Chapter 03 · Vectors & Motion in a Plane · Full book

Motion in a Plane: Breaking Free from the Line

Core intuition: Chapter Roadmap

Vectors as a language — why a single signed number cannot describe motion in a plane, and how an arrow (geometric) or a pair of components (algebraic) repairs that.
Position, velocity, and acceleration as vectors — the three kinematic quantities promoted from numbers to \(\hat{\imath},\hat{\jmath}\) form, with the constant-acceleration equations written once as a single vector equation and then split into independent component equations.
Independence of perpendicular motions — the structural backbone of the chapter: horizontal and vertical motions evolve separately, coupled only by the shared clock \(t\).
Projectile motion — trajectory, time of flight, maximum height, and range derived rather than memorised, then extended to horizontal and oblique launches, motion on an incline, and the meeting of two projectiles.
Relative motion in two dimensions — velocity composition and the moving-frame picture, with classics such as the tilted umbrella and the river crossing.
Vector algebra — components, magnitude and unit vectors, and the dot product as the tool that extracts work, projections, and the angle between two directions.

Perplexing Questions

Perplexing questions

Vectors are not just arrows; they are a way of thinking. Let these questions simmer in your mind.

The Lazy Walk: Is it possible to walk \(10\,\mathrm{km}\) and end up exactly where you started? If yes, what was your average velocity?
The Impossible Sum: Can the sum of two vectors of magnitude \(3\) and \(4\) ever be equal to \(1\)? Can it be equal to \(8\)?
The Wind Trap: An airplane flies North at \(100\,\mathrm{km/h}\). A wind blows from the West at \(100\,\mathrm{km/h}\). Does the wind make the plane reach its destination faster, slower, or at the same time? (Careful!)
The Component Paradox: Can a vector have a magnitude of zero if one of its components is non-zero?
Relative Rain: If rain is falling vertically, why do you have to tilt your umbrella forward when you walk? Does the rain actually change direction, or is it an illusion?

Why 1D Was Boring (And Why 2D is Real)

In Chapter 1, our world was a straight line. Everything was trapped on a railway track. You could go forward (\(+\)) or backward (\(-\)), but you could never turn.

But nature hates straight lines.

A cricket ball curves through the air.
A planet orbits in an ellipse.
A car navigates a winding mountain road.

Welcome to 2D Motion. Now, we are off the rails. We are on a football field. You can run North, East, or even North-East (to meditate in Himalayas!?).

The Price of Freedom

This freedom creates a problem. In 1D, a simple “plus” or “minus” sign was enough to tell the direction.

\(+5\,\mathrm{m/s}\) meant Right.
\(-5\,\mathrm{m/s}\) meant Left.

In 2D, a sign is not enough. If I say “I am walking at \(-5\,\mathrm{m/s}\),” you don't know if I'm walking West, South, or somewhere in between.

To survive in this new world, we need a new mathematical tool. A tool that carries direction as a permanent part of its identity. That tool is the Vector.

A ball leaves your hand. It moves forward and falls simultaneously.
Neither story alone describes the flight; both run in parallel, coupled only by the clock.
This chapter teaches you to decompose, solve, and reassemble planar motion.

Learning Objectives. After working through this chapter you will be able to:

Write the position, velocity, and acceleration of a particle in vector form and interpret each geometrically.
Apply the constant-acceleration kinematic equations as a single vector equation, then separate them into independent component equations.
State and exploit the principle that perpendicular components of motion are independent.
Derive the trajectory, time of flight, maximum height, and range of a projectile from the vector equations—not from memorised formulas.
Analyse horizontal projection and oblique projection as special cases of the same vector framework.

Why 2D Is Not Just “1D Twice”

In one dimension, a particle can only advance or retreat along a fixed line; a sign (\(+\) or \(-\)) encodes the entire directional content. In a plane, the particle is free to curve, loop, and change heading continuously. A single signed number is no longer enough—we need a vector at every instant to specify where the particle is, how fast it is moving, and how its velocity is changing.

The essential new idea is not that there are more equations (there are), but that the equations along perpendicular axes do not talk to each other. Gravity pulls a projectile downward; it does nothing to the horizontal motion. This independence is the structural backbone of everything in this chapter. It converts every 2D problem into two simultaneous but separate 1D problems, linked only by the shared parameter \(t\).

Forces push, velocities point, displacements connect. None of these can be captured by a single number. This chapter equips you with the language that makes direction precise.

Learning Objectives. After working through this chapter, you will be able to:

Explain why magnitude alone is insufficient to describe physical quantities like force and displacement.
Represent a 2D vector geometrically (as an arrow) and algebraically (in component form).
Add and subtract vectors using both the head-to-tail rule and component arithmetic.
Compute the magnitude and direction of any 2D vector from its components.
Use the dot product to extract physically meaningful quantities—work, projection, and the angle between two directions.

Why Vectors Are Necessary in Physics

Consider a simple question: you push a heavy crate with a force of \(50\;\mathrm{N}\). Does the crate accelerate to the right? Upward? At \(37^\circ\) to the floor? The number \(50\) tells you how hard you push, but without a direction the physics is incomplete. Newton's second law, \(\vec{F}=m\vec{a}\), is a vector equation—both sides carry direction. If we throw direction away, we lose the ability to predict where things go.

The same issue appears with displacement. Suppose you walk \(5\;\mathrm{km}\) and then \(3\;\mathrm{km}\). How far are you from the starting point? It depends entirely on the directions of the two walks. If both are due north, the answer is \(8\;\mathrm{km}\). If the second walk reverses direction, the answer is \(2\;\mathrm{km}\). At right angles, it is \(\sqrt{34}\;\mathrm{km}\). Scalars (pure numbers with units) cannot encode this directional dependence; vectors can.

Vectors are not a mathematical luxury imported into physics for elegance. They are the minimum language required to state the laws of mechanics honestly. Every equation we write from this point onward—displacement, velocity, acceleration, force, momentum—will be a statement about vectors.

The Geometric Meaning of a Vector

◈ Vector addition — triangle & parallelogram

live bench

Drag either glowing handle to reshape a and b; the parallelogram completes itself and the resultant glows. Flip to subtraction and watch a−b reach from head to head.

◈ loading the interactive bench…

↗ open in the Simulations page

A vector is a quantity that possesses both a magnitude (a non-negative real number) and a direction in space. Geometrically, we represent it as a directed line segment—an arrow—characterised by four things: a tail (starting point), a head (ending point), a length (magnitude), and an orientation (direction).

Equality of Vectors

Two vectors \(\vec{A}\) and \(\vec{B}\) are equal if and only if they have the same magnitude and the same direction. Their physical locations in space are irrelevant. A displacement of \(3\;\mathrm{m}\) due east is the same vector whether it starts at Delhi or at Tokyo. This is the defining property of free vectors: they are characterised entirely by length and direction, not by where they sit.

You can slide a free vector anywhere in the plane without changing the physics. This is what makes the parallelogram law of addition possible: we are allowed to reposition arrows tail-to-tail or head-to-tail because position does not alter identity.

Free Vectors vs. Bound Vectors

Most vectors you encounter in introductory mechanics—displacement, velocity, acceleration—are free vectors. However, some physical situations demand that we fix the point of application. A torque about a pivot, or a force applied at a specific point on a rigid body, cannot be freely relocated without changing the physical outcome. Such vectors are called bound (or localised) vectors.

For the remainder of this chapter, all vectors are free unless stated otherwise.

Notation

Textbooks use boldface (\(\mathbf{A}\)) for vectors; in handwriting, an arrow is placed overhead (\(\vec{A}\)). The magnitude is written as \(|\vec{A}|\) or simply \(A\). The statement \(\vec{A}=5\) is meaningless—a vector cannot equal a pure number. Write \(|\vec{A}|=5\;\mathrm{m}\) or \(\vec{A}=5\;\mathrm{m}\), east.

Component Representation

◈ Resolution into components

live bench

Spin a vector and watch its shadows fall on the axes: aₓ = a cosθ, a_y = a sinθ. The components ARE the vector, rewritten as arithmetic.

◈ loading the interactive bench…

↗ open in the Simulations page

Geometry gives us pictures; components give us numbers. The bridge between the two is a pair of basis vectors: \(\hat{i}\), a unit vector pointing along \(+x\), and \(\hat{j}\), a unit vector pointing along \(+y\). Any vector in the plane can be written as a unique linear combination of these two directions: \[ \vec{A} = A_x\,\hat{i} + A_y\,\hat{j}. \] Here \(A_x\) and \(A_y\) are the components of \(\vec{A}\)—signed real numbers that tell us how much of the vector lies along each axis. If \(\vec{A}\) makes an angle \(\theta\) with the positive \(x\)-axis, then \[ A_x = A\cos\theta, \qquad A_y = A\sin\theta, \] where \(A = |\vec{A}|\) is the magnitude.

The geometric picture and the algebraic one carry exactly the same information. The arrow is the hypotenuse of a right triangle whose legs are \(A_x\) and \(A_y\).

Vector Addition: Geometric vs. Algebraic

Geometrically, to add \(\vec{A}+\vec{B}\) we place the tail of \(\vec{B}\) at the head of \(\vec{A}\); the resultant \(\vec{R}\) runs from the tail of \(\vec{A}\) to the head of \(\vec{B}\) (the triangle law, also called head-to-tail addition).

Equivalently, if the two vectors share a common tail, the resultant is the diagonal of the parallelogram they form (parallelogram law).

Both laws are visually powerful, but calculations are far easier in component form. If \(\vec{A}=A_x\,\hat{i}+A_y\,\hat{j}\) and \(\vec{B}=B_x\,\hat{i}+B_y\,\hat{j}\), then \[ \vec{R} = \vec{A}+\vec{B} = (A_x+B_x)\,\hat{i} + (A_y+B_y)\,\hat{j}. \] Each axis is handled independently—the \(x\)-components add among themselves and the \(y\)-components add among themselves. There is no mixing. This is the computational engine that makes 2D physics tractable: a single vector equation becomes two ordinary scalar equations.

Vector subtraction follows immediately: \[ \vec{A}-\vec{B} = (A_x - B_x)\,\hat{i} + (A_y - B_y)\,\hat{j}. \] Geometrically, \(\vec{A}-\vec{B}\) is the vector from the head of \(\vec{B}\) to the head of \(\vec{A}\) when both are drawn from a common tail.

Misconception alert: Adding Magnitudes

A persistent error on JEE and AP exams: students compute \(|\vec{A}+\vec{B}|\) by writing \(A+B\). This is wrong unless the two vectors are perfectly parallel. In general, \[ |\vec{A}+\vec{B}| = \sqrt{A^2+B^2+2AB\cos\theta}, \] where \(\theta\) is the angle between \(\vec{A}\) and \(\vec{B}\). The magnitude of a sum depends on the angle; magnitudes do not add like ordinary numbers. At \(\theta=90^\circ\) you recover Pythagoras; at \(\theta=180^\circ\) the vectors partially cancel. The only safe route is: resolve into components, add components, then find the magnitude.

How big can a sum be? (Perplexing Question 2: the Impossible Sum.) Head-to-tail addition immediately bounds the resultant's magnitude. Stretching the two arrows into a straight line (parallel) gives the largest possible sum, \(|\vec{A}|+|\vec{B}|\); folding one back on the other (antiparallel) gives the smallest, \(\bigl||\vec{A}|-|\vec{B}|\bigr|\). Hence \[ \bigl||\vec{A}|-|\vec{B}|\bigr| \;\le\; |\vec{A}+\vec{B}| \;\le\; |\vec{A}|+|\vec{B}|. \] For magnitudes \(3\) and \(4\) the sum is therefore confined to \([1,\,7]\). A resultant of \(1\) is attainable (the vectors point opposite ways), but \(8\) is impossible—it exceeds the maximum \(3+4=7\).

Example 1 · Two Forces on a Crate

Two forces act on a crate: \(\vec{F}_1\) has magnitude \(10\;\mathrm{N}\) at \(30^\circ\) above the \(+x\)-axis, and \(\vec{F}_2\) has magnitude \(8\;\mathrm{N}\) at \(120^\circ\) from the \(+x\)-axis. Find the resultant force.

Solution. Resolve into components. \[ F_{1x} = 10\cos 30^\circ = 5\sqrt{3}\;\mathrm{N},\quad F_{1y} = 10\sin 30^\circ = 5\;\mathrm{N}. \] \[ F_{2x} = 8\cos 120^\circ = -4\;\mathrm{N},\quad F_{2y} = 8\sin 120^\circ = 4\sqrt{3}\;\mathrm{N}. \] Add component-wise. \[ R_x = 5\sqrt{3}-4 \approx 4.66\;\mathrm{N},\qquad R_y = 5+4\sqrt{3}\approx 11.93\;\mathrm{N}. \] Reconstruct. \[ R = \sqrt{R_x^2+R_y^2}\approx 12.81\;\mathrm{N},\qquad \phi = \tan^{-1}\!\left(\frac{R_y}{R_x}\right) \approx 68.6^\circ. \] The resultant force has magnitude \(\approx 12.8\;\mathrm{N}\), directed at about \(69^\circ\) above the \(+x\)-axis. Notice that \(12.8\;\mathrm{N}\) is neither \(10+8=18\) nor \(10-8=2\); the angle matters. Check: The two forces are \(120^\circ-30^\circ=90^\circ\) apart—perpendicular—so the resultant magnitude must be \(\sqrt{F_1^2+F_2^2}=\sqrt{100+64}=\sqrt{164}\approx12.81\,\mathrm{N}\) ✓, independent of the component bookkeeping. It also sits inside the allowed band \(|10-8|=2\le R\le 10+8=18\) ✓.

✓ Checkpoint · Vectors: Geometry and Components

Can the magnitude of a vector ever be smaller than the magnitude of one of its components?
No. Since \(|\vec{A}| = \sqrt{A_x^2 + A_y^2} \geq \sqrt{A_x^2} = |A_x|\) (and likewise \(\geq |A_y|\)), the magnitude is always at least as large as either component. Equality holds only when the other component is zero — the vector lies along an axis.
Two vectors have magnitudes \(3\) and \(5\). Without computing anything, state the full range of possible magnitudes of their sum.
From \(|5-3| = 2\) (antiparallel) up to \(5+3 = 8\) (parallel): \(2 \leq |\vec{R}| \leq 8\). The exact value depends on the angle \(\theta\) between them through \(|\vec{R}| = \sqrt{34 + 30\cos\theta}\).
A displacement of magnitude \(10\;\mathrm{m}\) points \(37^\circ\) above the \(+x\)-axis (take \(\sin 37^\circ = 0.6\), \(\cos 37^\circ = 0.8\)). Find its components.
\(A_x = 10\cos 37^\circ = 8\;\mathrm{m}\), \(A_y = 10\sin 37^\circ = 6\;\mathrm{m}\). Check: \(\sqrt{8^2+6^2} = 10\;\mathrm{m}\), and each component is smaller than the magnitude, as it must be.
Adding two vectors head-to-tail and adding them component-by-component give the same resultant. Why must the two methods agree?
Component addition is the head-to-tail rule projected onto each axis. Projection is linear, so the \(x\)-projection of the resultant equals the sum of the \(x\)-projections (likewise for \(y\)). The geometric picture and the algebra are one operation seen two ways.

Solved examples

Five fully-worked problems from this chapter, free — solution and answer shown in full. The complete set of worked examples is in the full book.

Example 1 · Resultant of Two Displacements

A hiker walks \(\vec{A} = 5\,\hat{i} + 3\,\hat{j}\;\mathrm{km}\) and then \(\vec{B} = 2\,\hat{i} - 7\,\hat{j}\;\mathrm{km}\) (with \(\hat{i}\) east, \(\hat{j}\) north). Find the magnitude and direction of the net displacement.

Solution. Setup. Net displacement adds component-by-component—the two legs share no coupling, so \(x\) and \(y\) are handled independently. \[ \vec{R} = \vec{A} + \vec{B} = (5+2)\,\hat{i} + (3-7)\,\hat{j} = 7\,\hat{i} - 4\,\hat{j}\;\mathrm{km}. \] Magnitude. \(|\vec{R}| = \sqrt{7^2 + (-4)^2} = \sqrt{65} \approx 8.06\;\mathrm{km}\). Direction. \(\theta = \arctan\!\dfrac{|{-4}|}{7} = \arctan(0.571) \approx 29.7^\circ\) south of east (the \(y\)-component is negative). Answer: \(\boxed{|\vec{R}| \approx 8.06\;\mathrm{km},\ 29.7^\circ \text{ S of E}}\). Check: The resultant must close the tip-to-tail triangle in the figure. It also lies inside the triangle-inequality band \(\bigl||\vec{A}|-|\vec{B}|\bigr| \le |\vec{R}| \le |\vec{A}|+|\vec{B}|\): here \(|\vec{A}|=\sqrt{34}\approx5.83\), \(|\vec{B}|=\sqrt{53}\approx7.28\), so \(1.45 \le 8.06 \le 13.11\) ✓.

Fig 2.10 Tip-to-tail addition: the resultant \(\vec{R}=\vec{A}+\vec{B}\) closes the triangle.

Example 2 · Projectile Basics on Level Ground

A ball is launched from level ground at \(30^\circ\) above the horizontal with speed \(20\;\mathrm{m/s}\). Taking \(g = 10\;\mathrm{m/s^2}\), find the time of flight, the maximum height, and the horizontal range.

Solution. Setup. Resolve once; then the vertical motion alone fixes the timing, and the (constant) horizontal motion rides along for that same time. \[ v_{0x} = 20\cos 30^\circ = 10\sqrt{3}\approx17.3\;\mathrm{m/s}, \qquad v_{0y} = 20\sin 30^\circ = 10\;\mathrm{m/s}. \] Time of flight (returns to launch height): \(T = \dfrac{2v_{0y}}{g} = \dfrac{2(10)}{10} = 2\;\mathrm{s}\). Maximum height: \(H = \dfrac{v_{0y}^2}{2g} = \dfrac{100}{20} = 5\;\mathrm{m}\). Range: \(R = v_{0x}\,T = 10\sqrt{3}\,(2) = 20\sqrt{3} \approx 34.6\;\mathrm{m}\). Answer: \(\boxed{T = 2\;\mathrm{s},\ H = 5\;\mathrm{m},\ R = 20\sqrt{3}\approx34.6\;\mathrm{m}}\). Check: The apex is reached at \(T/2 = 1\;\mathrm{s}\); falling from rest (vertically) for \(1\;\mathrm{s}\) gives \(\tfrac12 g (1)^2 = 5\;\mathrm{m}\), the same \(H\) ✓. Limiting case: as \(\theta \to 90^\circ\), \(v_{0x}\to0\) so \(R\to0\), as a vertical throw should ✓.

Fig 2.11 Velocity resolved at launch; the vertical component sets the time, the horizontal component sets the range.

Example 3 · Dot Product Detects Perpendicularity

Find the angle between \(\vec A = 3\,\hat i + 4\,\hat j\) and \(\vec B = 4\,\hat i - 3\,\hat j\).

Solution. Given: \(\vec A = 3\,\hat i + 4\,\hat j\), \(\vec B = 4\,\hat i - 3\,\hat j\).
Find: the angle \(\theta\) between them.
Setup: Use \(\vec A \cdot \vec B = |\vec A||\vec B|\cos\theta\), so \(\cos\theta = (\vec A \cdot \vec B)/(|\vec A||\vec B|)\).
Solve: \[ \vec A \cdot \vec B = (3)(4) + (4)(-3) = 12 - 12 = 0 \;\Rightarrow\; \cos\theta = 0 \;\Rightarrow\; \theta = 90^\circ. \] Answer: \(\boxed{\theta = 90^\circ\ (\vec A \perp \vec B)}\)
Check: Magnitudes: \(|\vec A| = |\vec B| = 5\), both nonzero, so the zero dot product is not an artefact of a zero vector—the vectors are genuinely perpendicular ✓. Geometry: \(\vec B\) is \(\vec A\) rotated \(90^\circ\) clockwise (\(x,y \to y,-x\)), confirming the right angle ✓.

Example 4 · Unit Vector and Scaling

Find the unit vector along \(\vec A = 6\,\hat i + 8\,\hat j\), and a vector of magnitude \(20\) pointing in the same direction.

Solution. Given: \(\vec A = 6\,\hat i + 8\,\hat j\).
Find: \(\hat A\) and \(20\,\hat A\).
Setup: \(\hat A = \vec A/|\vec A|\); scaling a unit vector by \(20\) gives the required vector.
Solve: \[ |\vec A| = \sqrt{6^2 + 8^2} = 10, \qquad \hat A = 0.6\,\hat i + 0.8\,\hat j, \qquad 20\,\hat A = 12\,\hat i + 16\,\hat j. \] Answer: \(\boxed{\hat A = 0.6\,\hat i + 0.8\,\hat j,\quad 20\,\hat A = 12\,\hat i + 16\,\hat j}\)
Check: Unit length: \(\sqrt{0.6^2 + 0.8^2} = 1\) ✓. Magnitude: \(\sqrt{12^2 + 16^2} = \sqrt{400} = 20\) ✓.

Example 5 · Horizontal Projectile off a Table

A ball rolls off a table \(1.25\,\mathrm{m}\) high with a horizontal speed of \(3\,\mathrm{m/s}\). Find (a) the time to land, (b) the horizontal distance from the table, and (c) the landing speed. (\(g = 10\,\mathrm{m/s^2}\).)

Solution. Given: \(h = 1.25\,\mathrm{m}\), \(u = 3\,\mathrm{m/s}\) horizontal, \(v_{0y} = 0\).
Find: time \(t\), range \(x\), landing speed \(v\).
Setup: Vertical and horizontal motions are independent. The vertical fall (from rest) fixes the time; the horizontal motion is at constant \(u\).
Solve: \[ t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2.5}{10}} = 0.5\,\mathrm{s}, \qquad x = u t = 3(0.5) = 1.5\,\mathrm{m}. \] \[ v_y = g t = 5\,\mathrm{m/s}, \qquad v = \sqrt{u^2 + v_y^2} = \sqrt{9 + 25} = \sqrt{34} \approx 5.83\,\mathrm{m/s}. \] Answer: \(\boxed{t = 0.5\,\mathrm{s},\ x = 1.5\,\mathrm{m},\ v \approx 5.83\,\mathrm{m/s}}\)
Check: Independence: the time depends only on the height, not on \(u\)—a ball dropped and a ball rolled off land together ✓. Energy: \(v = \sqrt{u^2 + 2gh} = \sqrt{9 + 25} = \sqrt{34}\), matching part (c) ✓.

Problem bank

Five questions from this chapter’s 50-question bank, free — attempt each one before you reveal the answer. The rest of the bank, and the timed test that draws on all of it, are in the full book.

✎ Free practice questions

A particle starts from the origin with velocity \(\vec{v}_0 = 6\,\hat{i} + 8\,\hat{j}\;\mathrm{m/s}\) and constant acceleration \(\vec{a} = -2\,\hat{i} + 1\,\hat{j}\;\mathrm{m/s^2}\). Find the speed at \(t = 4\;\mathrm{s}\).
\(v_x = 6-8=-2\;\mathrm{m/s}\), \(v_y = 8+4=12\;\mathrm{m/s}\). Speed \(= \sqrt{4+144} = 2\sqrt{37} \approx 12.2\;\mathrm{m/s}\).
A particle has initial position \(\vec{r}_0 = 2\,\hat{i} - 3\,\hat{j}\;\mathrm{m}\), initial velocity \(\vec{v}_0 = 5\,\hat{j}\;\mathrm{m/s}\), and constant acceleration \(\vec{a} = 4\,\hat{i}\;\mathrm{m/s^2}\). Find the distance from the origin at \(t = 3\;\mathrm{s}\).
\(x = 2 + 18 = 20\;\mathrm{m}\), \(y = -3+15 = 12\;\mathrm{m}\). Distance \(= \sqrt{400+144} = 4\sqrt{34} \approx 23.3\;\mathrm{m}\).
A particle starts from the origin with \(\vec{v}_0 = 3\,\hat{i} + 4\,\hat{j}\;\mathrm{m/s}\) under \(\vec{a} = -1\,\hat{i} - 2\,\hat{j}\;\mathrm{m/s^2}\). At what time does it move parallel to the \(x\)-axis?
Parallel to \(x\)-axis when \(v_y = 0\): \(4 - 2t = 0 \Rightarrow t = 2\;\mathrm{s}\).
A stone is thrown horizontally at \(15\;\mathrm{m/s}\) from a \(45\;\mathrm{m}\) tower. Find the horizontal distance from the base where it lands.
\(t = \sqrt{2h/g} = 3\;\mathrm{s}\). Distance \(= 45\;\mathrm{m}\).
Particle \(A\) at origin is launched with \(\vec{v}_A = 8\,\hat{i}+15\,\hat{j}\;\mathrm{m/s}\); particle \(B\) at \((32,0)\;\mathrm{m}\) is launched simultaneously. They collide at \(t = 2\;\mathrm{s}\). Find \(\vec{v}_B\).
Position of \(A\) at \(t=2\): \((16,10)\;\mathrm{m}\). \(32 + 2v_{Bx} = 16 \Rightarrow v_{Bx} = -8\;\mathrm{m/s}\). \(2v_{By} - 20 = 10 \Rightarrow v_{By} = 15\;\mathrm{m/s}\). \(\vec{v}_B = -8\,\hat{i}+15\,\hat{j}\;\mathrm{m/s}\).

Chapter test

A paper drawn at random from this chapter's bank. Choose the exam you are training for — the marking scheme, pace and difficulty mix follow the real pattern. Work on paper; when you finish (or the clock runs out), the answers are revealed and you mark yourself honestly.

Pattern

Questions

Clock

Free practice questions

A ball is launched from level ground toward a vertical wall \(24\;\mathrm{m}\) away and strikes it horizontally at height \(7.2\;\mathrm{m}\). Find the launch speed and angle.
Horizontal strike means apex: \(v_{0y}^2 = 2g\cdot7.2 = 144\), so \(v_{0y} = 12\;\mathrm{m/s}\). Time to apex \(= 1.2\;\mathrm{s}\); \(v_{0x} = 24/1.2 = 20\;\mathrm{m/s}\). Speed \(= \sqrt{544} \approx 23.3\;\mathrm{m/s}\); angle \(= \arctan(12/20) \approx 31^\circ\).
A stone passes through two points both at height \(15\;\mathrm{m}\), at \(t = 1\;\mathrm{s}\) and \(t = 3\;\mathrm{s}\), with total range \(80\;\mathrm{m}\). Find the maximum height and the horizontal distance between the two points.
From \(15 = v_{0y} - 5\) at \(t=1\): \(v_{0y} = 20\;\mathrm{m/s}\). \(H = 20\;\mathrm{m}\). \(T = 4\;\mathrm{s}\); \(v_{0x} = 20\;\mathrm{m/s}\). Horizontal separation \(= 20 \times 2 = 40\;\mathrm{m}\).
A target on level ground moves at \(5\;\mathrm{m/s}\) directly away; at the instant it is \(40\;\mathrm{m}\) away, a projectile with \(v_{0x} = 15\;\mathrm{m/s}\) is fired. Find the required \(v_{0y}\) to hit the target on the ground.
Hit condition: \(15t = 40 + 5t \Rightarrow t = 4\;\mathrm{s}\). This must equal the time of flight: \(T = 2v_{0y}/g \Rightarrow v_{0y} = 20\;\mathrm{m/s}\).
Projectile \(A\) is launched horizontally at \(10\;\mathrm{m/s}\) from a \(45\;\mathrm{m}\) cliff. Simultaneously, projectile \(B\) is launched vertically from the ground at \(30\;\mathrm{m/s}\). Find the initial horizontal separation \(D\) and the height of collision.
Vertical equations: \(45 - 5t^2 = 30t - 5t^2 \Rightarrow t = 1.5\;\mathrm{s}\). Height \(= 45 - 5(2.25) = 33.75\;\mathrm{m}\). \(D = 10 \times 1.5 = 15\;\mathrm{m}\).
A particle is projected from level ground with \(\vec{v}_0 = 15\,\hat{i} + 20\,\hat{j}\;\mathrm{m/s}\). At what time \(t \gt 0\) is the velocity perpendicular to \(\vec{v}_0\)?
\(\vec{v}(t)\cdot\vec{v}_0 = 0\): \(225 + (20-10t)(20) = 0 \Rightarrow 200t = 625 \Rightarrow t = 3.125\;\mathrm{s}\).
Two ships \(P\) and \(Q\): \(P\) moves North at \(20\;\mathrm{km/h}\); \(Q\) starts \(15\;\mathrm{km}\) East of \(P\) and moves West at \(15\;\mathrm{km/h}\). Find their closest distance of approach.
Relative velocity \(= -15\,\hat{i} - 20\,\hat{j}\), magnitude \(25\;\mathrm{km/h}\). Relative initial position \(= 15\,\hat{i}\). Closest distance \(= |15\,\hat{i} \times (-15\,\hat{i}-20\,\hat{j})| / 25 = 300/25 = 12\;\mathrm{km}\).