Motion in a Plane

1. What this chapter covers

The chapter splits cleanly in half: sections 3.2 to 3.6 build the mathematics, and 3.7 to 3.10 spend it.

2. Read this before you revise: one gap and one removal

The gap — scalar and vector products

CBSE lists "Scalar and Vector product of vectors" under Chapter 3. The NCERT chapter never mentions them.

This is not a matter of emphasis. The words "scalar product", "vector product", "dot" and "cross" appear zero times in the 2026-27 Chapter 3. The single occurrence of the word "product" in the whole chapter is about the dimensions of $\lambda \mathbf{A}$.

Where NCERT actually teaches them:

Chapter 5 says so in as many words: "We have learnt about vectors and their use in Chapter 3… We now need to know how vectors are multiplied."

So you are examinable on material your chapter does not contain. Section 7 below covers it, because leaving it out would cost you marks.

The removal — relative velocity in two dimensions

Older editions carried a section on relative velocity in a plane. It is gone from the 2026-27 chapter, and CBSE does not list it here either — so unlike Chapter 2's case, nothing is orphaned.

The idea survives only inside worked examples that combine two velocities by ordinary vector addition: a boy deciding how to hold his umbrella in a crosswind, and a motorboat crossing a current. Both are in section 6 below.

3. Scalars and vectors

A scalar is fully specified by one number and its unit. Distance, mass, temperature, the time an event happened. Scalars combine by ordinary arithmetic.

A vector means nothing until you give it both a magnitude and a direction. Displacement, velocity, acceleration, force.

The consequence is that vectors do not obey ordinary arithmetic. Walk 3 m east, then 4 m north:

Both numbers are right. They answer different questions. Every difficulty in this chapter traces back to that one distinction.

Multiplying a vector by a real number

Multiply $\mathbf{A}$ by a real number $\lambda$ and you get $\lambda \mathbf{A}$:

• Its magnitude is $|\lambda |$ times the original.
• Its direction is unchanged if $\lambda >0$, and exactly reversed if $\lambda <0$.

Multiplying by $-1$ therefore gives a vector of the same size pointing the opposite way — which is exactly what makes subtraction possible.

4. Adding and subtracting vectors graphically

Head-to-tail (triangle) method. Draw $\mathbf{A}$. From its head, draw $\mathbf{B}$. The vector from the tail of $\mathbf{A}$ to the head of $\mathbf{B}$ is $\mathbf{A}+\mathbf{B}$.

Parallelogram method. Draw $\mathbf{A}$ and $\mathbf{B}$ from a common tail, complete the parallelogram, and the diagonal from that shared tail is the resultant.

The two methods always agree. They are the same construction seen from different angles.

Two properties are worth stating rather than assuming:

Subtraction is defined, not invented. $\mathbf{A}-\mathbf{B}$ means $\mathbf{A}+(-\mathbf{B})$. Reverse $\mathbf{B}$, then add. There is no separate rule.

The null vector has zero magnitude, and no direction is specified, since a thing of zero length has no direction to speak of. It satisfies

$\mathbf{A}+0=\mathbf{A}\lambda 0=00\mathbf{A}=0$

You get it by adding a vector to its own negative — the formal way of saying that walking somewhere and back leaves zero displacement.

5. Resolution, and the analytical method

Graphical construction is good for understanding and hopeless for precision. The fix is to resolve each vector into perpendicular components.

For a vector $\mathbf{A}$ at angle $\theta$ to the x-axis:

$A_x=A\cos \theta A_y=A\sin \theta$

Running it backwards recovers the vector:

$A=\sqrt{A_x^2+A_y^2}\theta ={\tan }^{-1}\left(\frac{A_y}{A_x}\right)$

Once every vector is in component form, addition becomes arithmetic — add the x-components, add the y-components, done. All the geometry has been converted into two independent sums, and that conversion is the entire point of the method.

The trap here is the quadrant. An inverse tangent cannot tell a vector pointing up-and-right from one pointing down-and-left — both give the same ratio. Always check the signs of the components against your answer.

When the vectors are not at right angles

For two vectors separated by angle $\theta$, the resultant's magnitude follows the law of cosines:

$R=\sqrt{A^2+B^2+2AB\cos \theta }$

and its direction follows the law of sines.

Note the plus sign before the $2AB\cos \theta$ term. It is there because $\theta$ is the angle between the two vectors drawn from a common tail. Setting $\theta =90°$ collapses this to Pythagoras, which is a quick check that you have written it the right way round.

6. Two worked examples that replace the removed section

Which way to hold the umbrella (Example 3.1)

Rain falls vertically at 35 m/s. A wind blows at 12 m/s from east to west. Which way should a boy at a bus stop hold his umbrella?

Treat both as vectors and add them. They are perpendicular, so:

$R=\sqrt{35^2+12^2}=\sqrt{1225+144}=\sqrt{1369}=37\text{m/s}$

The direction is tilted from the vertical by ${\tan }^{-1}(12/35)\approx 19°$, toward the west.

The physical lesson: rain "falling vertically" only falls vertically for someone standing still in still air. Combine it with any horizontal motion and it arrives at a slant.

The motorboat and the current (Example 3.3)

A motorboat heads north at 25 km/h; the current runs at 10 km/h, 60° east of south. Find the resultant velocity.

These are not perpendicular, so the law of cosines earns its keep. The angle between them is 120°, and $\cos 120°=-\frac{1}{2}$:

$R=\sqrt{25^2+10^2+2(25)(10)(-\frac{1}{2})}=\sqrt{625+100-250}\approx 22\text{km/h}$

Notice the resultant (22 km/h) is less than the boat's own speed (25 km/h). The current has a southward component that partly cancels the boat's progress. Adding two vectors can decrease the total — something that never happens with scalars.

7. Scalar and vector products — examinable, but not in this chapter

CBSE lists both under Chapter 3, so here they are. Neither appears in the NCERT Chapter 3 text.

The scalar (dot) product

$\mathbf{A}\cdot \mathbf{B}=AB\cos \theta$

It takes two vectors and returns a scalar. In component form:

$\mathbf{A}\cdot \mathbf{B}=A_x{B}_x+A_y{B}_y+A_z{B}_z$

It is commutative: $\mathbf{A}\cdot \mathbf{B}=\mathbf{B}\cdot \mathbf{A}$.

This is how work is defined in Chapter 5 — only the part of the force along the displacement does work, and $\cos \theta$ is what extracts that part.

The vector (cross) product

$\mathbf{A}\times \mathbf{B}=AB\sin \theta \hat{\mathbf{n}}$

It takes two vectors and returns a vector, perpendicular to both, with direction given by the right-hand rule.

Order matters, and this is the classic slip:

$\mathbf{A}\times \mathbf{B}=-\mathbf{B}\times \mathbf{A}$

The dot product does not care about order. The cross product reverses sign.

This is how torque and angular momentum are defined in Chapter 6.

Remember the two zeros the opposite way round: the dot product vanishes when the vectors are perpendicular; the cross product vanishes when they are parallel.

8. Motion in a plane, and the idea that makes it easy

With the algebra in place, the kinematics carries straight over from Chapter 2. Every quantity simply becomes a vector.

For constant acceleration, the Chapter 2 equations hold in vector form:

$\mathbf{v}={\mathbf{v}}_0+\mathbf{a}t\mathbf{r}={\mathbf{r}}_0+{\mathbf{v}}_{0}t+\frac{1}{2}\mathbf{a}{t}^2$

Here is the idea that makes two-dimensional problems tractable at all. Because a vector equation holds component by component, the x-motion and the y-motion are completely independent. What happens horizontally has no influence on what happens vertically.

So a two-dimensional problem is not twice as hard as a one-dimensional one. It is two one-dimensional problems that happen to share a clock. Every projectile question in this chapter is solved by exploiting exactly that.

9. Projectile motion

A projectile is given an initial velocity and then left alone, with gravity the only force and air resistance neglected. Taking upward as positive:

$a_x=0a_y=-g$

That single line is the whole physical content. Horizontally there is no acceleration, so the horizontal velocity never changes. Vertically it is ordinary free fall.

Launching at speed $v_0$ and angle ${\theta }_0$:

$x=(v_0\cos {\theta }_0)ty=(v_0\sin {\theta }_0)t-\frac{1}{2}g{t}^2$

Eliminate $t$ between them and $y$ becomes a quadratic in $x$. That is why the path is a parabola — not because anyone assumed it, but because it falls out of two independent motions.

The three standard results

Since $\sin 2{\theta }_0$ is largest when $2{\theta }_0=90°$, the range is maximum at ${\theta }_0=45°$.

Galileo's symmetry result (Example 3.6)

Galileo stated in Two New Sciences that elevations exceeding or falling short of 45° by equal amounts give equal ranges.

The proof is short. For angles $(45°+\alpha )$ and $(45°-\alpha )$, the quantity $2{\theta }_0$ becomes $(90°+2\alpha )$ and $(90°-2\alpha )$. But

$\sin (90°+2\alpha )=\sin (90°-2\alpha )=\cos 2\alpha$

Range depends on the angle only through $\sin 2{\theta }_0$, so the two ranges are identical. A launch at 30° and a launch at 60° travel exactly as far.

The horizontal throw (Example 3.7)

A hiker on a 490 m cliff throws a stone horizontally at 15 m/s. Find the time to reach the ground and the impact speed.

The initial vertical velocity is zero, so the vertical motion is a plain drop from rest — the horizontal 15 m/s makes no difference to how long the fall takes.

From $-490=-\frac{1}{2}(9.8)t^2$ we get $t=10$ s.

At impact the horizontal velocity is still 15 m/s, since nothing ever changed it. The vertical velocity is $0-9.8\times 10=-98$ m/s. So:

$v=\sqrt{15^2+98^2}\approx 99\text{m/s}$

This is independence at its most vivid: a stone thrown horizontally and a stone simply dropped from the same cliff hit the ground at the same moment.

The cricket ball (Example 3.8)

Thrown at 28 m/s, 30° above the horizontal, with $g=9.8$ m/s²:

10. Uniform circular motion

An object moving on a circle at constant speed is in uniform circular motion. "Uniform" refers to the speed, and only to the speed.

This motion is accelerated. Velocity is a vector; its magnitude is constant here, but its direction changes continuously, since velocity always points along the tangent. A changing vector has a non-zero rate of change, so there is acceleration — even though the speedometer never moves.

Anyone who says "constant speed means no acceleration" has quietly swapped a vector for a scalar.

Which way does that acceleration point?

Take the object at two nearby points, P and P′.

1. The path is circular, so $\mathbf{v}$ is perpendicular to $\mathbf{r}$, and ${\mathbf{v}}^{\prime }$ is perpendicular to ${\mathbf{r}}^{\prime }$.
2. It follows that $\Delta \mathbf{v}$ is perpendicular to $\Delta \mathbf{r}$.
3. Average acceleration points along $\Delta \mathbf{v}$, and $\Delta \mathbf{r}$ points along the motion.
4. So the acceleration is perpendicular to the motion.

Shrink the interval to zero and that perpendicular direction becomes precisely toward the centre — which is why it is called centripetal, meaning centre-seeking.

$a_c=\frac{v^2}{R}={\omega }^{2}R$

with the related quantities:

$\omega =\frac{2\pi }{T}=2\pi \nu v=\omega R=2\pi R\nu a_c=4{\pi }^2{\nu }^{2}R$

This chapter gives the centripetal acceleration only. Centripetal force, the $m{v}^2/R$ that causes it, needs Newton's laws and belongs to Chapter 4. It is not part of this chapter's kinematics.

The insect in a groove (Example 3.9)

An insect completes 7 revolutions in 100 s in a circular groove of radius 12 cm.

The example then asks the better question: is the acceleration vector constant?

No. Its magnitude is fixed at 2.3 cm/s², but it always points at the centre, and that direction turns continuously as the insect goes round. A vector of fixed length and rotating direction is not a constant vector.

Constant magnitude and constant vector are two different claims. This example exists to keep them apart.

Summary

• Along a line, signs were enough. In a plane you need magnitude and direction, so vectors replace signed numbers.
• Path length and displacement answer different questions: 3 m east then 4 m north is 7 m walked, 5 m displaced.
• $\lambda \mathbf{A}$ scales the magnitude by $|\lambda |$ and reverses the direction when $\lambda$ is negative — which is what makes subtraction possible.
• Vector addition is commutative and associative; subtraction is defined as adding the negative.
• Resolve into components and addition becomes two independent sums. Check the quadrant — an inverse tangent alone cannot.
• Not at right angles: $R=\sqrt{A^2+B^2+2AB\cos \theta }$. The plus sign is there because $\theta$ is measured between the vectors.
• CBSE examines scalar and vector products under this chapter, but NCERT teaches them in 5.1.1 and 6.5. Dot gives a scalar and vanishes at 90°; cross gives a vector and vanishes at 0°.
• A vector equation holds component by component, so horizontal and vertical motion are independent and share only a clock.
• For a projectile $a_x=0$ and $a_y=-g$; the path is a parabola because it must be.
• $T=2v_0\sin {\theta }_0/g$, $H=v_0^2{\sin }^2{\theta }_0/2g$, $R=v_0^2\sin 2{\theta }_0/g$, with range greatest at 45°.
• Angles equally above and below 45° give equal ranges — 30° and 60° travel the same distance.
• A horizontally thrown stone and a dropped stone land together.
• Uniform circular motion is accelerated: $a_c=v^2/R={\omega }^{2}R$, directed at the centre. Constant speed is not constant velocity.
• The acceleration in circular motion has constant magnitude but is not a constant vector.