Is Motion in a Plane part of the EAPCET Physics syllabus?

Yes. Motion in a Plane is taught in Intermediate first year Physics, and the EAPCET Engineering syllabus is based on the Intermediate first- and second-year syllabus. Check the official notification each year for any changes.

How many Physics questions are there in EAPCET?

40 of the 160 questions in the Engineering paper are Physics, spread across 30 chapters.

What mistakes do students make in Motion in a Plane?

Using sin θ instead of sin 2θ in the range formula, or sin θ instead of sin²θ in the height formula. Saying the velocity is zero at the top of the path — only the vertical component is zero; the speed there is u cos θ. Adding vector magnitudes directly — use the components or R² = A² + B² + 2AB cos θ. Thinking uniform circular motion has zero acceleration — the speed is constant but the velocity changes direction, giving v²/r towards the centre.

EAPCET Physics · Intermediate first year

Motion in a Plane

Motion in a Plane uses vectors to describe two-dimensional motion, with projectile motion and uniform circular motion as the main applications.

Subject: Physics
Taught in: Intermediate first year
Physics in the paper: 40 of 160 questions

Key formulas and results

Range $R = \frac{u^{2}\sin 2\theta}{g}$ , maximum at $\theta = 45^\circ$
Maximum height $H = \frac{u^{2}\sin^{2}\theta}{2g}$
Time of flight $T = \frac{2u\sin\theta}{g}$
Projections at $\theta$ and $90^\circ - \theta$ have the same range
Centripetal acceleration $a = \frac{v^{2}}{r} = \omega^{2}r$
Equation of trajectory: $y = x\tan\theta - \frac{gx^{2}}{2u^{2}\cos^{2}\theta} = x\tan\theta\left(1 - \frac{x}{R}\right)$
Horizontal projection from height $h$ : time $T = \sqrt{2h/g}$ , horizontal distance $= u\sqrt{2h/g}$
Range and height from components: $R = u_{x}T = \frac{2u_{x}u_{y}}{g}$ ; $H = \frac{u_{y}^{2}}{2g}$
Projectile thrown perpendicular to an incline of angle $\alpha$ : time of flight $T = \frac{2u}{g\cos\alpha}$
Rain and umbrella: $v_{\text{rain}}$ relative to man $= v_{\text{rain}} - v_{\text{man}}$ ; tilt the umbrella from the vertical by $\tan\theta = v_{\text{man}}/v_{\text{rain}}$
River crossing (boat speed $v$ in still water, river speed $u$ , width $d$ ): shortest time $d/v$ with drift $ud/v$ ; zero drift needs $\sin\theta = u/v$ (upstream) and takes $\frac{d}{\sqrt{v^{2} - u^{2}}}$
Non-uniform circular motion: $a = \sqrt{a_{c}^{2} + a_{t}^{2}}$ , with $a_{c} = \frac{v^{2}}{r}$ and $a_{t} = \frac{dv}{dt}$
Uniform circular motion through angle $\theta$ : average velocity $= \frac{2v\sin(\theta/2)}{\theta}$

What this chapter covers

Scalars and vectors
Resolution of vectors
Projectile motion
Relative velocity in two dimensions
Uniform circular motion

How it is usually asked

Find the range, height or time of flight of a projectile
Resolve and add vectors
River–boat and rain–man problems
Uniform circular motion

Understanding Motion in a Plane

Two-dimensional motion becomes simple once you resolve vectors into perpendicular components. A vector of magnitude A at angle θ to the x-axis has components A cos θ and A sin θ, and vectors add component by component. Because the x and y motions are independent, a plane problem splits into two one-dimensional problems that share the same time.

In projectile motion the horizontal velocity u cos θ stays constant, while the vertical motion is free fall with acceleration g downwards. From this you get the time of flight, maximum height and range. At the highest point only the vertical velocity is zero; the body still moves horizontally. In uniform circular motion the speed is constant, but the direction changes, so there is a centripetal acceleration v²/r towards the centre.

EAPCET mostly asks for range, maximum height or time of flight, the equal ranges of complementary angles, river-crossing and rain–umbrella problems, and the addition or resolution of vectors. In river problems, draw the velocity triangle first: to cross in the shortest time, head straight across; to reach the point directly opposite, aim upstream so that the drift of the river is cancelled.

Worked example

EAPCET-style question

A ball is projected with a speed of 20 m/s at 30° above the horizontal. Taking $g = 10\ \text{m/s}^{2}$ , its horizontal range is

(1) $20\sqrt{3}\ \text{m}$
(2)40 m
(3) $10\sqrt{3}\ \text{m}$
(4)20 m

Solution

Use $R = \frac{u^{2}\sin 2\theta}{g}$ .
$\sin 2\theta = \sin 60^\circ = \frac{\sqrt{3}}{2}$ .
$R = \frac{20^{2} \times \sqrt{3}/2}{10} = \frac{400 \times \sqrt{3}/2}{10}$ .
$R = 20\sqrt{3}\ \text{m} \approx 34.6\ \text{m}$ .

Answer: $20\sqrt{3}\ \text{m}$

Shortcuts and results to remember

For complementary angles $\theta$ and $90^\circ - \theta$ : $t_{1}t_{2} = \frac{2R}{g}$ and $R = 4\sqrt{h_{1}h_{2}}$ .
The horizontal velocity of a projectile never changes, so the angle $\alpha$ of the velocity at any instant is given by $\tan\alpha = \frac{v_{y}}{u\cos\theta}$ .
Use $\sin 37^\circ = 3/5$ and $\cos 37^\circ = 4/5$ : 50 m/s at 53° splits into 30 m/s horizontal and 40 m/s vertical.
Two vectors of equal size $P$ at angle $\theta$ : $|\vec{A} + \vec{B}| = 2P\cos(\theta/2)$ and $|\vec{A} - \vec{B}| = 2P\sin(\theta/2)$ ; at 60° these are $\sqrt{3}P$ and $P$ , at 120° they are $P$ and $\sqrt{3}P$ .
On an inclined plane, resolve $g$ into $g\sin\alpha$ along the plane and $g\cos\alpha$ perpendicular to it.

Common mistakes to avoid

Using sin θ instead of sin 2θ in the range formula, or sin θ instead of sin²θ in the height formula.
Saying the velocity is zero at the top of the path — only the vertical component is zero; the speed there is u cos θ.
Adding vector magnitudes directly — use the components or R² = A² + B² + 2AB cos θ.
Thinking uniform circular motion has zero acceleration — the speed is constant but the velocity changes direction, giving v²/r towards the centre.

How to prepare it

Treat horizontal and vertical motion separately; only time is shared between them.

Once the formulas feel familiar, test them the way the exam does: practise Motion in a Plane questions chapter-wise, then attempt a full previous year paper to see whether it holds up under a 3-hour timer.

Related Physics chapters

See every chapter in the EAPCET Physics syllabus.

Practise Motion in a Plane chapter-wise

Chapter-wise quizzes show exactly which topics in this chapter you still get wrong — before a full paper does.

Frequently asked questions

Is Motion in a Plane part of the EAPCET Physics syllabus?: Yes. Motion in a Plane is taught in Intermediate first year Physics, and the EAPCET Engineering syllabus is based on the Intermediate first- and second-year syllabus. Check the official notification each year for any changes.
How should I prepare Motion in a Plane for EAPCET?: Treat horizontal and vertical motion separately; only time is shared between them.
How many Physics questions are there in EAPCET?: 40 of the 160 questions in the Engineering paper are Physics, spread across 30 chapters.
What mistakes do students make in Motion in a Plane?: Using sin θ instead of sin 2θ in the range formula, or sin θ instead of sin²θ in the height formula. Saying the velocity is zero at the top of the path — only the vertical component is zero; the speed there is u cos θ. Adding vector magnitudes directly — use the components or R² = A² + B² + 2AB cos θ. Thinking uniform circular motion has zero acceleration — the speed is constant but the velocity changes direction, giving v²/r towards the centre.