Motion in 2D

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Motion in 2D — Key Facts

When an object moves in a plane, its position, velocity, and acceleration can be described using components in x and y directions.

Position Vector: $$\vec{r} = x\hat{i} + y\hat{j}$$

Velocity: $$\vec{v} = \frac{d\vec{r}}{dt} = v_x\hat{i} + v_y\hat{j}$$ $$v_x = \frac{dx}{dt}, \quad v_y = \frac{dy}{dt}$$

Acceleration: $$\vec{a} = \frac{d\vec{v}}{dt} = a_x\hat{i} + a_y\hat{j}$$ $$a_x = \frac{dv_x}{dt}, \quad a_y = \frac{dv_y}{dt}$$

⚡ JEE Exam Tip: In 2D motion, the horizontal component (x-direction) and vertical component (y-direction) are completely independent when acceleration has no component in a direction. This is the principle of superposition for motion.

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🟡 Standard — Regular Study (2d–2mo)

For students who want genuine understanding…

Projectile Motion:

Projectile motion is the motion of an object under constant gravitational acceleration (g = 9.8 m/s²) with no other forces.

Initial conditions: Object launched with speed u at angle θ to horizontal $$u_x = u\cos\theta, \quad u_y = u\sin\theta$$

At time t: $$x = u\cos\theta \cdot t$$ $$y = u\sin\theta \cdot t - \frac{1}{2}gt^2$$

Key Quantities:

Quantity	Formula
Time of flight	$T = \frac{2u\sin\theta}{g}$
Maximum height	$H = \frac{u^2\sin^2\theta}{2g}$
Horizontal range	$R = \frac{u^2\sin 2\theta}{g}$

Important Properties:

• Range is maximum when θ = 45°
• Complementary angles (θ and 90°-θ) give the same range but different trajectories
• At highest point: v_y = 0, v = u_x = u cosθ (purely horizontal)

Trajectory Equation: $$y = x\tan\theta - \frac{gx^2}{2u^2\cos^2\theta}$$

This is the equation of a parabola.

⚡ JEE Exam Tip: For projectile thrown from a height h (not ground level), the time to hit ground is found by solving $h + u\sin\theta \cdot t - \frac{1}{2}gt^2 = 0$. The range formula is different.

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Projectile Motion on Inclined Plane:

When projectile is launched up an inclined plane (angle α):

Resolve axes along and perpendicular to the plane:

• Along plane: acceleration = -g sinα
• Perpendicular to plane: acceleration = -g cosα

Equations become more complex; trajectory is not symmetric.

Relative Motion in 2D:

Velocity of A relative to B: $$\vec{v}_{AB} = \vec{v}_A - \vec{v}_B$$

Relative acceleration: $\vec{a}_{AB} = \vec{a}_A - \vec{a}_B$

Applications:

River Boat Problems:

• Downstream speed: $v_d = v_b + v_r$ (current assists)
• Upstream speed: $v_u = v_b - v_r$ (current opposes)
• Time to cross river (width d): $t = d/v_b$ (perpendicular component unchanged)

Aircraft in Wind:

• Wind affects aircraft’s ground velocity, not airspeed
• Ground velocity = air velocity + wind velocity

Circular Motion:

When an object moves in a circle of radius r with speed v:

Quantity	Formula
Angular displacement	$\theta = s/r$
Angular velocity	$\omega = d\theta/dt = v/r$
Time period	$T = 2\pi r/v = 2\pi/\omega$
Centripetal acceleration	$a_c = v^2/r = \omega^2 r$
Centripetal force	$F_c = mv^2/r = m\omega^2 r$

Non-Uniform Circular Motion:

When speed varies along the circular path: $$\vec{a} = \vec{a}_r + \vec{a}_t$$

• Radial (centripetal): $a_r = v^2/r$ toward centre
• Tangential: $a_t = dv/dt$ (change in speed)

Banking of Roads:

For a banked curve without friction: $$\tan\theta = \frac{v^2}{rg}$$

where θ = angle of banking, v = design speed

This allows vehicles to navigate curves without relying on friction.

Conical Pendulum:

A mass attached to a string moving in a horizontal circle:

• String makes angle θ with vertical
• $T\cos\theta = mg$, $T\sin\theta = mv^2/r$
• $v = \sqrt{rgtan\theta}$
• Period: $T = 2\pi\sqrt{\frac{r}{gtan\theta}}$

⚡ JEE Advanced 2022 Analysis: Questions on projectile motion from heights, relative velocity in 2D, and banking of roads appeared in recent papers. For projectile on inclined plane, use coordinate axes along and perpendicular to the plane to simplify calculations.

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