Kinematics

A hunter aims a rifle directly at a monkey hanging from a branch and fires. At the exact instant of the shot, the monkey lets go and falls. Does the bullet hit it?

The instinct is that it cannot. The bullet drops below the line of sight, and the monkey has moved.

It hits — every time, for any launch speed fast enough to reach the tree, and at any range.

Both bullet and monkey have the same acceleration, $\vec{g}$. So the acceleration of one relative to the other is zero:

${\vec{a}}_{BM}={\vec{a}}_B-{\vec{a}}_M=\vec{g}-\vec{g}=\vec{0}$

In the monkey's frame the bullet therefore travels in a straight line at constant velocity — straight along the original line of sight, which is exactly where the monkey is.

That is the whole method of this chapter. JEE Main kinematics asks you to substitute into $v=u+at$. JEE Advanced asks you to choose the frame in which the problem becomes easy, and to handle acceleration that refuses to stay constant.

1. When acceleration is not constant: three different problems

"Variable acceleration" is not one situation. It is three, and each needs a different first move. Reaching for the wrong one is the commonest way to lose a whole question.

The middle row is the one candidates forget exists. It comes from the chain rule and nothing else:

$a=\frac{dv}{dt}=\frac{dv}{dx}\cdot \frac{dx}{dt}=v\frac{dv}{dx}$

The bottom row needs a decision. With $a=f(v)$ you get time by integrating $dv/f(v)$ and position by integrating $vdv/f(v)$. Both are available; picking the one the question does not want costs you the question.

Illustration 1

A body moving at $v_0$ enters a resistive medium where $a=-kv$. Find $v(t)$, $x(t)$, and the total distance it travels before stopping.

Acceleration depends on velocity, and we want time first:

$\frac{dv}{dt}=-kv\Rightarrow {\int }_{v_0}^v\frac{dv}{v}=-k{\int }_0^{t}dt$

$v=v_0{e}^{-kt}$

Integrating once more for position:

$x=\frac{v_0}{k}\left(1-e^{-kt}\right)$

The body never actually stops — $v$ reaches zero only as $t\to \infty$. Yet the distance is finite:

$x_{total}={\lim }_{t\to \infty }x=\frac{v_0}{k}$

Get the same answer without touching time, which is the cleaner route and the one worth learning:

$v\frac{dv}{dx}=-kv\Rightarrow \frac{dv}{dx}=-k\Rightarrow v=v_0-kx$

Setting $v=0$ gives $x=v_0/k$ directly, in two lines. Infinite time, finite distance — a combination that looks paradoxical until you notice the exponential.

Illustration 2

A particle starts from rest at $x=a$ with $a_{acc}=-\frac{\mu }{x^2}$, directed toward the origin. Find its speed when it reaches $x=a/2$.

Acceleration is a function of position, so the middle row applies:

$v\frac{dv}{dx}=-\frac{\mu }{x^2}$

${\int }_0^{v}vdv=-\mu {\int }_a^{a/2}\frac{dx}{x^2}=-\mu {\left[-\frac{1}{x}\right]}_a^{a/2}=-\mu \left(-\frac{2}{a}+\frac{1}{a}\right)=\frac{\mu }{a}$

$\frac{v^2}{2}=\frac{\mu }{a}\Rightarrow v=\sqrt{\frac{2\mu }{a}}$

Recognise what this was. With $\mu =GM$ it is a body falling radially toward a mass — and the answer is exactly the energy-conservation result. Kinematics with $a=f(x)$ is the work–energy theorem, arrived at without ever using the word energy.

2. Relative motion as a change of frame

${\vec{r}}_{AB}={\vec{r}}_A-{\vec{r}}_B,{\vec{v}}_{AB}={\vec{v}}_A-{\vec{v}}_B,{\vec{a}}_{AB}={\vec{a}}_A-{\vec{a}}_B$

Read the subscript as "of A as seen from B". The three equations are one equation differentiated twice, so nothing new is being assumed.

The single most useful line in the chapter: if two bodies have the same acceleration, then ${\vec{a}}_{AB}=0$ and the relative motion is uniform velocity in a straight line — however complicated each individual path looks.

Two projectiles, a projectile and a falling body, two bodies in the same gravitational field: all of them see each other move in straight lines at constant speed. That is what killed the monkey question in one line, and it is what makes the next section possible.

Illustration 3

Two balls are projected simultaneously from the same point, one at $20\text{m/s}$ at $60°$ and the other at $20\text{m/s}$ at $30°$ to the horizontal. Find the distance between them after $2\text{s}$.

Both are in free fall, so ${\vec{a}}_{rel}=0$ and the separation grows linearly at the relative speed.

${\vec{u}}_1=20(\cos 60°,\sin 60°)=(10.0,17.32)$

${\vec{u}}_2=20(\cos 30°,\sin 30°)=(17.32,10.0)$

${\vec{u}}_{12}=(-7.32,7.32)\Rightarrow |{\vec{u}}_{12}|=10.35\text{m/s}$

$d=10.35(2)=20.7\text{m}$

Notice what never appeared: $g$. Computing two full trajectories and subtracting would have taken half a page and produced the same number, because every $g$ term cancels. The relative frame is not a shortcut here — it is the correct way to see the problem.

3. Closest approach

Two bodies move with constant velocities. How near do they get, and when?

Work in B's frame. There, A moves in a straight line at constant ${\vec{v}}_{AB}$, starting from ${\vec{r}}_{AB}$. The problem becomes the perpendicular distance from a point to a line:

$d_{min}=\frac{|{\vec{r}}_{AB}\times {\vec{v}}_{AB}|}{|{\vec{v}}_{AB}|},t_{min}=-\frac{{\vec{r}}_{AB}\cdot {\vec{v}}_{AB}}{{|{\vec{v}}_{AB}|}^2}$

Read the dot product's sign before computing anything. If ${\vec{r}}_{AB}\cdot {\vec{v}}_{AB}>0$ they are already separating, $t_{min}$ comes out negative, and the closest approach was in the past — the answer is the present separation. Missing this is the standard trap.

They collide if $d_{min}$ is zero, i.e. if ${\vec{r}}_{AB}$ and ${\vec{v}}_{AB}$ are antiparallel. That is the collision condition, and it is a statement about directions, not distances.

Illustration 4

At $t=0$, ship A is at the origin moving at $(3,4)\text{m/s}$, and ship B is at $(200,100)\text{m}$ moving at $(-1,4)\text{m/s}$. Find their least separation and when it occurs.

${\vec{r}}_{AB}=(0,0)-(200,100)=(-200,-100)$

${\vec{v}}_{AB}=(3,4)-(-1,4)=(4,0)$

The relative velocity is purely horizontal — the vertical components were identical and cancelled, which is the whole simplification.

${\vec{r}}_{AB}\cdot {\vec{v}}_{AB}=-800<0$, so they are still approaching. Good.

$t_{min}=-\frac{-800}{16}=50\text{s}$

$\left|{\vec{r}}_{AB}\times {\vec{v}}_{AB}\right|=|(-200)(0)-(-100)(4)|=400$

$d_{min}=\frac{400}{4}=100\text{m}$

Check it directly: after 50 s, A is at $(150,200)$ and B at $(150,300)$. Separation $100$ m, and the two are level horizontally — exactly the moment the horizontal gap closes, since the vertical gap never changes at all.

4. River crossing, done properly

Width $d$, current $u$ along the bank, boat speed $v$ in still water, steered at angle $\theta$ upstream from the normal. Resolve:

Minimum time. $t$ falls as $\cos \theta$ rises, so $\theta =0$: point the boat straight across.

$t_{min}=\frac{d}{v},\text{drift}=\frac{ud}{v}$

You accept drift to buy time. Note that steering upstream always costs time — there is no free lunch.

Zero drift, possible only if $v>u$: kill the along-bank component with $v\sin \theta =u$.

$\sin \theta =\frac{u}{v},t=\frac{d}{\sqrt{v^2-u^2}}$

When $u>v$ the boat cannot reach the point directly opposite at all. No steering angle makes $v\sin \theta$ equal $u$. The question then changes to minimum drift, and this is the part JEE Main never asks.

Minimise $\frac{u-v\sin \theta }{v\cos \theta }$ by differentiating and setting to zero. The condition comes out remarkably clean:

$\sin \theta =\frac{v}{u},{\text{drift}}_{min}=\frac{d\sqrt{u^2-v^2}}{v}$

The two conditions are reciprocals of each other — $\sin \theta =u/v$ when the boat is faster, $\sin \theta =v/u$ when the river is. Worth remembering as a pair, because mixing them up is easy and produces an impossible angle.

Illustration 5

A river $100\text{m}$ wide flows at $5\text{m/s}$. A boat does $3\text{m/s}$ in still water. Find the minimum drift and the crossing time in that case.

Here $u>v$, so landing directly opposite is impossible. Use the second condition:

$\sin \theta =\frac{v}{u}=\frac{3}{5}\Rightarrow \theta =37°$ upstream of the normal

${\text{drift}}_{min}=\frac{100\sqrt{25-9}}{3}=\frac{100(4)}{3}=133\text{m}$

Crossing time: $t=\frac{d}{v\cos \theta }=\frac{100}{3(0.8)}=41.7\text{s}$

Compare with pointing straight across: $t=100/3=33.3$ s with a drift of $5(33.3)=167$ m. Steering at $37°$ cuts the drift by 34 m and costs 8.4 extra seconds. Both cannot be optimised at once, and Advanced questions are usually explicit about which one they want.

Illustration 6

The same river, but now the boat can do $10\text{m/s}$. Compare the least-time crossing with the zero-drift crossing.

Now $v>u$, so both are available.

Eliminating all 50 m of drift costs just 1.5 seconds, because $\cos 30°=0.866$ is close to 1. The penalty for steering is second-order in $\theta$ while the benefit is first-order, which is why aiming a little upstream is almost always worth it in practice.

5. Rain, and what the umbrella is really pointing at

An umbrella must point along ${\vec{v}}_{rain,man}$, not along ${\vec{v}}_{rain}$. The faster you walk, the further forward you tilt it — which is why rain seems to come at you from the front on a still day.

${\vec{v}}_{rm}={\vec{v}}_r-{\vec{v}}_m$

Illustration 7

Rain falls vertically at $4\text{m/s}$. A man walks horizontally at $3\text{m/s}$. At what angle must he hold his umbrella, and at what speed does the rain strike him? What if he doubles his walking speed?

${\vec{v}}_r=(0,-4)$, ${\vec{v}}_m=(3,0)$, so ${\vec{v}}_{rm}=(-3,-4)$.

The rain appears to come from ahead and above, at

$\tan \alpha =\frac{3}{4}\Rightarrow \alpha =37°$ from the vertical, tilted forward

Speed relative to him: $\sqrt{9+16}=5\text{m/s}$ — faster than the $4$ m/s the rain actually falls at.

Walking at $6\text{m/s}$: $\tan \alpha =6/4=1.5\Rightarrow \alpha =56.3°$, and the relative speed rises to $7.2$ m/s.

The trap: doubling the speed does not double the angle. $37°$ becomes $56°$, not $74°$, because the arctangent flattens out. Any question offering a doubled angle among the options is testing exactly this.

6. Projectile on an inclined plane

The horizontal–vertical split is now the wrong split, because the landing condition is "back on the incline", which is not a horizontal line. Rotate the axes onto the slope. Gravity then has two components:

$a_{\parallel }=-g\sin \beta (\text{along the incline}),a_{\perp }=-g\cos \beta$

Launch at speed $u$ at angle $\alpha$ measured from the incline surface. The perpendicular motion is now the one that starts and ends at zero, so it sets the flight time:

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

Substituting into the along-slope motion, which has both an initial velocity and a deceleration:

$R=\frac{2u^2\sin \alpha \cos (\alpha +\beta )}{g{\cos }^2\beta }$

Setting $dR/d\alpha =0$ and using the product-to-sum identity gives a memorable optimum:

${\alpha }_{max}=\frac{90°-\beta }{2},R_{max}=\frac{u^2}{g(1+\sin \beta )}$

Down the incline, replace $\beta$ by $-\beta$ throughout:

${\alpha }_{max}=\frac{90°+\beta }{2},R_{max}^{down}=\frac{u^2}{g(1-\sin \beta )}$

Sanity-check both against $\beta =0$. Each collapses to $\alpha =45°$ and $R=u^2/g$, the flat-ground result. Any version of these formulas that fails that check has been misremembered.

The optimum launch direction bisects the angle between the incline and the vertical — up or down the slope. That single geometric statement replaces both formulas.

Illustration 8

A particle is projected at $20\text{m/s}$ up a $30°$ incline, at $30°$ to the incline surface. Take $g=10$. Find the time of flight and the range along the incline.

$T=\frac{2u\sin \alpha }{g\cos \beta }=\frac{2(20)(0.5)}{10(0.866)}=2.31\text{s}$

$R=\frac{2u^2\sin \alpha \cos (\alpha +\beta )}{g{\cos }^2\beta }=\frac{2(400)(0.5)\cos 60°}{10(0.75)}=\frac{400(0.5)}{7.5}=26.7\text{m}$

Was this the best angle? The optimum is $\alpha =(90°-30°)/2=30°$ — it was. And the maximum range formula agrees: $\frac{400}{10(1.5)}=26.7$ m. Two independent routes to the same number.

Illustration 9

From the same $30°$ incline and the same $20\text{m/s}$, compare the greatest range up the slope with the greatest range down it.

$R_{max}^{up}=\frac{u^2}{g(1+\sin \beta )}=\frac{400}{10(1.5)}=26.7\text{m}$

$R_{max}^{down}=\frac{u^2}{g(1-\sin \beta )}=\frac{400}{10(0.5)}=80.0\text{m}$

Three times as far downhill, from an identical launch. The ratio is $\frac{1+\sin \beta }{1-\sin \beta }$, which blows up as $\beta \to 90°$ — the limit in which "down the incline" becomes a straight drop off a cliff and the range is unbounded.

7. Radius of curvature of a trajectory

Every smooth path is locally a circle. At any instant, split the acceleration into a piece along the velocity, which changes the speed, and a piece across it, which bends the path:

$a_{\perp }=\frac{v^2}{R}⟹R=\frac{v^2}{a_{\perp }}=\frac{v^3}{|\vec{v}\times \vec{a}|}$

For a projectile $\vec{a}=\vec{g}$ is vertical, so the cross product picks out only the horizontal velocity component:

$\left|\vec{v}\times \vec{g}\right|=g{v}_x⟹R=\frac{v^3}{g{v}_x}=\frac{v^3}{gu\cos \alpha }$

This is the bridge between kinematics and circular dynamics, and Advanced papers use it constantly — a question that looks like projectile motion suddenly asks for a normal reaction, and the answer runs through $v^2/R$.

Illustration 10

A projectile is launched at $u$ and angle $\alpha$. Find the radius of curvature at the launch point and at the highest point.

At launch: $v=u$ and $v_x=u\cos \alpha$:

$R=\frac{u^3}{gu\cos \alpha }=\frac{u^2}{g\cos \alpha }$

Check by components: at launch the angle between $\vec{v}$ and $\vec{g}$ is $90°+\alpha$, so $a_{\perp }=g\cos \alpha$, giving $R=u^2/(g\cos \alpha )$. Agrees.

At the top: the velocity is horizontal, $v=v_x=u\cos \alpha$, and the whole of $g$ is perpendicular:

$R=\frac{(u\cos \alpha {)}^3}{gu\cos \alpha }=\frac{u^2{\cos }^2\alpha }{g}$

Read the ratio: $\frac{R_{launch}}{R_{top}}=\frac{1}{{\cos }^3\alpha }$. For $\alpha =60°$ that is a factor of 8 — the path is dramatically more sharply curved at the apex than at launch, which is exactly what the picture of a parabola shows.

8. Choosing the frame: a checklist

Most Advanced kinematics questions are won or lost in the first thirty seconds, on this decision alone.

The ground frame is always legal and often stupid. Every one of these problems can be done from the ground with enough algebra. The examiner knows how long that takes, and sets the time limit accordingly.

Summary

• Same acceleration for two bodies means zero relative acceleration, so each sees the other move in a straight line at constant speed. This settles the monkey-and-hunter, two-projectile and closest-approach families in one line.
• Variable acceleration is three problems, not one: $a=f(t)$ integrates directly, $a=f(x)$ needs $vdv/dx$, $a=f(v)$ needs a choice between $t$ and $x$.
• $a=-kv$ gives infinite stopping time but the finite distance $v_0/k$.
• $a=f(x)$ integrated is the work–energy theorem in disguise.
• Closest approach: $d_{min}=\frac{|{\vec{r}}_{AB}\times {\vec{v}}_{AB}|}{|{\vec{v}}_{AB}|}$. Check the sign of ${\vec{r}}_{AB}\cdot {\vec{v}}_{AB}$ first — positive means they are already separating.
• River: least time at $\theta =0$; zero drift needs $\sin \theta =u/v$ and only exists if $v>u$; when $u>v$ the least drift is at $\sin \theta =v/u$. The two conditions are reciprocals.
• Inclined-plane projectile: rotate the axes. $T=\frac{2u\sin \alpha }{g\cos \beta }$, and the optimum launch bisects the angle between the incline and the vertical.
• $R_{max}^{up}=\frac{u^2}{g(1+\sin \beta )}$ and $R_{max}^{down}=\frac{u^2}{g(1-\sin \beta )}$; both reduce to $u^2/g$ at $\beta =0$.
• Umbrella points along ${\vec{v}}_{rain}-{\vec{v}}_{man}$, and doubling your speed does not double the angle.
• $R=\frac{v^3}{|\vec{v}\times \vec{a}|}$; for a projectile this is $\frac{v^3}{g{v}_x}$, giving $\frac{u^2}{g\cos \alpha }$ at launch and $\frac{u^2{\cos }^2\alpha }{g}$ at the top.