A sin θ and Ay = A cos θ.
→ The x-component is adjacent to the angle θ with the x-axis, so it uses cos, not sin.
✓ Write Ax = A cos θ and Ay = A sin θ.
⚠ Forgetting the negative sign for gravity in projectile motion
✗ Writing y = (v0 sin θ0) t + ½ g t².
→ Gravity acts downward, so acceleration in the vertical direction is -g (taking upward as positive).
✓ Write y = (v0 sin θ0) t - ½ g t².
⚠ Mixing up time of flight and time to maximum height
✗ Using Tf = (v0 sin θ0)/g for total time of flight.
→ Time of flight is twice the time to reach maximum height, because the projectile goes up and then down.
✓ Use Tf = 2(v0 sin θ0)/g for total time of flight, and tm = (v0 sin θ0)/g for time to maximum height.
⚠ Using wrong angle in range formula
✗ Writing R = (v0² sin θ0)/g.
→ The range formula involves sin 2θ0, not sin θ0.
✓ Write R = (v0² sin 2θ0)/g.
⚠ Confusing centripetal acceleration with tangential acceleration
✗ Stating that acceleration in uniform circular motion is tangential.
→ In uniform circular motion, speed is constant, so there is no tangential acceleration; the acceleration is centripetal (towards the centre).
✓ The acceleration in uniform circular motion is centripetal, directed towards the centre, with magnitude v²/R.
Compare & Contrast
Direction
Scalar quantity
Vector quantity
A scalar has only magnitude, while a vector has both magnitude and direction and obeys triangle/parallelogram law of addition.
Time interval
Average velocity
Instantaneous velocity
Average velocity is total displacement divided by total time interval, while instantaneous velocity is the limit of average velocity as the time interval approaches zero, giving velocity at a specific instant.
Acceleration
Projectile motion
Uniform circular motion
In projectile motion, acceleration is constant (g, downward), while in uniform circular motion, acceleration is variable in direction (always towards centre) but constant in magnitude (v²/R).
Exam Focus
Marks pattern: 2 MCQs (1 mark each), 1 SA-I (2 marks), 1 SA-II (3 marks), 1 LA (5 marks) — total 12 marks
Most asked: Resolution of vectors into components · Projectile motion: time of flight, maximum height, range · Uniform circular motion: centripetal acceleration · Vector addition using analytical method
Don't skip: Derivation of range and maximum height of a projectile; Numerical problems on projectile motion (e.g., finding time of flight, range, speed at impact); Derivation of centripetal acceleration
One-Line Revision (read aloud before the exam)
  1. Scalars have only magnitude; vectors have magnitude and direction and obey triangle/parallelogram law of addition.
  2. A vector multiplied by a positive scalar changes magnitude but not direction; a negative scalar reverses direction.
  3. Vector addition is commutative (A + B = B + A) and associative [(A + B) + C = A + (B + C)].
  4. A null vector has zero magnitude and unspecified direction.
  5. A unit vector has magnitude 1 and specifies direction only.
  6. Any vector A can be resolved into components: A = Ax î + Ay ĵ, where Ax = A cos θ and Ay = A sin θ.
  7. The magnitude of a vector from its components is A = √(Ax² + Ay²).
  8. The direction of a vector from its components is given by tan θ = Ay / Ax.
  9. The resultant of two vectors A and B has magnitude R = √(A² + B² + 2AB cos θ).
  10. Motion in a plane with constant acceleration can be treated as two independent 1D motions.
  11. The path of a projectile is a parabola given by y = x tan θ0 - (g x²)/(2 v0² cos² θ0).
  12. Time of flight of a projectile is Tf = 2 v0 sin θ0 / g.
  13. Maximum height of a projectile is hm = (v0² sin² θ0) / (2g).
  14. Horizontal range of a projectile is R = (v0² sin 2θ0) / g, maximum at θ0 = 45°.
  15. In uniform circular motion, centripetal acceleration a = v² / R, directed towards the centre.
  16. For a given initial speed, ranges are equal for projection angles that are complementary (e.g., 30° and 60°).

MCQs — 10 Questions with Answers

Q1. Which of the following is a vector quantity?

  • A. Mass of an object
  • B. Temperature of a body
  • C. Displacement of a car ✓
  • D. Distance travelled along a road

Answer: C — Displacement has both magnitude and direction (vector), while mass, temperature, and distance are scalars with magnitude only.

Q2. Two vectors A and B have equal magnitudes. Under what condition will their resultant have magnitude equal to A?

  • A. When A and B are perpendicular to each other
  • B. When A and B are at 120° to each other ✓
  • C. When A and B are parallel (same direction)
  • D. When A and B are anti-parallel (opposite directions)

Answer: B — For equal magnitude vectors, R = √(A² + B² + 2AB cos θ); setting R = A gives cos θ = −1/2, so θ = 120°.

Q3. A vector A of magnitude 10 units is multiplied by −0.5. What is the magnitude and direction of the resultant vector?

  • A. Magnitude 5 units, same direction as A
  • B. Magnitude 5 units, opposite direction to A ✓
  • C. Magnitude 10 units, same direction as A
  • D. Magnitude 20 units, opposite direction to A

Answer: B — Multiplying by negative scalar −0.5 reverses direction and scales magnitude: |−0.5A| = 0.5 × 10 = 5 units in opposite direction.

Q4. A projectile is launched at an angle of 30° to the horizontal with initial speed 20 m/s. What is the horizontal component of velocity during its flight?

  • A. 10 m/s
  • B. 10√3 m/s ✓
  • C. 20 m/s
  • D. 5 m/s

Answer: B — Horizontal velocity vₓ = u cos 30° = 20 × (√3/2) = 10√3 m/s, and remains constant throughout flight as no horizontal acceleration exists.

Q5. A stone is thrown horizontally from the top of a cliff with velocity 15 m/s. After 2 seconds, what is the magnitude of its velocity? (Take g = 10 m/s²)

  • A. 15 m/s
  • B. 20 m/s
  • C. 25 m/s ✓
  • D. 30 m/s

Answer: C — vₓ = 15 m/s (constant), vᵧ = gt = 10 × 2 = 20 m/s; total velocity v = √(15² + 20²) = √(225 + 400) = 25 m/s.

Q6. A projectile has maximum range when its angle of projection is 45°. Which of the following statements is NOT correct regarding this condition?

  • A. The initial velocity components are equal: u cos 45° = u sin 45°
  • B. The time of flight equals the time to reach maximum height
  • C. The range is maximum only if the launch and landing heights are equal
  • D. The maximum range is independent of the initial velocity ✓

Answer: D — Range R = (u² sin 2θ)/g directly depends on u²; greater initial velocity gives greater maximum range (R ∝ u²).

Q7. In uniform circular motion, a particle moves in a circle at constant speed. Which statement best describes the acceleration?

  • A. Acceleration is zero because speed is constant
  • B. Acceleration is directed toward the center and has constant magnitude v²/r ✓
  • C. Acceleration is tangent to the circle in the direction of motion
  • D. Acceleration is directed away from the center (centrifugal)

Answer: B — Although speed is constant, velocity direction changes continuously; centripetal acceleration aᶜ = v²/r points toward center with constant magnitude.

Q8. Assertion (A): In projectile motion, the velocity at maximum height is zero. Reason (R): At maximum height, the vertical component of velocity becomes zero. Which of the following is correct?

  • A. Both A and R are true, and R is the correct explanation of A
  • B. Both A and R are true, but R is not the correct explanation of A
  • C. A is true, but R is false
  • D. A is false, but R is true ✓

Answer: D — The vertical component vᵧ = 0 at maximum height, but horizontal component vₓ ≠ 0, so total velocity is NOT zero; only R is correct.

Q9. Two projectiles are launched with the same initial speed but at angles 30° and 60° to the horizontal. The ratio of their maximum heights is:

  • A. 1 : 1
  • B. 1 : 3 ✓
  • C. 3 : 1
  • D. √3 : 1

Answer: B — H = (u² sin² θ)/(2g); H₁/H₂ = sin² 30° / sin² 60° = (1/2)² / (√3/2)² = (1/4) / (3/4) = 1/3.

Q10. A car moves in a horizontal circle of radius 50 m with constant speed 20 m/s. The centripetal force required is 4000 N. What is the mass of the car?

  • A. 500 kg ✓
  • B. 1000 kg
  • C. 2000 kg
  • D. 250 kg

Answer: A — From Fᶜ = mv²/r: 4000 = m × 20² / 50 = m × 400/50 = 8m; solving gives m = 500 kg.

Flashcards

What is the key difference between a scalar quantity and a vector quantity?

A scalar has only magnitude (e.g., mass, temperature), while a vector has both magnitude and direction (e.g., displacement, velocity, force).

State the condition for two vectors A and B to be equal.

Two vectors are equal if and only if they have the same magnitude and the same direction.

If a vector A is multiplied by a negative number −λ, what happens to the vector?

The resultant vector has magnitude λ times |A| but points in the opposite direction to A.

Why is displacement vector independent of the path taken by an object?

Because displacement is defined as the straight-line vector joining initial and final positions, regardless of the actual route followed.

What does the parallelogram law of vector addition state?

If two vectors are represented by adjacent sides of a parallelogram, their resultant is represented by the diagonal of that parallelogram.

In projectile motion, why does horizontal velocity remain constant?

Because there is no acceleration in the horizontal direction; only gravity acts vertically downward.

At the maximum height of a projectile, what is the value of vertical velocity component vᵧ?

The vertical velocity component is zero (vᵧ = 0), but the horizontal component vₓ remains non-zero.

Define centripetal acceleration in uniform circular motion.

Centripetal acceleration is the acceleration directed toward the center of the circular path, with magnitude aᶜ = v²/r = ω²r.

In uniform circular motion, is the speed of the object constant or changing?

The speed (magnitude of velocity) is constant, but velocity direction changes continuously, so acceleration is non-zero.

What is the relationship between angular velocity ω and linear velocity v in circular motion?

The linear velocity v is related to angular velocity ω by the equation v = ωr, where r is the radius of the circular path.

Important Board Questions

Define displacement vector. Why is displacement independent of the path taken between two points? Give one example. [2 marks]

Displacement = straight-line vector from initial to final position only. Use diagram showing same displacement for different paths (e.g., walking PABCQ vs direct line PQ).

A projectile is fired with initial velocity 50 m/s at an angle of 37° above the horizontal. Calculate (i) the time to reach maximum height, (ii) the maximum height reached, and (iii) the horizontal range. (Take g = 10 m/s², sin 37° = 0.6, cos 37° = 0.8) [5 marks]

Resolve velocity into components: uₓ = u cos 37°, uᵧ = u sin 37°. At max height, vᵧ = 0; use vᵧ = uᵧ − gt. For range, use R = (u² sin 2θ)/g or separate x and y equations.

Derive the equation for centripetal acceleration in uniform circular motion. Show that aᶜ = v²/r = ω²r. Explain why this acceleration is always directed toward the center and why it does not change the speed of the particle. [6 marks]

Use velocity change method: Δv points toward center; as Δt → 0, aᶜ = lim(Δv/Δt) = v²/r. Centripetal force ⊥ to velocity means no work done; kinetic energy (speed) unchanged. Connect to angular velocity ω = v/r.

Assertion-Reason Questions for Motion in a Plane

Q1. Assertion (A):

Temperature and density are scalar quantities because they can be completely specified by a single number with proper unit.

Reason (R):

Scalar quantities obey the rules of ordinary algebra and can be added, subtracted, multiplied and divided.

Answer: B — Both A and R are true, but R is NOT the correct explanation of A — The assertion is correct—temperature (35.6°C) and density (2.7×10³ kg/m³) are scalars with magnitude only. The reason is also true (scalars follow algebraic rules), but it doesn't explain WHY they are scalars. The real reason is lack of direction, not the algebraic rules. Hence, reason does not explain the assertion.

Q2. Assertion (A):

A displacement vector from P to Q has the same value regardless of the actual path taken by the moving object.

Reason (R):

Displacement is defined as the straight-line vector joining initial and final positions, which is independent of the trajectory followed.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both assertion and reason are true. The reason directly explains the assertion—the definition of displacement as a straight-line vector (not path length) is precisely why it remains the same for all paths connecting P to Q. This is a fundamental concept of vectors in kinematics.

Q3. Assertion (A):

Two vectors A and B with |A| = |B| are always equal.

Reason (R):

Equality of vectors requires both the same magnitude and the same direction.

Answer: D — A is false but R is true — The assertion is false. Two vectors with equal magnitudes but different directions are NOT equal. For example, a 10 N force to the east and 10 N force to the north have |A| = |B| but A ≠ B. The reason is true but irrelevant since it contradicts the assertion.

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Case-Based Questions (CBQ) for Motion in a Plane

Case Study:

A surveyor in New Delhi uses GPS technology to map a new residential plot. The surveyor marks the initial corner at position O (origin). Moving 400 m due east reaches point A, then 300 m due north reaches point B (final corner of the plot). A construction vehicle travels from O to B via a different route: O to A (400 m east) then from A to B (300 m north). Calculate both the displacement from O to B and the total distance traveled by the vehicle to understand why property boundaries use straight-line displacement rather than curved paths.

Q1. What is the magnitude of the displacement vector from O to B?

  • A. 700 m
  • B. 500 m ✓
  • C. 1200 m
  • D. 100 m

Answer: B

Q2. The total distance traveled by the vehicle from O to B via point A is:

  • A. 500 m
  • B. 400 m
  • C. 700 m ✓
  • D. 300 m

Answer: C

Q3. Why is the displacement (straight-line OB) used for property boundaries rather than the actual path traveled?

  • A. Displacement is shorter and therefore costs less for fencing
  • B. Displacement depends only on initial and final positions, not on the path, making it unique and independent
  • C. Distance changes with different routes, but displacement remains constant
  • D. Both B and C are correct ✓

Answer: D

Q4. If the vehicle could travel diagonally from O directly to B instead of via A, how would the distance compare to the path O→A→B?

  • A. The diagonal distance would be greater than 700 m
  • B. The diagonal distance would be exactly 500 m, equal to displacement magnitude ✓
  • C. The diagonal distance would be 400 m
  • D. The diagonal and path distance would be equal

Answer: B

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