Motion in a Plane

• Sections 3.2 and 3.3 lay the foundation — what makes a quantity a scalar or a vector, and what it means to multiply a vector by a real number, including reversing it, which is exactly how vector subtraction is defined.
• Sections 3.4 to 3.6 build the vector-algebra toolkit — graphical addition and subtraction using the triangle and parallelogram laws, resolving a vector into components along two directions, unit vectors i-hat and j-hat, and the analytical component method that replaces careful ruler-and-protractor diagrams with plain algebra.
• Section 3.7 applies that toolkit to motion itself — position vector, displacement, average and instantaneous velocity, and acceleration, all written in component form.
• Section 3.8 states the central simplification of the whole chapter — under constant acceleration, motion in a plane is nothing more than two independent straight-line motions happening at once along perpendicular axes.
• Section 3.9 uses that independence to derive projectile motion in full — the parabolic path, time of flight, maximum height and horizontal range.
• Section 3.10 turns to uniform circular motion, where speed stays constant but direction does not, producing a centripetal acceleration of magnitude $v^2/R$ that always points toward the centre.

Scalars and vectors

In physics every quantity is either a scalar or a vector. A scalar quantity is specified completely by a single number together with its unit — distance, mass, temperature, time, speed and density are all scalars, and they combine by ordinary algebra: add, subtract, multiply, divide. A vector quantity needs both a magnitude and a direction, and vectors obey the triangle law of addition (equivalently the parallelogram law), not ordinary arithmetic. Displacement, velocity, acceleration and force are vectors. A vector is written in bold face, such as $\vec A$, with its magnitude denoted by the light-face symbol $A$ or $|\vec A|$. Two vectors are equal only if they have the same magnitude and the same direction; two vectors can have identical length and still be unequal if they point differently. A null vector (written $\vec 0$) has zero magnitude, so its direction cannot be specified; it arises whenever a vector is added to its own negative, $\vec A+(-\vec A)=\vec 0$, or when displacement is zero because an object returns exactly to its starting point. Note the important asymmetry between path length and displacement: the displacement vector joins only the initial and final positions and is independent of the actual path taken, so the magnitude of displacement is always less than or equal to the path length — equal only when the motion never changes direction.

Multiplying a vector by a real number, and adding vectors graphically

Multiplying a vector $\vec A$ by a positive real number $\lambda$ gives a new vector $\lambda\vec A$ with magnitude $\lambda A$ and the same direction as $\vec A$; multiplying by a negative number $-\lambda$ reverses the direction while scaling the magnitude by $\lambda$. This is the basis of vector subtraction: $\vec A-\vec B$ is defined as $\vec A+(-\vec B)$, adding $\vec A$ to the reversed vector $-\vec B$. To add two vectors $\vec A$ and $\vec B$ graphically by the triangle (head-to-tail) method, place the tail of $\vec B$ at the head of $\vec A$; the vector from the tail of $\vec A$ to the head of $\vec B$ is the resultant $\vec R=\vec A+\vec B$. Vector addition is commutative, $\vec A+\vec B=\vec B+\vec A$, and associative, $(\vec A+\vec B)+\vec C=\vec A+(\vec B+\vec C)$, so the order in which several vectors are added or grouped never matters. The equivalent parallelogram method brings both vectors' tails to a common origin, completes the parallelogram, and takes the diagonal through that origin as the resultant — the two constructions always agree.

Resolving a vector: unit vectors and components

A vector $\vec A$ lying in a plane can be resolved into two component vectors along any two non-parallel directions in that plane; the most convenient choice is a rectangular x-y coordinate system, using the unit vectors $\hat i$ and $\hat j$, each of magnitude one and pointing along the x- and y-axes respectively, with $|\hat i|=|\hat j|=1$. Any vector in the plane can then be written

where: $A_x$ and $A_y$ are the x- and y-components of $\vec A$ — real numbers, not vectors themselves, though $A_x\hat i$ and $A_y\hat j$ are vectors. If $\vec A$ makes angle $\theta$ with the x-axis, simple trigonometry gives $A_x=A\cos\theta$ and $A_y=A\sin\theta$; conversely, given the components, the magnitude and direction follow from $A=\sqrt{A_x^2+A_y^2}$ and $\theta=\tan^{-1}(A_y/A_x)$. This gives two equally valid ways to specify a vector — magnitude and angle, or the pair of components — and problems routinely require switching between them. In three dimensions the same idea extends with a third unit vector $\hat k$ along the z-axis, giving $\vec A=A_x\hat i+A_y\hat j+A_z\hat k$ and magnitude $A=\sqrt{A_x^2+A_y^2+A_z^2}$.

Adding vectors analytically: components and the law of cosines

Graphical addition is easy to visualise but tedious and limited in accuracy. The analytical method is faster and exact: write each vector in component form and add the corresponding components separately. If $\vec A=A_x\hat i+A_y\hat j$ and $\vec B=B_x\hat i+B_y\hat j$, then

where: each component of the resultant is simply the sum of the corresponding components of the vectors being added — the physical reading is that motion, force or velocity along x can be tracked completely independently of what happens along y. The same idea extends to any number of vectors and to three dimensions without any change in method. When only the magnitudes $A$, $B$ and the angle $\theta$ between two vectors are known, the equivalent geometric result is the law of cosines, $R=\sqrt{A^2+B^2+2AB\cos\theta}$, derived in full below.

Position, displacement, velocity and acceleration in a plane

The position vector of a particle relative to a chosen origin O is $\vec r=x\hat i+y\hat j$, where $x$ and $y$ are the particle's coordinates. If the particle moves from a point with position vector $\vec r$ at time $t$ to a point with position vector $\vec r\,'$ at time $t'$, its displacement is

where: $\Delta x = x'-x$ and $\Delta y=y'-y$ are the changes in the two coordinates. The average velocity over that interval is $\vec v_{avg}=\Delta\vec r/\Delta t$, pointing in the same direction as $\Delta\vec r$; the instantaneous velocity is the limit as $\Delta t\to0$,

where: the crucial physical reading is that as $\Delta t\to0$, the chord $\Delta\vec r$ collapses onto the tangent to the path, so the instantaneous velocity of a particle is always directed tangent to its trajectory, in the direction of motion. The magnitude is $v=\sqrt{v_x^2+v_y^2}$ and its direction follows $\tan\theta=v_y/v_x$. Acceleration follows the identical pattern: average acceleration $\vec a_{avg}=\Delta\vec v/\Delta t$, and instantaneous acceleration $\vec a=d\vec v/dt=a_x\hat i+a_y\hat j$, with $a_x=dv_x/dt$ and $a_y=dv_y/dt$. In straight-line motion, velocity and acceleration are always along the same line; in a plane, the angle between $\vec v$ and $\vec a$ can be anything between 0 degrees and 180 degrees, which is exactly why circular motion — where that angle is always 90 degrees — feels so different from straight-line speeding up or slowing down.

Motion in a plane with constant acceleration

Suppose acceleration $\vec a$ is constant. If the velocity is $\vec v_0$ at $t=0$ and $\vec v$ at time $t$, then by the definition of average acceleration, $\vec a=(\vec v-\vec v_0)/t$, so

where: $\vec r_0$ and $\vec v_0$ are the position and velocity at $t=0$. Written in components, these become $v_x=v_{0x}+a_xt$, $v_y=v_{0y}+a_yt$, and $x=x_0+v_{0x}t+\tfrac12a_xt^2$, $y=y_0+v_{0y}t+\tfrac12a_yt^2$. The single most important reading of this result is that the motions along x and along y can be analysed completely independently of each other — motion in a plane under constant acceleration is nothing but two simultaneous one-dimensional motions with constant acceleration, one along each perpendicular axis. This independence is the working principle behind both projectile motion and every two-dimensional constant-acceleration problem in this chapter, and it extends without modification to three dimensions.

Projectile motion — the flight of a thrown object

An object in flight after being thrown or launched, moving only under gravity once released, is called a projectile — a cricket ball, a shot put, a stone from a sling. Galileo first stated, in 1632, that the horizontal and vertical parts of a projectile's motion are independent; this chapter's vector machinery makes that claim precise. Neglecting air resistance, once launched with speed $v_0$ at angle $\theta_0$ to the horizontal, the only acceleration is gravity, straight down: $a_x=0$, $a_y=-g$. Taking the launch point as origin, the initial velocity components are $v_{0x}=v_0\cos\theta_0$ and $v_{0y}=v_0\sin\theta_0$, and the constant-acceleration equations above immediately give

where: the x-motion has zero acceleration, so $v_x=v_0\cos\theta_0$ never changes during flight, while the y-motion is a vertical free fall superposed on an initial upward speed $v_0\sin\theta_0$. Eliminating $t$ between the two equations gives the trajectory — a parabola, derived below — and from it follow the three quantities examiners ask about most: time to the top, maximum height, and range. At the topmost point, $v_y=0$ momentarily while $v_x$ stays unchanged, so the velocity there is purely horizontal.

Uniform circular motion — constant speed, changing direction

When an object travels a circular path at constant speed, the motion is called uniform circular motion; "uniform" refers only to the speed, which does not change, while the direction of motion changes continuously. Because velocity is a vector, a change in direction alone is a change in velocity, so the object is accelerating even though its speed never varies. A geometric argument, carried out in full in the derivations below, shows that this acceleration always points toward the centre of the circle and has magnitude

where: $v$ is the constant speed and $R$ is the radius. This is called centripetal acceleration, from the Greek for "centre-seeking," a term coined by Newton; a full analysis was first published by Christiaan Huygens in 1673. Its direction keeps changing even though its magnitude is fixed, so it is not a constant vector. Circular motion also has an angular description: angular speed $\omega$ is angular distance swept per unit time, related to linear speed by $v=R\omega$, so $a_c=\omega^2R$ equally. With period $T$ (time per revolution) and frequency $\nu=1/T$, $\omega=2\pi\nu=2\pi/T$, $v=2\pi R\nu$ and $a_c=4\pi^2\nu^2R$.

Board level - Holding an umbrella against rain and wind (NCERT Example 3.1)

Rain is falling vertically at $35\ \text{m/s}$. A wind then begins blowing from east to west at $12\ \text{m/s}$. In which direction should a boy standing at a bus stop tilt his umbrella?

Step 1. Represent the two velocities as vectors from a common point: $\vec v_r$, magnitude $35\ \text{m/s}$, pointing straight down, and $\vec v_w$, magnitude $12\ \text{m/s}$, pointing horizontally (east to west). Since a raindrop actually experiences both motions together, the boy must tilt the umbrella to face the resultant velocity $\vec v_r+\vec v_w$, not just the vertical part.

Step 2. Because $\vec v_r$ and $\vec v_w$ are perpendicular, the resultant magnitude follows from Pythagoras directly, a special case of the general law of cosines with $\theta=90$ degrees: $R=\sqrt{v_r^2+v_w^2}=\sqrt{35^2+12^2}=\sqrt{1225+144}=\sqrt{1369}=37\ \text{m/s}$.

Step 3. The direction, measured from the vertical (the direction of $\vec v_r$) toward the wind, satisfies $\tan\theta=v_w/v_r=12/35=0.343$, so $\theta=\tan^{-1}(0.343)\approx19^\circ$. Answer: the boy should tilt the umbrella about 19 degrees from the vertical, toward the east (into the wind), since that is the direction the rain is effectively falling from.

Board level - The flight of a cricket ball (NCERT Example 3.8)

A cricket ball is thrown at $28\ \text{m/s}$ at $30^\circ$ above the horizontal. Calculate (a) the maximum height, (b) the time taken to return to the same level, and (c) the horizontal distance travelled before it lands (take $g=9.8\ \text{m/s}^2$).

Step 1. Maximum height uses $h_m=\dfrac{v_0^2\sin^2\theta_0}{2g}$. With $v_0\sin\theta_0=28\times0.5=14\ \text{m/s}$, $h_m=\dfrac{14^2}{2\times9.8}=\dfrac{196}{19.6}=10.0\ \text{m}$.

Step 2. Time of flight uses $T_f=\dfrac{2v_0\sin\theta_0}{g}=\dfrac{2\times14}{9.8}=\dfrac{28}{9.8}=2.9\ \text{s}$.

Step 3. Range uses $R=\dfrac{v_0^2\sin2\theta_0}{g}$, with $2\theta_0=60^\circ$ and $\sin60^\circ\approx0.866$: $R=\dfrac{28^2\times0.866}{9.8}=\dfrac{784\times0.866}{9.8}=\dfrac{679.1}{9.8}\approx69\ \text{m}$. Answer: maximum height 10.0 m, time of flight 2.9 s, range about 69 m.

Competitive level - Locating a particle under constant 2-D acceleration (NCERT Example 3.5, extended)

A particle starts from the origin at $t=0$ with velocity $5.0\hat i\ \text{m/s}$ and moves under a constant acceleration $(3.0\hat i+2.0\hat j)\ \text{m/s}^2$. (a) Find the y-coordinate when $x=84\ \text{m}$. (b) Find the speed at that instant. (c) Find the magnitude and direction of the displacement vector at that instant, and the angle between the displacement and the velocity.

Step 1. From $x=x_0+v_{0x}t+\tfrac12a_xt^2$ with $x_0=0$, $v_{0x}=5.0$, $a_x=3.0$: $x(t)=5.0t+1.5t^2$. Setting $x=84$ gives $1.5t^2+5.0t-84=0$, which factors to $t=6.0\ \text{s}$ (rejecting the negative root). Then $y(t)=v_{0y}t+\tfrac12a_yt^2=0+1.0t^2$, so at $t=6.0\ \text{s}$, $y=1.0(6.0)^2=36\ \text{m}$.

Step 2. Velocity components at $t=6.0\ \text{s}$: $v_x=5.0+3.0(6.0)=23.0\ \text{m/s}$, $v_y=2.0(6.0)=12.0\ \text{m/s}$, giving speed $v=\sqrt{23.0^2+12.0^2}=\sqrt{529+144}=\sqrt{673}\approx25.9\ \text{m/s}$.

Step 3. Displacement vector is $\vec r=(84,\,36)$, so its magnitude is $r=\sqrt{84^2+36^2}=\sqrt{7056+1296}=\sqrt{8352}\approx91.4\ \text{m}$, making angle $\tan^{-1}(36/84)\approx23.2^\circ$ with the x-axis. The velocity direction is $\tan^{-1}(12.0/23.0)\approx27.6^\circ$ with the x-axis, so the angle between $\vec r$ and $\vec v$ is only about $27.6^\circ-23.2^\circ\approx4.4^\circ$. Answer: y = 36 m, speed ≈ 25.9 m/s, displacement ≈ 91.4 m at 23.2 degrees, nearly parallel (about 4.4 degrees apart) to the velocity vector at that instant. Timing note: two to three minutes. The near-parallelism of $\vec r$ and $\vec v$ at $t=6\ \text{s}$ is not a coincidence to chase further — it is a useful sanity check that both are dominated by the same growing x-component, and a mismatch of tens of degrees would flag an arithmetic slip.

Competitive level - An insect in a circular groove (NCERT Example 3.9, extended)

An insect trapped in a circular groove of radius $R=12\ \text{cm}$ completes 7 revolutions in 100 s at steady speed. (a) Find the angular speed and linear speed. (b) Find the centripetal acceleration in cm/s² and in m/s². (c) Express the acceleration as a fraction of $g=9.8\ \text{m/s}^2$, and find the arc length covered and the period of one revolution.

Step 1. Angular speed: $\omega=2\pi/T$, where $T=100/7\approx14.3\ \text{s}$ is the time per revolution, so $\omega=2\pi(7)/100\approx0.44\ \text{rad/s}$.

Step 2. Linear speed: $v=\omega R=0.44\times12\approx5.3\ \text{cm/s}$. Since both $v$ and $R$ are constant, the magnitude of $a_c$ is constant even though its direction keeps changing, so it is not a constant vector: $a_c=\omega^2R=(0.44)^2\times12\approx2.3\ \text{cm/s}^2=0.023\ \text{m/s}^2$.

Step 3. As a fraction of $g$: $\dfrac{0.023}{9.8}\approx0.0023$, roughly $1/425$ of the acceleration due to gravity — a useful reminder that everyday circular motions can involve accelerations far smaller than $g$. The arc length in one full revolution is the circle's circumference, $2\pi R=2\pi(12)\approx75.4\ \text{cm}$, covered in the period $T=100/7\approx14.3\ \text{s}$. Answer: ω ≈ 0.44 rad/s, v ≈ 5.3 cm/s, a_c ≈ 2.3 cm/s² ≈ 0.023 m/s², about 1/425 of g, with circumference ≈ 75.4 cm covered every T ≈ 14.3 s. Timing note: about two minutes. The trap is unit consistency — keep R in the same unit throughout a sub-part, and only convert cm/s² to m/s² at the very last step when comparing with g, which is conventionally quoted in m/s².

Vector questions lose marks in very specific, recurring ways. Students often add magnitudes of two vectors directly (as if they were scalars) instead of using the triangle law or components — this only works when the vectors happen to be parallel. Equally common is forgetting that two vectors are equal only when both magnitude and direction match; two vectors of the same length pointing in different directions are not equal, and treating them as such silently breaks the rest of a calculation. In resolving a vector, a frequent slip is mixing up $A\cos\theta$ and $A\sin\theta$ for the x- and y-components, especially when $\theta$ is measured from the y-axis instead of the x-axis in a particular figure — always check which axis the angle is measured from before writing components. In projectile problems, a very common error is assuming the time of flight equals the time to maximum height, rather than twice it, or forgetting that the range formula $R=v_0^2\sin2\theta_0/g$ is valid only when the projectile lands at the same height it was launched from — a ball thrown off a cliff or into a well needs the general y(t) equation solved directly, not the shortcut formula. In circular motion, the most damaging error is assuming that constant speed means zero acceleration; direction is changing continuously, so the object is always accelerating, just never speeding up or slowing down. Finally, students sometimes carry angular speed in degrees per second instead of radians per second into formulas like $v=R\omega$ or $a_c=\omega^2R$, which silently produces numbers off by a factor connected to $\pi/180$.

• Classify any given physical quantity correctly as a scalar or a vector, and explain why.
• Add and subtract two or more vectors both graphically (triangle and parallelogram methods) and analytically (by components).
• Resolve a vector into its x- and y-components given its magnitude and angle, and recover the magnitude and angle from given components.
• Write position, velocity and acceleration in unit-vector (i-hat, j-hat) form and differentiate one to obtain the next.
• Explain why velocity is always tangent to the path of motion, and why velocity and acceleration need not be along the same line in a plane.
• Set up and solve constant-acceleration problems in two dimensions by splitting them into independent x- and y-equations.
• Derive and apply the projectile-motion results — equation of trajectory, time of flight, maximum height and range — and identify when the standard range formula does not apply.
• State why an object in uniform circular motion accelerates despite constant speed, and compute centripetal acceleration from $v^2/R$, $\omega^2R$ or $4\pi^2\nu^2R$.
• Convert cleanly between angular speed, linear speed, period and frequency for circular motion.
• Recognise and avoid the chapter's classic traps — treating unequal-direction vectors as equal, misapplying the range formula, and assuming constant speed means zero acceleration.

Ask the tutor: If you are stuck on resolving a vector, setting up a projectile problem, or seeing why centripetal acceleration points inward, ask the tutor to walk through your specific numbers step by step.