Kinematics — Easy Practice (Set 1)

Quick Recap — Motion Basics

A short refresher before you practise.

• Distance (scalar, total path length) vs displacement (vector, shortest start-to-finish); always $|\text{displacement}| \le \text{distance}$.
• Speed (scalar) vs velocity (vector). Average velocity $= \dfrac{\text{displacement}}{\text{time}}$.
• Acceleration $a = \dfrac{\Delta v}{\Delta t}$, SI unit m/s$^2$.
• Equations of motion (uniform $a$): $v = u + at$, $\ s = ut + \tfrac{1}{2}at^2$, $\ v^2 = u^2 + 2as$.
• Free fall: take $a = g \approx 9.8$ m/s$^2$ (often 10 m/s$^2$ in problems), directed downward.

Beyond-NCERT JEE Formulae

These are the high-yield formulae that JEE Main loves but NCERT barely touches. The equations of motion, free fall and basic projectile relations are assumed (see the practice sets); here we go one level deeper. For each result: when to use it and a [JEE Tip].

1. River-Boat (Relative Velocity in 2D)

Let the boat's speed in still water be $v_b$, the current $v_r$ (parallel to the banks) and the river width $d$.

(a) Cross in shortest TIME — head straight across. Point the boat perpendicular to the banks; the whole of $v_b$ then drives it across while the current only sweeps it downstream. The crossing time is $t=\dfrac{d}{v_b}$ and the downstream drift is

$$\text{drift}=v_r\,t=\dfrac{v_r\,d}{v_b}$$

the ground speed being $\sqrt{v_b^2+v_r^2}$.

When to use: any "minimum time" or "quickest" crossing.

[JEE Tip] The current has no component across the river, so the minimum time is independent of $v_r$ — a stronger current increases the drift but never the crossing time.

(b) Shortest PATH (land directly opposite, zero drift) — aim upstream. Steer at an angle $\theta$ to the straight-across direction so the upstream component cancels the current, $v_b\sin\theta=v_r$. Hence

$$\sin\theta=\dfrac{v_r}{v_b},\qquad t=\dfrac{d}{\sqrt{v_b^2-v_r^2}}$$

When to use: "reach the point exactly opposite", "no drift", or "shortest path".

[JEE Tip] This is possible only if $v_b>v_r$. If instead $v_b<v_r$ the drift cannot be killed; the least drift is $\dfrac{d\sqrt{v_r^2-v_b^2}}{v_b}$, obtained by heading upstream at $\sin\theta=\dfrac{v_b}{v_r}$.

2. Rain-Man Problem

Let the rain's velocity in the ground frame be $\vec{v}_r$ and the man's velocity be $\vec{v}_m$. The rain the man actually feels is the relative velocity

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

and he must tilt the umbrella along $\vec{v}_{rm}$. If the rain falls vertically at speed $v_r$ while he walks horizontally at $v_m$, the apparent speed is $\sqrt{v_r^2+v_m^2}$ and the tilt from the vertical is $\tan\theta=\dfrac{v_m}{v_r}$.

When to use: apparent speed or direction of rain for any moving observer; umbrella tilt.

[JEE Tip] Tilt the umbrella FORWARD, into your motion, and always measure $\theta$ from the vertical. Two classic traps: measuring from the horizontal, and adding the speeds instead of combining them by Pythagoras. If the rain looks vertical to the moving man, then $\vec{v}_r=\vec{v}_{rm}+\vec{v}_m$ recovers the true, slanted rainfall.

3. Projectile on an Inclined Plane

Take an incline of angle $\alpha$; the projectile is fired up the slope with speed $u$ at angle $\theta$ to the HORIZONTAL. Rotate the axes so that $x$ runs along the incline and $y$ perpendicular to it: the motion becomes an ordinary projectile with "gravity" $g\cos\alpha$ perpendicular to the slope and a retardation $g\sin\alpha$ along it. The range measured UP the incline is

$$R=\dfrac{2u^2\sin(\theta-\alpha)\cos\theta}{g\cos^2\alpha}$$

and the maximum range up the incline, reached at $\theta=45^\circ+\dfrac{\alpha}{2}$, is

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

When to use: a ball thrown up (or down) a hillside and landing back on the slope.

[JEE Tip] For projection DOWN the incline simply flip the sign of $\alpha$: the range becomes $\dfrac{2u^2\sin(\theta+\alpha)\cos\theta}{g\cos^2\alpha}$ and the maximum range becomes $\dfrac{u^2}{g\,(1-\sin\alpha)}$. The time of flight along the slope is $T=\dfrac{2u\sin(\theta-\alpha)}{g\cos\alpha}$.

4. Variable (Non-Uniform) Acceleration

The constant-$a$ formulae collapse the moment $a$ changes. Choose the differential form by what $a$ depends on. The bridge relation, from the chain rule, is

$$a=\dfrac{dv}{dt}=v\,\dfrac{dv}{dx}$$

• If $a=a(t)$: integrate in time, $v=u+\displaystyle\int a\,dt$, then $x=\displaystyle\int v\,dt$.
• If $a=a(x)$: use $v\,dv=a\,dx$, that is $\displaystyle\int_{u}^{v}v\,dv=\int_{x_0}^{x}a\,dx$.
• If $a=a(v)$: for the speed-versus-position relation use $\dfrac{v\,dv}{a(v)}=dx$; for speed-versus-time use $\dfrac{dv}{a(v)}=dt$.

When to use: drag laws such as $a=-kv$ or $a=-kv^2$, a position-dependent $a=a(x)$, or any acceleration that is not constant.

[JEE Tip] If a question wants speed as a function of DISTANCE, reach for $v\,dv=a\,dx$; if it wants speed as a function of TIME, use $dv=a\,dt$. Never push a varying $a$ into $v^2=u^2+2as$ — that is the single most common trap in this topic.

5. Relative Acceleration and Minimum Separation

For two bodies A and B write $\vec{r}_{BA}=\vec{r}_B-\vec{r}_A$, $\vec{v}_{BA}=\vec{v}_B-\vec{v}_A$ and $\vec{a}_{BA}=\vec{a}_B-\vec{a}_A$. Work in A's frame: B then sets out from $\vec{r}_0$ with velocity $\vec{v}_{BA}$, moving in a straight line whenever $\vec{a}_{BA}=0$. The closest approach is the perpendicular distance from A to that line,

$$d_{\min}=\dfrac{|\vec{r}_0\times\vec{v}_{BA}|}{|\vec{v}_{BA}|},\qquad t^\ast=-\dfrac{\vec{r}_0\cdot\vec{v}_{BA}}{|\vec{v}_{BA}|^2}$$

When to use: the least distance between two cars, ships or particles, and whether they collide.

[JEE Tip] Two bodies in free fall share the same $g$, so $\vec{a}_{BA}=0$ and their separation changes at the CONSTANT rate $|\vec{v}_{BA}|$ — the $\tfrac12 g t^2$ terms cancel exactly. They collide only if $\vec{r}_0$ is anti-parallel to $\vec{v}_{BA}$, i.e. $d_{\min}=0$.

Extra JEE Gems

• Velocity perpendicular to the launch (level ground): the projectile's velocity turns perpendicular to $\vec{u}$ at $t=\dfrac{u}{g\sin\theta}$.
• Radius of curvature of a path: $\rho=\dfrac{v^2}{a_\perp}$, where $a_\perp$ is the acceleration component perpendicular to $\vec{v}$. At a projectile's highest point this gives $\rho=\dfrac{u^2\cos^2\theta}{g}$.

Solved Examples — Beyond-NCERT Formulae

Example 1 — River crossing in shortest time (drift). A river $200$ m wide flows at $v_r=3$ m/s. A boat can do $v_b=4$ m/s in still water and the boatman heads straight across for the quickest crossing. Find the crossing time, the drift and the ground speed.

Solution. Heading straight across, the entire $v_b$ is spent going across the stream, so the crossing time is $t=\dfrac{d}{v_b}=\dfrac{200}{4}=50$ s. The current sweeps the boat downstream for this whole time, giving a drift $=v_r\,t=3\times 50=150$ m. The ground speed is $\sqrt{v_b^2+v_r^2}=\sqrt{16+9}=5$ m/s. Note the $50$ s would be the same for ANY current — only the drift depends on $v_r$.

Example 2 — Boat on the shortest path (zero drift). A boat with $v_b=5$ m/s must land at the point directly opposite, across a river $300$ m wide flowing at $v_r=3$ m/s. Find the heading and the crossing time.

Solution. To cancel the drift the upstream component must equal the current, so $\sin\theta=\dfrac{v_r}{v_b}=\dfrac{3}{5}=0.6$, giving $\theta\approx 37^\circ$ measured from the straight-across direction (aimed upstream). The useful across-speed is $\sqrt{v_b^2-v_r^2}=\sqrt{25-9}=4$ m/s, so the crossing time is $t=\dfrac{d}{\sqrt{v_b^2-v_r^2}}=\dfrac{300}{4}=75$ s. This is longer than the $\dfrac{300}{5}=60$ s of a straight-across crossing — you trade time for zero drift.

Example 3 — Rain-man umbrella. Rain falls vertically at $10$ m/s. A man runs along a horizontal road at $10$ m/s. At what angle to the vertical should he hold his umbrella, and how fast does the rain strike him?

Solution. In the man's frame the rain has velocity $\vec{v}_{rm}=\vec{v}_r-\vec{v}_m$. Taking east as $\hat{i}$ and up as $\hat{j}$, $\vec{v}_r=-10\,\hat{j}$ and $\vec{v}_m=10\,\hat{i}$, so

$$\vec{v}_{rm}=-10\,\hat{i}-10\,\hat{j}$$

The tilt from the vertical is $\tan\theta=\dfrac{v_m}{v_r}=\dfrac{10}{10}=1$, so $\theta=45^\circ$, and the apparent speed is $\sqrt{10^2+10^2}=10\sqrt{2}\approx 14.1$ m/s. He leans the umbrella forward, into his run.

Example 4 — Projectile up an incline (range). A ball is thrown up a $30^\circ$ incline with speed $u=15$ m/s at $\theta=60^\circ$ to the horizontal, with $g=10$ m/s$^2$. Find the range measured along the incline.

Solution. Here $\alpha=30^\circ$, so $\theta-\alpha=30^\circ$. Using the up-incline range,

$$R=\dfrac{2u^2\sin(\theta-\alpha)\cos\theta}{g\cos^2\alpha}=\dfrac{2(15)^2\sin 30^\circ\cos 60^\circ}{10\cos^2 30^\circ}$$

Numerically $R=\dfrac{2(225)(0.5)(0.5)}{10(0.75)}=\dfrac{112.5}{7.5}=15$ m. The along-slope time of flight is $T=\dfrac{2u\sin(\theta-\alpha)}{g\cos\alpha}=\dfrac{2(15)(0.5)}{10\cos 30^\circ}=\sqrt{3}\approx 1.73$ s.

Example 5 — Maximum range on an incline. A projectile is fired up the same $30^\circ$ incline, now with $u=30$ m/s ($g=10$ m/s$^2$). Find the greatest possible range along the incline and the launch angle that achieves it.

Solution. The maximum up-incline range is

$$R_{\max}=\dfrac{u^2}{g\,(1+\sin\alpha)}=\dfrac{(30)^2}{10\,(1+\sin 30^\circ)}=\dfrac{900}{15}=60$$

that is $60$ m, reached at $\theta=45^\circ+\dfrac{\alpha}{2}=45^\circ+15^\circ=60^\circ$ to the horizontal. As a check, substituting $\theta=60^\circ$ into the general range formula gives $\dfrac{2(900)\sin 30^\circ\cos 60^\circ}{10\cos^2 30^\circ}=\dfrac{450}{7.5}=60$ m — consistent.

Example 6 — Variable acceleration using $a=v\,\dfrac{dv}{dx}$. A particle on the $x$-axis has a position-dependent acceleration $a=-4x$ (SI units) and is released from rest at $x=5$ m. Find its speed as it passes $x=3$ m and the points where it momentarily stops.

Solution. Because $a$ depends on position, write $a=v\dfrac{dv}{dx}$, so $v\,dv=-4x\,dx$. Integrating from rest at $x=5$ to a general $x$,

$$\dfrac{v^2}{2}=-2x^2+2(5)^2\quad\Longrightarrow\quad v^2=100-4x^2$$

At $x=3$ m this gives $v^2=100-36=64$, so $v=8$ m/s. The particle stops when $v=0$, i.e. $4x^2=100$, giving $x=\pm 5$ m. It therefore oscillates between $x=+5$ m and $x=-5$ m (this is SHM with $\omega=2$ rad/s), sweeping through the origin at its maximum speed $\sqrt{100}=10$ m/s.

Example 7 — Closest approach of two particles. At $t=0$ particle A is at the origin moving east at $4$ m/s, while particle B is $20$ m due east of A and moving north at $3$ m/s; both keep constant velocity. Find their least separation and the instant it occurs.

Solution. The velocity of B relative to A is

$$\vec{v}_{BA}=\vec{v}_B-\vec{v}_A=(0,3)-(4,0)=(-4,3)$$

with magnitude $|\vec{v}_{BA}|=5$ m/s, and B starts at $\vec{r}_0=(20,0)$ m relative to A. The least separation is the perpendicular distance from A to B's relative path,

$$d_{\min}=\dfrac{|\vec{r}_0\times\vec{v}_{BA}|}{|\vec{v}_{BA}|}=\dfrac{|(20)(3)-(0)(-4)|}{5}=\dfrac{60}{5}=12$$

that is $12$ m, occurring at $t^\ast=-\dfrac{\vec{r}_0\cdot\vec{v}_{BA}}{|\vec{v}_{BA}|^2}=-\dfrac{(20)(-4)+(0)(3)}{25}=\dfrac{80}{25}=3.2$ s. A direct check: the separation squared is $(20-4t)^2+(3t)^2=25t^2-160t+400$, least at $t=\dfrac{160}{50}=3.2$ s, where it equals $144$, i.e. $12$ m.