Motion in a Plane · Lesson 13 of 13

Chapter Summary and Practice

“Connect vector algebra, plane kinematics, projectile motion and circular motion through worked and conceptual practice.”

Learning Objectives

• Choose an appropriate vector or component method. • Match a motion formula to its assumptions. • Check direction, dimensions, limiting cases and geometry. • Explain the differences among speed, velocity and acceleration in a plane.

A practical route through a plane-motion problem

Begin with a sketch, a reference frame, axes and a clock zero. Mark the initial position and velocity, then identify the acceleration vector or the frame velocities being combined. Resolve vectors into components and keep one time for each event. Calculate the requested coordinate, vector or magnitude, and state the physical direction of any vector answer. The sketch is part of the model: a ceiling, landing height, current or observation geometry can change which formula is valid.

Idea	Relationship	Condition or reminder
Vector magnitude	A = √(A_x² + A_y² + A_z²)	Perpendicular Cartesian components
Direction in a plane	tanθ = A_y/A_x	Choose quadrant using signs
Vector addition	R_i = A_i + B_i	Common axes and compatible physical quantities
Relative velocity	v_A/B = v_A/G − v_B/G	Both right-hand velocities in the same frame
Instantaneous velocity	v = dr/dt	Tangent to a smooth path at nonzero speed
Instantaneous acceleration	a = dv/dt	Can change speed, direction or both
Constant acceleration	v = u + at; r = r₀ + ut + ½at²	The full acceleration vector must be constant

Quantity	SI unit	Dimension
Position, displacement, path length	m	[L]
Velocity and speed	m/s	[L T⁻¹]
Acceleration	m/s²	[L T⁻²]
Time and period	s	[T]
Angle	rad (dimensionless)	[1]
Frequency and angular speed	Hz and rad/s respectively	[T⁻¹]
Unit vector	No physical unit	[1]

Projectile equations and their domain

Projectile component modelLaTeX

x = x_{0} + u_{x} t, y = y_{0} + u_{y} t - \frac{1}{2} g t^{2}, v_{x} = u_{x}, v_{y} = u_{y} - g t

Constant downward gravity and negligible drag. These equations handle unequal launch and landing heights.

Return to launch heightLaTeX

T = \frac{2 u sin θ}{g}, H = \frac{u ^{2} sin ^{2} θ}{2 g}, R = \frac{u ^{2} sin 2 θ}{g}

T and R here require landing at the launch height. H is the height gained above launch, not necessarily the height above ground.

Circular-motion equations and their domain

Uniform circular motionLaTeX

v = R ω, ω = 2 π f = \frac{2 π}{T}, a_{c} = \frac{v ^{2}}{R} = R ω^{2}

Velocity is tangent. With constant speed, total acceleration is radial inward. The acceleration vector changes direction around the circle.

Mixed worked practice

Practice: combine and normalise vectors

Problem
Let A = 6 i-hat − 2 j-hat m and B = −2 i-hat + 5 j-hat m. Find their sum and its unit direction.

1.R = 4 i-hat + 3 j-hat m. Its magnitude is 5 m.
2.The unit direction is (4/5)i-hat + (3/5)j-hat, which is dimensionless.
3.Direction from +x is arctan(3/4) = 36.87°. Its squared unit components add to 16/25 + 9/25 = 1.

Practice: a constant-acceleration event

Problem
A particle starts at r₀ = 2 i-hat + 3 j-hat m with u = 4 i-hat m/s and a = −2 j-hat m/s². Find r and v after 2 s.

1.Position is r = (2 + 4 × 2)i-hat + (3 − ½ × 2 × 2²)j-hat = 10 i-hat − j-hat m.
2.Velocity is 4 i-hat − 4 j-hat m/s. Speed is 4√2 = 5.657 m/s.
3.The displacement is 8 i-hat − 4 j-hat m, different from the final position vector because the initial position was nonzero.

Practice: reconstruct a projectile

Problem
A ball is launched with components u_x = 15 m/s and u_y = 20 m/s from ground level. Take g = 10 m/s² and level landing.

1.Launch speed is √(15² + 20²) = 25 m/s and angle is 53.13° above horizontal.
2.The peak time is 20/10 = 2 s and height is 20²/(2 × 10) = 20 m.
3.Flight time is 4 s and range is 15 × 4 = 60 m.
4.At 2 s, velocity is 15 i-hat m/s, not zero. At 4 s it is 15 i-hat − 20 j-hat m/s, with the same speed as at launch.

Practice: circle averages and instantaneous values

Problem
A particle goes around a 2 m radius circle at 4 m/s. Compare one-revolution average velocity, average acceleration and instantaneous acceleration.

1.The period is 2πR/v = π s.
2.One revolution returns both position and velocity to their initial values, so average velocity and average acceleration are zero vectors.
3.At each instant, acceleration magnitude is 4²/2 = 8 m/s² inward. Average speed is 4 m/s.
4.Zero vector averages do not mean that the particle was at rest or unaccelerated during the interval.

Concept checks with reasons

Claim	Assessment
A vector magnitude is a scalar	True: it has no spatial direction.
Every vector component is a scalar physical quantity independent of axis orientation	False: a signed scalar component generally changes when axes rotate; a component vector is itself a vector.
Three noncoplanar vectors can sum to zero	False: the negative sum of two lies in their plane.
A scalar can never be negative	False: potential energy relative to a chosen zero can be negative.
The acceleration of a projectile vanishes at its peak	False: gravity remains downward.
All circular-motion acceleration is purely radial	False: varying speed adds a tangential component.
Constant acceleration magnitude justifies r = r₀ + ut + ½at²	False: its direction must remain fixed as well.
Velocity addition and relative velocity use the same arithmetic operation	False: connected-frame addition and common-frame subtraction answer different questions.

Check before accepting an answer

For a resultant, compare the magnitude with the sum and difference bounds. For a projectile, confirm that its calculated landing time satisfies the actual vertical equation. For a circular motion result, check that the acceleration units are length per time squared and its direction is stated. In relative motion, write frame labels before combining velocities. A calculation can be arithmetically correct and still describe the wrong motion if its assumptions or geometry do not match the question.

To practise, solve the worked questions again while covering the solutions. Draw one arrow diagram or spatial trajectory for each. Then verify with a second representation: components versus a triangle for vectors, vertical-motion timing versus a trajectory equation for a projectile, or v²/R versus ω²R for circular motion. Explaining why two methods agree is a useful check of understanding.

Quick check

Which situation permits the constant-acceleration vector position equation over the full interval?

Key Takeaways

• Resolve vectors into consistent components and reconstruct their magnitudes and directions. • Use one common time for simultaneous coordinate motions. • Check assumptions before using range, flight-time or constant-acceleration formulas. • Uniform circular motion has changing velocity and a rotating inward acceleration. • State the reference frame and geometry whenever they affect the result.

Previous · Lesson 12

Angular Speed, Period and Circular-Motion Problems

End of chapter