JEE Mains Physics · Motion in a Plane

Velocity in a Plane and Relative Velocity

Velocity and acceleration in a plane are the time derivatives of the position vector, taken one component at a time; the velocity of A as seen from B is v_A − v_B, which settles river crossings, rain and umbrellas, and anything watched from a moving vehicle.

All Motion in a Plane notes JEE Mains Physics strategy

Why this matters

Nineteen PYQs, fourteen of them multiple choice, and four from 2026. Ten give a position or a force as a function of time and ask for a velocity, a direction or the shape of the path. Four are river crossings and five change the frame: rain seen by a runner, a bomb seen from its plane, a ball thrown from a moving boat. Every one of them is component-by-component work; say which axis an angle is measured from before you write it.

Concept 1 of 3: Position, velocity and acceleration vectors

Motion in a plane is two straight-line motions running side by side, one along x and one along y. Differentiate each coordinate on its own to get the velocity components, and again to get the acceleration. The direction of motion is the direction of the velocity, and the net force always points along the acceleration, which need not be the direction of motion.

Definition

$\vec{v} = \dfrac{d\vec{r}}{dt}$ , $\vec{a} = \dfrac{d\vec{v}}{dt}$ , component by component.
Speed $|\vec{v}| = \sqrt{v_x^{2} + v_y^{2}}$ ; direction $\tan\theta = v_y/v_x$ from the x-axis. From the y-axis it is $\tan^{-1}(v_x/v_y)$ .
Constant acceleration: $\vec{v} = \vec{u} + \vec{a}t$ , $\vec{r} = \vec{r}_0 + \vec{u}t + \tfrac{1}{2}\vec{a}t^{2}$ , and $v_x^{2} = u_x^{2} + 2a_x x$ along each axis.
Net force is along $\vec{a}$ : $\vec{F} = m\vec{a}$ . A force that varies with time is integrated, starting from the stated initial velocity and position.
Shape of the path: eliminate t. One coordinate linear and the other quadratic in t gives a parabola; $x = a\cos\omega t$ , $y = a\sin\omega t$ gives a circle of radius a.

Motion in two dimensions

\vec{v} = \frac{d\vec{r}}{dt},\ \vec{a} = \frac{d\vec{v}}{dt} \qquad \vec{r} = \vec{r}_0 + \vec{u}t + \tfrac{1}{2}\vec{a}t^{2} \qquad \tan\theta = \frac{v_y}{v_x}

Worked example

A particle's position is

\vec{r} = (t^{3}\hat{i} + 6t\hat{j})

m. Find its speed and direction of motion at t = 2 s, and the direction of the net force on it.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 24 Jan 2025 · Q83Moderate

Example 1 · Motion in a Plane · Velocity in a Plane and Relative Velocity

The position vector of a moving body at any instant of time is given as

\overrightarrow{r}=\left( 5t^{2}\widehat{i}- 5t\widehat{j} \right)m

. The magnitude and direction of velocity at

t= 2\text{ }s

is,

Quoting an angle without its axis

A velocity 4i − j m/s is at tan⁻¹(1/4) below the +x axis and at tan⁻¹ 4 from the −y axis. Options often give the right number against the wrong axis.

Force along the motion

The net force points along the acceleration. For r = 2t i + t² j the particle moves diagonally, but the force is along +y only.

Drill 9 more on position, velocity and acceleration vectors

Concept 2 of 3: Crossing a river

The boat's velocity relative to the water and the river's velocity add as vectors. Only the part of the boat's own velocity that points across the river carries it to the far bank, so that part alone fixes the crossing time. Whatever along-river velocity is left over, the river's plus the boat's own, carries it downstream for that time.

Definition

River width d, flow u, boat speed v in still water.
Heading straight across: least time $t = d/v$ ; drift downstream $= ut$ .
Heading at angle θ to the flow: across speed $v\sin\theta$ , so $t = \dfrac{d}{v\sin\theta}$ ; along-river speed $u + v\cos\theta$ .
To land directly opposite (shortest path), head upstream at α from straight across with $\sin\alpha = u/v$ ; then $t = \dfrac{d}{\sqrt{v^{2} - u^{2}}}$ . This needs $v > u$ .
Swimmer as fast as the river: the resultant bisects the angle between the heading and the flow.
A round trip is two crossings, and the drift adds up over both.

Crossing time and drift

t_{\min} = \frac{d}{v} \qquad x_{drift} = \frac{ud}{v} \qquad \sin\alpha = \frac{u}{v},\ t = \frac{d}{\sqrt{v^{2} - u^{2}}}

Worked example

A river 120 m wide flows at 3 m/s. A boat moves at 5 m/s in still water. (a) It heads straight across: how long does it take and how far downstream does it land? (b) Which way must it head to land directly opposite, and how long does that take?

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 21 Jan 2026 Shift 2 · Q14Moderate

Example 2 · Motion in a Plane · Velocity in a Plane and Relative Velocity

A river of width 200 m is flowing from west to east with a speed of

18\text{ }km/h

. A boat, moving with speed of

36\text{ }km/h

in still water, is made to travel one-round trip (bank to bank of the river). Minimum time taken by the boat for this journey and also the displacement along the river bank are

\_\_\_\_

and

\_\_\_\_

respectively.

Dividing the width by the resultant speed

The crossing time is the width divided by the ACROSS component of the boat's own velocity. The river's flow is along the banks, so it never shortens or lengthens the crossing.

Least time is not shortest path

Heading straight across gives the least time but lands downstream. Landing directly opposite needs an upstream heading and always takes longer.

Drill 3 more on crossing a river

Concept 3 of 3: Relative velocity and change of frame

What an observer sees is the object's velocity minus the observer's own. Rain falling straight down looks slanted to a runner because the runner's velocity is subtracted from it. A bomb released from a plane keeps the plane's horizontal speed, so the pilot sees it fall straight down. Work out the motion in whichever frame makes it simplest, then subtract or add the frame's velocity.

Definition

$\vec{v}_{AB} = \vec{v}_A - \vec{v}_B$ : velocity of A as seen from B.
Rain and umbrella: $\vec{v}_{rain,man} = \vec{v}_{rain} - \vec{v}_{man}$ ; hold the umbrella along this relative velocity, tilted towards it.
Rain that appears vertical: the man's velocity equals the rain's horizontal component.
A body released from a moving vehicle starts with the vehicle's velocity. Seen from the vehicle (no acceleration of its own), it falls straight down.
A ball thrown straight up from a cart moving at V lands back in the cart; seen from the ground it moves sideways at V for its whole time in the air, $2u/g$ for an upward throw at u, so it lands $2uV/g$ from where it was thrown.
To hit a plane flying overhead at speed V, a shell of speed u needs $u\cos\theta = V$ , so both cover the same horizontal distance.

Relative velocity

\vec{v}_{AB} = \vec{v}_A - \vec{v}_B \qquad \tan\theta_{umbrella} = \frac{v_{man}}{v_{rain}}\ (\text{vertical rain})

Worked example

Rain falls vertically at 8 m/s. A man walks east at 6 m/s. Find the speed of the rain relative to the man and how he should hold his umbrella.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 27 June 2022 · Q2Moderate

Example 3 · Motion in a Plane · Velocity in a Plane and Relative Velocity

A girl standing on road holds her umbrella at

45^{\circ}

with the vertical to keep the rain away. If she starts running without umbrella with a speed of

15\sqrt{2}kmh^{- 1}

, the rain drops hit her head vertically. The speed of rain drops with respect to the moving girl is:

Subtracting in the wrong order

The rain's velocity relative to the man is v_rain − v_man. Reversing it gives the man's velocity relative to the rain, which points the other way.

Taking a released object to start from rest

An object let go from a moving plane, boat or balloon starts with that vehicle's velocity. Taking it to start from rest gives the wrong range and the wrong path.

Drill 4 more on relative velocity and change of frame

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Position, velocity and acceleration vectors
Motion in two dimensions
$\vec{v} = \frac{d\vec{r}}{dt},\ \vec{a} = \frac{d\vec{v}}{dt} \qquad \vec{r} = \vec{r}_0 + \vec{u}t + \tfrac{1}{2}\vec{a}t^{2} \qquad \tan\theta = \frac{v_y}{v_x}$
Crossing a river
Crossing time and drift
$t_{\min} = \frac{d}{v} \qquad x_{drift} = \frac{ud}{v} \qquad \sin\alpha = \frac{u}{v},\ t = \frac{d}{\sqrt{v^{2} - u^{2}}}$
Relative velocity and change of frame
Relative velocity
$\vec{v}_{AB} = \vec{v}_A - \vec{v}_B \qquad \tan\theta_{umbrella} = \frac{v_{man}}{v_{rain}}\ (\text{vertical rain})$

Watch out for (6)

Quoting an angle without its axis
→ Position, velocity and acceleration vectors
Force along the motion
→ Position, velocity and acceleration vectors
Dividing the width by the resultant speed
→ Crossing a river
Least time is not shortest path
→ Crossing a river
Subtracting in the wrong order
→ Relative velocity and change of frame
Taking a released object to start from rest
→ Relative velocity and change of frame

Test yourself on Motion in a Plane

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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