Why motion in a plane needs vectors

In two-dimensional motion, an object changes position along two perpendicular axes, often labelled x and y. A scalar has magnitude alone (for example speed, distance or mass); a vector has both magnitude and direction (for example displacement, velocity, acceleration or force).

Vectors can be represented by directed arrows and decomposed along perpendicular directions. Two vectors are equal when they have the same magnitude and direction irrespective of their starting points in a diagram.

• An eastward 3 m displacement and a northward 4 m displacement combine to a 5 m northeast resultant.
• A negative component simply indicates direction opposite to the positive axis.
• A vector’s magnitude is non-negative; its individual Cartesian components may have either sign.

Vector addition, resolution and resultant magnitude

For a vector A making an angle θ with the positive x-axis, its components are A_x = A cosθ and A_y = A sinθ. Two vectors add by summing their corresponding components. The resultant R has components R_x = A_x + B_x and R_y = A_y + B_y.

The magnitude is |R| = √(R_x² + R_y²). Its direction depends on both signs and can be found using the appropriate quadrant of tan⁻¹(R_y/R_x). For two vectors of magnitudes A and B separated by angle θ, |R| = √(A² + B² + 2AB cosθ).

• For perpendicular vectors, resultant magnitude is √(A² + B²).
• For parallel, same-direction vectors, magnitude is A + B; opposite-direction vectors, |A − B|.
• The ordinary arithmetic sum of magnitudes generally is not the resultant magnitude.

Unit vectors, dot product and cross product

Unit vectors specify direction and have magnitude one. The Cartesian basis vectors î and ĵ indicate the positive x- and y-directions. A vector can be written A = A_xî + A_yĵ.

The scalar (dot) product A·B = AB cosθ gives a scalar; it is zero for nonzero perpendicular vectors. The magnitude of the vector (cross) product is |A×B| = AB sinθ, directed perpendicular to the plane according to the right-hand rule. Dot products appear in work, and cross products in torque and angular momentum.

• A·B is maximum in magnitude for aligned or anti-aligned vectors, depending on the sign.
• |A×B| is zero for parallel vectors and maximal for perpendicular vectors.
• Do not confuse A·B with A×B: they produce different kinds of quantities.

Independent horizontal and vertical motions

For projectile motion without air resistance near Earth’s surface, the horizontal acceleration is zero and the vertical acceleration is −g when upward is positive. These two components evolve independently but share a single time variable. Horizontal velocity stays constant, whereas vertical velocity changes steadily due to gravity.

If a projectile launches with speed u at angle θ above the horizontal, u_x = u cosθ and u_y = u sinθ. Its position after time t from the launch point is x = u cosθ·t and y = u sinθ·t − ½gt².

• At maximum height the vertical component v_y = 0, but the horizontal component remains u cosθ.
• Trajectory is parabolic under the ideal constant-g, no-drag assumptions.
• Launching horizontally is a special case with initial vertical speed zero.

Projectile time of flight, maximum height and range

For a projectile that lands at the same vertical level as it launched, the total flight time is T = 2u sinθ/g, maximum height H = u²sin²θ/(2g), and horizontal range R = u²sin(2θ)/g. These expressions are not universal for launch and landing at different heights.

With fixed launch speed and equal launch/landing elevation, the ideal maximum range occurs at 45°. Complementary angles θ and (90° − θ) give equal ranges because sin(2θ) is the same. Real-world air resistance changes the results.

• At 30° and 60°, ideal ranges are equal at fixed speed and level ground.
• Maximum height is reached halfway through total flight time only under the level-launch/landing assumptions.
• If θ = 0° for a horizontal launch, the level-ground range formula would give zero because it assumes return to the same height; horizontal launch from a cliff must be treated separately.

Horizontal projection from a height

For an object launched horizontally from height h with horizontal speed u, the vertical displacement obeys h = ½gt², giving t = √(2h/g). The horizontal range to the ground is x = ut = u√(2h/g), neglecting air resistance.

The fall time depends on vertical motion, while the horizontal distance depends on both fall time and initial horizontal speed. Doubling launch speed doubles horizontal range, but does not change fall time from a fixed height in this ideal model.

• A horizontally fired object and a dropped object from the same height hit level ground together if vertical starting velocities are equal and drag is negligible.
• Horizontal speed does not create an upward force that delays free fall.
• At impact, the velocity direction is generally downward and forward.

Uniform circular motion and centripetal acceleration

A particle moving around a circle at constant speed still accelerates because its velocity direction continuously changes. The acceleration points radially inward toward the centre and has magnitude a_c = v²/r = ω²r, where r is radius, v tangential speed and ω angular speed.

The required net inward force is F_c = mv²/r. Centripetal force is not a separate fundamental force; it is the name for the inward resultant provided by forces such as friction, gravity or tension. In ideal uniform circular motion, instantaneous velocity is tangential and perpendicular to acceleration.

• For constant speed, doubling speed increases centripetal acceleration fourfold.
• At fixed speed, a larger radius means smaller centripetal acceleration.
• If the inward net force disappears, the object moves along the instantaneous tangent, not radially outward.

Relative velocity in two dimensions

The vector velocity of A observed from B is v_A/B = v_A − v_B. For river-boat questions, ground velocity = boat velocity relative to water + water velocity relative to ground. Resolve each velocity along and across the stream.

If a boat can move at 5 m/s straight across still water while a river flows downstream at 3 m/s, then pointed exactly across the river its ground-speed magnitude is √(5² + 3²) ≈ 5.83 m/s and it drifts downstream. To arrive exactly opposite, it must aim upstream at an appropriate angle, possible only when its still-water speed exceeds the stream speed.

• With straight-across pointing, crossing time = river width/boat-across-water speed.
• Downstream drift = river speed × crossing time when flow is uniform.
• For zero drift, the boat’s upstream component relative to water must cancel the downstream river speed.

Essential formula sheet

Use these equations only when their physical assumptions are satisfied. Symbols and units are explained in the chapter.

Solved numericals and applications

Work each question once without looking at the solution; then compare your method with the steps.

UPSC-style practice MCQs with explanations

These are original practice questions written for competitive-examination revision. They are not presented as verbatim UPSC previous-year questions.

Last-minute revision and common traps

• A vector’s magnitude cannot be negative, though components can.
• At a projectile’s apex, vertical velocity is zero but horizontal velocity is generally nonzero.
• The 45° maximum-range rule assumes same launch and landing height and neglects air resistance.
• Centripetal force is the inward net force, not an additional independent force.
• Constant speed does not mean zero acceleration in circular motion.
• A boat’s ground velocity is a vector sum of the boat-through-water and river-current velocities.