Unit 2: Kinematics

Kinematics is the language of mechanics. Two or three questions appear directly in JEE Main every year, and projectile/relative-motion problems form the basis for the more advanced Newtonian dynamics and energy problems later. In JEE Advanced, kinematics is hidden inside multi-concept questions: projectile + collisions, river-boat + work-energy, calculus-based motion under variable forces.

Typical question types:

• JEE Main: equations of motion, $v$-$t$/$x$-$t$ graphs, projectile range and max-height, relative velocity (boat/rain).
• JEE Advanced: variable acceleration with $a = f(v)$ or $a=f(x)$, projectile on incline (max range derivation), two-particle simultaneous projectile problems, the relative-velocity trick that turns a 2-particle problem into a 1-particle problem.

Concept Map

• 1D motion
  • Position, displacement, distance
  • Average vs instantaneous velocity & speed
  • Calculus relations: $v = dx/dt$, $a = dv/dt = v\,dv/dx$
  • Equations of uniform acceleration; motion under gravity
  • Graphs ($x$-$t$, $v$-$t$, $a$-$t$); area under curves
  • Variable acceleration: $a=f(t)$, $f(v)$, $f(x)$
• 2D motion
  • Position vector, velocity, acceleration in vector form
  • Independence of perpendicular motions
• Projectile motion
  • Horizontal range, max height, time of flight, equation of trajectory
  • Projectile from a height
  • Projectile on inclined plane (along & against slope)
  • Maximum range on incline
• Relative velocity
  • 1D and 2D relative velocity
  • River-boat: cross in minimum time vs minimum drift
  • Rain-man problems

Topic 1: One-Dimensional Motion

Sub-topic A: Position, displacement, distance

For motion along the $x$-axis, position is $x(t)$; displacement between $t_1, t_2$ is $\Delta x = x(t_2)-x(t_1)$. Distance is the total path length; it is always $\ge \vert \Delta x\vert$ and equal only when motion is in one direction.

Sub-topic B: Velocity and speed

• Average velocity $\bar v = \Delta x/\Delta t$ (vector).
• Average speed $=$ total distance / total time (scalar).
• Instantaneous velocity $v=dx/dt$.
• Instantaneous speed $\vert v\vert$.

Sub-topic C: Calculus relations

$v = \frac{dx}{dt},\qquad a = \frac{dv}{dt} = \frac{d^2x}{dt^2} = v\frac{dv}{dx}.$

The chain-rule identity $a = v\,dv/dx$ is invaluable when $a$ is a function of position.

Sub-topic D: Equations of motion (uniform acceleration)

Starting from $a = \text{const}$:

• $\int dv = \int a\,dt \Rightarrow v = u + at$.
• $\int dx = \int (u+at)\,dt \Rightarrow x = ut + \tfrac12 a t^2$.
• From $a\,dx = v\,dv$: $v^2 = u^2 + 2as$.
• Distance in $n$-th second: $s_n = u + \tfrac{a}{2}(2n-1)$.

Important: These hold only for constant $a$.

Sub-topic E: Motion under gravity

Take upward positive, $a=-g$:

• Freely falling body from rest: $v = gt$, $h = \tfrac12 g t^2$, $v^2 = 2gh$.
• Particle thrown up with speed $u$: time to top $= u/g$, max height $= u^2/(2g)$, returns to start in $2u/g$ with same speed $u$ (downward).
• Time of flight from height $h$ thrown up with $u$: $-h = ut-\tfrac12 g t^2\Rightarrow t = \dfrac{u+\sqrt{u^2+2gh}}{g}$.

Sub-topic F: Graphs

Key: the slope of an $x$-$t$ graph at any instant is $v$; if the slope is zero (horizontal tangent), the particle is momentarily at rest.

Sub-topic G: Variable acceleration

Case 1: $a = f(t)$. Integrate: $v = u + \int_0^t f(t')\,dt'$. Then $x = x_0 + \int_0^t v\,dt'$.

Case 2: $a = f(v)$. Use $a = dv/dt$ and separate variables:

$dt = \frac{dv}{f(v)},\quad t = \int \frac{dv}{f(v)}.$

Or use $a = v\,dv/dx$:

$dx = \frac{v\,dv}{f(v)}.$

Case 3: $a = f(x)$. Use $a = v\,dv/dx$:

$v\,dv = f(x)\,dx \Rightarrow \tfrac12 v^2 = \int f(x)\,dx + C.$

Worked Examples (JEE Main level)

Example 1.1. A particle moves along $x$ with $x(t)=t^3-6t^2+9t+1$ (SI). Find (a) velocity, (b) acceleration at $t=2$ s, (c) when is the particle at rest, (d) distance travelled in the first 3 s.

(a) $v=dx/dt=3t^2-12t+9$. At $t=2$: $v=12-24+9=-3$ m/s.

(b) $a=dv/dt=6t-12$. At $t=2$: $a=0$.

(c) $v=0\Rightarrow 3t^2-12t+9=0\Rightarrow t=1,3$ s.

(d) Distance = $\vert x(1)-x(0)\vert +\vert x(3)-x(1)\vert =\vert 5-1\vert +\vert 1-5\vert =4+4=8$ m.

Example 1.2. A ball is dropped from a height $H$. Find ratio of distances covered in 1st, 2nd, 3rd second.

$s_n = u + \tfrac g2(2n-1)$ with $u=0$. Ratio $1:3:5$ (Galileo's odd-numbers rule).

Example 1.3. A car accelerates from rest at $2\,\text{m/s}^2$ for $5$ s, then moves at constant velocity for $10$ s, then decelerates at $4\,\text{m/s}^2$ until stopping. Total distance?

Phase 1: $v_1 = 0+2\cdot5=10$ m/s; $s_1=\tfrac12\cdot2\cdot25=25$ m.

Phase 2: $s_2 = 10\cdot10=100$ m.

Phase 3: $0 = 10-4t\Rightarrow t=2.5$ s; $s_3 = 10\cdot2.5-\tfrac12\cdot4\cdot(2.5)^2 = 25-12.5=12.5$ m.

Total $=25+100+12.5 = 137.5$ m.

Worked Examples (JEE Advanced level)

Example 1.A1. Acceleration of a particle is $a=-kv^2$. Initial speed $u$ at $x=0$. Find $v(x)$.

$v\,dv/dx=-kv^2\Rightarrow dv/v=-k\,dx\Rightarrow \ln(v/u)=-kx\Rightarrow v=ue^{-kx}$.

Example 1.A2. A particle starts from rest at $x_0$ and moves with $a = -\omega^2 x$. Show that $x(t)=x_0\cos\omega t$.

$v\,dv/dx=-\omega^2 x\Rightarrow v^2 = \omega^2(x_0^2-x^2)$.

Then $dx/dt = -\omega\sqrt{x_0^2-x^2}$ (sign chosen so $x$ decreases initially). Integrating with $x(0)=x_0$ gives $x = x_0\cos\omega t$. (SHM about origin.)

Example 1.A3. Two trains, $120$ m and $80$ m long, run on parallel tracks in opposite directions at $54$ and $36$ km/h. Time to completely pass each other?

Relative speed = $90$ km/h $= 25$ m/s. Total length = $200$ m. Time = $200/25 = 8$ s.

Topic 2: Two-Dimensional Motion

Sub-topic A: Vector form

$\vec r(t) = x(t)\hat\imath + y(t)\hat\jmath,\quad \vec v = \dot{\vec r},\quad \vec a = \ddot{\vec r}.$

If $\vec a$ is constant:

$\vec v = \vec u + \vec a t,\qquad \vec r = \vec r_0 + \vec u t + \tfrac12 \vec a t^2.$

Each component obeys the 1D equations independently.

Sub-topic B: Independence of perpendicular motions

If $\vec a = (a_x, a_y)$ and $\vec u = (u_x, u_y)$, then:

$x = x_0 + u_x t + \tfrac12 a_x t^2,\qquad y = y_0 + u_y t + \tfrac12 a_y t^2.$

Solve each independently.

Worked Examples (JEE Main level)

Example 2.1. A particle has $\vec u = 3\hat\imath$ m/s and $\vec a = 4\hat\jmath$ m/s². Find $\vec v$ and $\vec r$ at $t=2$ s.

$\vec v = 3\hat\imath + 8\hat\jmath$ m/s. $\vec r = 6\hat\imath + 8\hat\jmath$ m.

Example 2.2. Position vector $\vec r = (3t)\hat\imath + (4t-5t^2)\hat\jmath$ m. Find speed at $t=0.5$ s.

$\vec v = 3\hat\imath + (4-10t)\hat\jmath$. At $t=0.5$: $\vec v = 3\hat\imath - 1\hat\jmath$. Speed $=\sqrt{10}$ m/s.

Topic 3: Projectile Motion

Sub-topic A: Horizontal ground

Launch from origin with speed $u$ at angle $\theta$ above horizontal; $g$ downward.

$u_x = u\cos\theta$, $u_y = u\sin\theta$. $a_x=0$, $a_y=-g$.

$x = u\cos\theta\,t,\qquad y = u\sin\theta\,t - \tfrac12 g t^2.$

Time of flight $T$: set $y=0$ (other than $t=0$): $T = \dfrac{2u\sin\theta}{g}$.

Range $R$: $R = u\cos\theta\,T = \dfrac{u^2\sin2\theta}{g}$. Maximum $R_{\max} = u^2/g$ at $\theta=45^\circ$.

Maximum height $H$: $H = \dfrac{u^2\sin^2\theta}{2g}$.

Equation of trajectory: eliminate $t$:

$y = x\tan\theta - \frac{g x^2}{2u^2\cos^2\theta} = x\tan\theta\left(1-\frac{x}{R}\right).$

Useful relations:

• $H/R = \tan\theta/4$.
• At max height, $v=u\cos\theta$ (horizontal).
• $\theta$ and $90^\circ-\theta$ give same range (complementary angles).

Sub-topic B: Projectile from a height

A ball thrown horizontally with speed $u$ from height $h$:

$y=h-\tfrac12 g t^2$ — hits ground at $t=\sqrt{2h/g}$. Horizontal range $R = u\sqrt{2h/g}$. Velocity at ground: $v_x=u, v_y=-\sqrt{2gh}$, speed $=\sqrt{u^2+2gh}$.

For projectile thrown at angle $\theta$ from height $h$ with speed $u$, time of flight from $y=0$ down to $-h$:

$-h = u\sin\theta\,t - \tfrac12 g t^2 \Rightarrow t = \dfrac{u\sin\theta+\sqrt{u^2\sin^2\theta+2gh}}{g}$.

Sub-topic C: Projectile on an inclined plane

Take axes along ($x'$) and perpendicular ($y'$) to incline of angle $\alpha$. Gravity components: $g_{x'}=-g\sin\alpha$ (along incline, downhill), $g_{y'}=-g\cos\alpha$ (into incline).

If projectile is launched from the foot of the incline at angle $\beta$ to incline with speed $u$ (and angle $\alpha+\beta$ with horizontal):

$u_{x'}=u\cos\beta$, $u_{y'}=u\sin\beta$.

Time of flight (returns to incline, $y'=0$):

$T = \frac{2u\sin\beta}{g\cos\alpha}.$

Range along incline:

$R = u\cos\beta\,T - \tfrac12 g\sin\alpha\,T^2.$

After simplification:

$\boxed{R = \frac{2u^2\sin\beta\cos(\alpha+\beta)}{g\cos^2\alpha}}\quad\text{(up the incline).}$

Down the incline (projected from top, angle $\beta$ below the line of incline reversed): replace $\sin\alpha\to-\sin\alpha$:

$R_{\downarrow} = \frac{2u^2\sin\beta\cos(\alpha-\beta)}{g\cos^2\alpha}.$

Maximum range up the incline. Maximize over $\beta$: derivative of $\sin\beta\cos(\alpha+\beta)$ vanishes at $\beta = \pi/4-\alpha/2$, giving

$R_{\max,\uparrow} = \frac{u^2}{g(1+\sin\alpha)}.$

Maximum range down the incline: $\beta = \pi/4+\alpha/2$,

$R_{\max,\downarrow} = \frac{u^2}{g(1-\sin\alpha)}.$

Derivation of max-range angle (up the incline). Set $f(\beta)=\sin\beta\cos(\alpha+\beta)$. Use product-to-sum:

$f(\beta)=\tfrac12[\sin(2\beta+\alpha) - \sin\alpha].$

Maximum when $\sin(2\beta+\alpha)=1$, i.e. $2\beta+\alpha=\pi/2$, so $\beta=\pi/4-\alpha/2$. Plugging back: $f_{\max}=\tfrac12(1-\sin\alpha)$. So $R_{\max,\uparrow}=\dfrac{u^2(1-\sin\alpha)}{g\cos^2\alpha}=\dfrac{u^2}{g(1+\sin\alpha)}$ (using $\cos^2\alpha = 1-\sin^2\alpha$).

Worked Examples (JEE Main level)

Example 3.1. A projectile is fired with $u=20$ m/s at $30^\circ$ above horizontal. Find $R, H, T$. ($g=10$.)

$T=2\cdot20\cdot\tfrac12/10=2$ s. $R=400\cdot\sin60^\circ/10=20\sqrt 3\approx34.6$ m. $H=400\cdot(1/4)/20=5$ m.

Example 3.2. A ball thrown horizontally from a $45$ m cliff with $u=20$ m/s. Where does it land? With what velocity?

$t=\sqrt{2\cdot45/10}=3$ s. $R=20\cdot3=60$ m. $v_y=30$ m/s, so $v=\sqrt{400+900}=\sqrt{1300}\approx 36$ m/s. Angle $\arctan(30/20)=56.3^\circ$ below horizontal.

Example 3.3. A stone is projected at $30^\circ$ above horizontal at $20$ m/s, from the top of a $25$ m tower. Find total time of flight.

$u_y=10$, $h=25$. $25 = -10 t + 5 t^2 \Rightarrow t^2-2t-5=0\Rightarrow t = 1+\sqrt 6\approx 3.45$ s.

Worked Examples (JEE Advanced level)

Example 3.A1. A projectile is fired from the bottom of an incline of angle $30^\circ$ at speed $u$ at angle $\beta$ above the incline. Find $\beta$ for which the range up the incline is maximum and find that range.

$\beta_{\max} = \pi/4 - \alpha/2 = 45^\circ - 15^\circ = 30^\circ$. $R_{\max}=u^2/(g(1+\sin30^\circ))=u^2/(1.5g)=2u^2/(3g)$.

Example 3.A2. Two particles are thrown simultaneously from the same point with speeds $u_1$ and $u_2$ at angles $\theta_1$ and $\theta_2$. Show that the path of one as seen from the other is a straight line.

Relative velocity $\vec v_{12} = \vec u_1 - \vec u_2$ is constant since both experience the same $\vec g$ (their accelerations cancel in the relative frame). Hence the relative motion is uniform → straight-line path.

Example 3.A3. A ball is thrown with speed $u$ at angle $\theta$ above horizontal. Find the radius of curvature at (i) the highest point, (ii) the point of projection.

At any point, $a_n = g\cos\phi$ where $\phi$ is the angle of velocity with horizontal; $a_n = v^2/R_c$.

(i) At highest point, $\phi=0$, $v=u\cos\theta$, so $R_c = u^2\cos^2\theta/g$.

(ii) At point of projection, $\phi=\theta$, $v=u$, so $R_c = u^2/(g\cos\theta)$.

Topic 4: Relative Velocity

Sub-topic A: Definition

For two particles $A, B$ with velocities $\vec v_A, \vec v_B$ measured in the ground frame, the velocity of $A$ as seen by $B$ is

$\vec v_{AB} = \vec v_A - \vec v_B.$

Similarly $\vec a_{AB} = \vec a_A - \vec a_B$ and $\vec r_{AB} = \vec r_A - \vec r_B$.

Sub-topic B: 1D problems

Two cars on a straight road moving at $v_1, v_2$ in the same direction: relative speed $= \vert v_1-v_2\vert$. Opposite directions: $v_1+v_2$.

Sub-topic C: River-boat problem

Let the river flow with speed $v_R$ along $+x$ (east). The boat speed in still water is $v_B$. Let $\theta$ be the angle of the boat's heading (in still water) measured from the bank-normal (north).

Velocity of boat w.r.t. ground:

$\vec v_{Bg} = (v_R - v_B\sin\theta)\hat\imath + (v_B\cos\theta)\hat\jmath.$

Crossing in minimum time (no constraint on drift): max $y$-component → $\theta=0$ (head straight across). $t_{\min}=W/v_B$ where $W$ is width. Drift $= v_R\cdot t_{\min}=v_R W/v_B$.

Crossing with zero drift (minimum drift = 0, only possible if $v_B>v_R$): require $v_B\sin\theta=v_R\Rightarrow \sin\theta=v_R/v_B$. Boat heads upstream at this angle. Time $=W/(v_B\cos\theta)=W/\sqrt{v_B^2-v_R^2}$.

If $v_B<v_R$, zero drift impossible. Minimum drift: set $dy/dt$ and $dx/dt$ such that the velocity vector is as close to the across-direction as possible. Drift $x = v_R t - v_B\sin\theta\,t$. Width crossed $W = v_B\cos\theta\,t$. So $x = W\cdot\tfrac{v_R - v_B\sin\theta}{v_B\cos\theta}$. Minimise over $\theta$: $dx/d\theta=0$ gives $\sin\theta = v_B/v_R$. Then $\cos\theta=\sqrt{1-v_B^2/v_R^2}$ and minimum drift $x_{\min} = W\sqrt{v_R^2-v_B^2}/v_B$.

Sub-topic D: Rain-man problem

A man walks east with speed $v_M$. Rain falls vertically with speed $v_R$. In the man's frame, $\vec v_{RM} = \vec v_R - \vec v_M = -v_M\hat\imath - v_R\hat\jmath$. So rain appears to come at an angle $\arctan(v_M/v_R)$ from the vertical, in the direction opposite to walking — he tilts his umbrella forward.

If the rain falls at angle $\alpha$ with vertical with speed $v_R$ in the ground frame, decompose and re-apply $\vec v_{RM} = \vec v_R - \vec v_M$.

Worked Examples (JEE Main level)

Example 4.1. A boat can move at $5$ m/s in still water. River flows at $3$ m/s. Width $200$ m. (a) Min time to cross & drift? (b) Min-drift case & time?

(a) Min time $= 200/5 = 40$ s. Drift $=3\cdot40 = 120$ m.

(b) Head upstream at $\sin\theta = 3/5\Rightarrow\theta=37^\circ$. Time $=200/(5\cdot4/5)=200/4=50$ s. Drift $=0$.

Example 4.2. A man walks at $4$ m/s in rain falling vertically at $3$ m/s. At what angle should he hold his umbrella?

$\tan\alpha = 4/3\Rightarrow \alpha=53^\circ$ from vertical, tilted forward.

Example 4.3. Two trains on parallel tracks $80$ m apart, moving north at $20$ m/s and south at $25$ m/s. Min distance during passage?

Tracks parallel, perpendicular separation always $80$ m; minimum distance $=80$ m. (Trap question — answer is the perpendicular distance.)

Worked Examples (JEE Advanced level)

Example 4.A1. A swimmer can swim at $v_B$ in still water. A river of width $W$ flows at $v_R>v_B$. Find the angle $\theta$ from the across-direction at which she must swim to land at the point directly opposite to the start, knowing that no such point exists; instead find the minimum drift point.

Already shown above: $\sin\theta = v_B/v_R$ (heading partly into the stream gives minimum drift). Min drift $=W\sqrt{v_R^2-v_B^2}/v_B$.

Example 4.A2. Two cars start simultaneously from points $A$ and $B$ on perpendicular roads, moving at $40$ km/h and $30$ km/h toward the junction $O$, which is $4$ km from $A$ and $3$ km from $B$. After how long is their separation minimum, and what is it?

Relative velocity of $B$ w.r.t. $A$: $\vec v_{BA} = (0,30) - (40,0) = (-40, 30)$ km/h (taking $A\to O$ as $+x$ and $B\to O$ as $+y$). Initial relative position $\vec r_{BA}(0)=(-4,-3)$ km (taking $O$ as origin: $A$ at $(4,0)$, $B$ at $(0,3)$, then $\vec r_{BA}= (0,3)-(4,0)=(-4,3)$). Recompute carefully.

Place $O$ at origin. $A$ at $(4,0)$, $B$ at $(0,3)$. $A$ moves with $\vec v_A = (-40,0)$ km/h (toward $O$). $B$ moves with $\vec v_B = (0,-30)$ km/h. Position at time $t$: $A: (4-40t,0)$, $B: (0,3-30t)$. Separation:

$d^2 = (4-40t)^2 + (3-30t)^2.$

$d(d^2)/dt = 2(4-40t)(-40)+2(3-30t)(-30) = -80(4-40t)-60(3-30t) = -320+3200t -180+1800t = 5000t - 500 = 0\Rightarrow t = 0.1$ h $=6$ min.

$d^2 = (4-4)^2+(3-3)^2=0+0=0$. Oops — they collide at $O$! Of course: both arrive at $O$ at $t=0.1$ h.

So if speeds are tuned correctly they meet at the junction. Otherwise $d_{\min}>0$. In general $t_{\min} = -\vec r_0\cdot\vec v_{\text{rel}}/v_{\text{rel}}^2$, $d_{\min}=\vert \vec r_0\times\vec v_{\text{rel}}\vert /v_{\text{rel}}$.

Example 4.A3. Rain falls vertically at $v_R$. A man on a bus moving at $v_M$ holds an umbrella at $30^\circ$ with vertical. Later when the bus stops, rain appears to be vertical. Find $v_R/v_M$.

In the bus frame, rain's horizontal velocity is $-v_M$ (forward by $v_M$). $\tan30^\circ = v_M/v_R\Rightarrow v_R/v_M = \sqrt 3$.

Topic 5: Calculus-heavy problems

Pure calculus problems are JEE Advanced staple. The trick is identifying which of $a=f(t), f(v), f(x)$ applies and using the right form of $a$.

Example 5.1. A particle of mass $m$ moves along $x$ under force $F=-kx$ (Hooke's law). Find $v(x)$ given $v=v_0$ at $x=0$.

$m\,v\,dv/dx = -kx$. Integrate: $\tfrac12 mv^2 - \tfrac12 mv_0^2 = -\tfrac12 kx^2\Rightarrow v = \sqrt{v_0^2 - (k/m)x^2}$.

Example 5.2. A particle starts from rest with acceleration $a = \alpha - \beta v$. Find $v(t)$ and terminal speed.

$dv/dt = \alpha-\beta v$. Separate: $dv/(\alpha-\beta v) = dt\Rightarrow -\tfrac{1}{\beta}\ln(\alpha-\beta v)=t+C$. With $v(0)=0$: $C=-\tfrac{1}{\beta}\ln\alpha$. So $v(t)=\tfrac{\alpha}{\beta}(1-e^{-\beta t})$. Terminal speed $=\alpha/\beta$.

Example 5.3. A particle's position is $x = A\sin\omega t$. Find $a$ as a function of $x$.

$v = A\omega\cos\omega t$, $a=-A\omega^2\sin\omega t = -\omega^2 x$. (SHM.)

Problem-Solving Heuristics

1. Choose axes wisely in 2D motion — for inclined-plane projectiles, use axes along and perpendicular to the incline; for horizontal projectiles, use horizontal/vertical.
2. For two-particle problems, switch to the relative frame — it converts a two-body kinematics problem into single-particle with constant relative velocity (since $\vec a_{rel}=\vec 0$ if both feel only $\vec g$).
3. For variable acceleration, identify whether $a$ depends on $t, v$ or $x$, and use:
   • $a=f(t)\Rightarrow v=\int f\,dt$.
   • $a=f(v)\Rightarrow dt = dv/f$ or $dx = v\,dv/f$.
   • $a=f(x)\Rightarrow v\,dv = f\,dx$.
4. For projectile-on-incline problems: rotate axes; remember $\beta$ is measured from the incline, not horizontal.
5. Symmetry in projectiles: horizontal time up = time down; horizontal range is symmetric in $\theta$ about $45^\circ$.
6. Use the equation of trajectory $y=x\tan\theta(1-x/R)$ when given a point through which the projectile passes.
7. Always verify with the dimensional check and a limit check (e.g. $\alpha=0$ should reduce incline formulas to horizontal-ground formulas).
8. River-boat: the boat angle is measured w.r.t. the bank-normal, not the bank. Drawing a clean diagram avoids 90°-mix-ups.

Common Traps & Mistakes

• Distance vs displacement: a particle making a round trip has zero displacement but non-zero distance.
• Confusing instantaneous and average velocity. They are equal only for uniform motion.
• Using $v = u+at$ for variable acceleration — wrong.
• $H/R = \tan\theta/4$ is for projectile on level ground, not for incline.
• For projectile from height $h$, the time is not $2u\sin\theta/g$; that's only for return to same height. Use the quadratic with $y=-h$.
• Relative velocity formula: $\vec v_{AB} = \vec v_A - \vec v_B$ (velocity of $A$ as observed by $B$). The order matters — sign errors lose marks.
• In river-boat problems, students often confuse "angle with the flow" vs "angle with normal to flow". Pick one and stick to it.
• In rain-man problems, the man tilts the umbrella in the direction of his motion, not against.
• A particle moving in a circle has non-zero acceleration (centripetal) even with constant speed.

Quick Revision Card

• $v=u+at$, $s=ut+\tfrac12 at^2$, $v^2=u^2+2as$, $s_n=u+\tfrac a2(2n-1)$.
• Projectile (level): $T=2u\sin\theta/g$, $R=u^2\sin2\theta/g$, $H=u^2\sin^2\theta/(2g)$.
• Equation of trajectory: $y=x\tan\theta(1-x/R)$.
• Max range up an incline: $\beta = 45^\circ-\alpha/2$, $R=u^2/[g(1+\sin\alpha)]$.
• Down an incline: $\beta=45^\circ+\alpha/2$, $R=u^2/[g(1-\sin\alpha)]$.
• River-boat min time: head straight across. Zero drift: $\sin\theta = v_R/v_B$ (only if $v_B>v_R$).
• Radius of curvature: $R_c = v^2/a_n$, where $a_n$ is normal component of acceleration.

Formula Sheet