JEE Main (Engineering) · Physics (JEE & NEET)

Kinematics, Newtons Laws and Projectile Motion

Motion in one and two dimensions using vectors, relative velocity, projectile trajectories and the three laws of motion with friction.

Eight concepts, from the three equations of uniform acceleration to a projectile fired up a slope. Almost every one of them is a bookkeeping skill — split the motion, choose the axes, write the constraint — and that is exactly where the marks are won and lost.

JEE Main (Engineering)
Medium level
8 concepts
5 practice questions

1Uniformly accelerated motion

For constant acceleration a the three standard equations are not separate facts — they follow from integrating a = dv/dt once, integrating v = ds/dt again, and then eliminating t between the two. That gives v = u + at, s = ut + ½at², and v² = u² + 2as. Every one of them assumes a is constant; none of them survives a changing acceleration.

The displacement during the n-th second, s_n = u + (a/2)(2n − 1), is the same algebra applied twice and subtracted. It is a distance in metres even though it is numerically the average velocity over that one-second interval.

Figure. v = u + at is a straight line: its intercept is u and its slope is a. The other two equations are the same line read differently — the area beneath it up to time t is s, and squaring the height gives the time-free relation.

Where they come from

Integrate oncea = dv/dt with a constant gives v = u + at — velocity is linear in time.
Integrate againv = ds/dt then gives s = ut + ½at², the area under that straight line.
Eliminate tSubstituting t = (v − u)/a leaves v² = u² + 2as, the one equation with no time in it.

Distance in the n-th second

A train starts from rest with a constant acceleration of 2 m s⁻². How far does it travel during the 5th second?

s(5) = ½ × 2 × 5²25 m
s(4) = ½ × 2 × 4²16 m
Distance in the 5th second = 25 − 169 m
Formula check: s_n = u + (a/2)(2n − 1) = 0 + 1 × 99 m ✓

Pro tip. From rest, the distances covered in successive seconds go 1 : 3 : 5 : 7 — Galileo's odd-number rule, and here 1, 3, 5, 7, 9 metres. If a question gives you a ratio of successive-second distances, that ratio alone tells you the motion started from rest.

A body starts from rest with constant acceleration. The ratio of the distance it covers in the 3rd second to the distance it covers in the first 3 seconds is

1 : 3
5 : 9
1 : 9

The 3rd second gives (a/2)(2×3 − 1) = 5a/2; the first three seconds give ½a(3)² = 9a/2. The ratio is 5 : 9 — the odd numbers 1, 3, 5 summing to 9.

2Reading motion graphs

A graph answers questions the equations cannot, because it does not care whether the acceleration is constant. Slopes go down the chain: the slope of x–t is v, the slope of v–t is a. Areas go up it: the area under a–t is Δv, the area under v–t is displacement.

The trap is the sign. Area below the time axis counts as negative displacement but as positive distance, so a body that comes back to where it started has zero displacement and a stubbornly non-zero distance. Average velocity uses the first number; average speed uses the second.

Figure. The line crosses the dashed v = 0 line at t = 2 s. The triangle above it is +8 m and the one below is −8 m: they cancel for displacement and add for distance. Nothing about the graph changes; only which of the two questions you were asked.

How to read one

Find the zerosWhere v crosses the axis the body turns round — that is where distance and displacement part company.
Take areas piecewiseCompute the area on each side of the crossing separately, then add them signed for displacement and unsigned for distance.
Read curvature, not valuesA curved v–t line means a is changing, so v² = u² + 2as is off the table; only the area still works.

Slopes down, areas up
Graph	Its slope gives	Its area gives
Position x vs t	velocity v	nothing useful
Velocity v vs t	acceleration a	displacement (signed)
Acceleration a vs t	rate of change of a	change in velocity Δv

Zero displacement, plenty of distance

A particle moves with v = 8 − 4t (SI units) from t = 0 to t = 4 s. Find its displacement, the distance it travels, and its average speed.

Turning point: 8 − 4t = 0t = 2 s
Area 0 → 2 s = ½ × 2 × 8+8 m
Area 2 → 4 s = ½ × 2 × (−8)−8 m
Displacement 0 m, distance 8 + 8 = 16 m, so average speed = 16/44 m s⁻¹

Pro tip. Average velocity here is exactly zero while average speed is 4 m s⁻¹. Any question that offers you both as options is testing this one distinction, and the giveaway is a velocity that changes sign inside the interval.

A particle's velocity–time graph is a straight line falling from +8 m s⁻¹ at t = 0 to −8 m s⁻¹ at t = 4 s. Over those 4 seconds its average velocity and average speed are

0 and 4 m s⁻¹
4 m s⁻¹ and 4 m s⁻¹
0 and 0

Displacement is the signed area, +8 − 8 = 0, so the average velocity is zero. Distance is the unsigned area, 8 + 8 = 16 m, giving an average speed of 16/4 = 4 m s⁻¹.

3Newton's three laws

The first law is a definition in disguise: it names the frames in which the other two hold. The second law is properly F = dp/dt, which collapses to F = ma only when the mass is constant — rockets and falling chains need the momentum form. The third law says forces come in equal and opposite pairs acting along the same line, at the same instant, on two different bodies.

That last clause is the whole examination. A book's weight and the normal force on it are equal in magnitude, but both act on the book, so they are not a third-law pair — they are merely balanced. Put the book in an accelerating lift and the equality breaks while the genuine pair stays exact.

Figure. N and N′ are the third-law pair: equal, opposite, on two different bodies. N and mg both act on the book, so they are not a pair — they are equal only because the book is in equilibrium. All three arrows are drawn the same length because all three magnitudes are equal. Set the table in an upward-accelerating lift and N exceeds mg while N and N′ stay exactly equal.

How to use each one

First — inertiaNo net force is needed to keep something moving. If a body's velocity is constant, the forces on it sum to zero; use that as an equation.
Second — momentumWrite F = Δp/Δt for impacts and jets, F = ma only once you are sure the mass is fixed.
Third — pairsName both bodies out loud: 'the wall pushes the ball' pairs with 'the ball pushes the wall'. If both forces land on the same body, it is not a pair.

Force from a rebound

A 150 g ball strikes a wall horizontally at 20 m s⁻¹ and rebounds at 15 m s⁻¹. The contact lasts 0.02 s. Find the average force the wall exerts on the ball, and the force on the wall.

Δp = m(v_f − v_i) = 0.15 × (−15 − 20)−5.25 kg m s⁻¹
F = Δp/Δt = −5.25 / 0.02−262.5 N
By the third law, the ball pushes the wall262.5 N outward

Pro tip. The minus sign on the rebound velocity is the entire question. Writing 20 − 15 = 5 instead of −15 − 20 = −35 gives 37.5 N — seven times too small. Momentum change on a rebound is always larger than on a dead stop, never smaller.

A book rests on a table. The third-law reaction to the book's weight is

The normal force the table exerts on the book
The gravitational pull the book exerts on the Earth
The weight of the table

Partners act on different bodies. The Earth pulls the book down, so the book pulls the Earth up. The normal force acts on the same body as the weight, so it cannot be its partner — it is equal here only because the book happens to be in equilibrium.

4Free-body diagrams and constraints

A free-body diagram is one body, isolated, carrying only the forces that act on it from outside. Draw one per body. Tension in a connecting string appears on both diagrams, pointing away from each body along the string, and cancels when you add the equations — which is exactly why adding them is the fastest route to the acceleration.

What closes the system is the constraint. An inextensible string means the two bodies' accelerations have equal magnitude; a block resting on a wedge means its motion perpendicular to the surface is zero. Count your unknowns, then make sure you have written that many equations — the constraint is usually the one people forget.

Figure. One free-body diagram per body, with the same T on both. The string does not appear as a drawn link — the constraint is the shared magnitude |a|, opposite in direction for an Atwood pair.

The routine

IsolateDraw a box round one body. Only forces crossing that box exist: weight, normal, tension, friction, applied.
Choose axes along aPoint one axis along the direction the body will actually accelerate — down the slope, along the string — so only one equation carries the acceleration.
Add the constraintInextensible string ⇒ |a₁| = |a₂|. That extra relation is what makes the number of equations match the number of unknowns.

Atwood machine

Blocks of 3 kg and 2 kg hang from the ends of a light inextensible string over a frictionless pulley. Take g = 10 m s⁻². Find the acceleration and the tension.

3 kg falling: 3g − T = 3a30 − T = 3a
2 kg rising: T − 2g = 2aT − 20 = 2a
Add — T cancels: 10 = 5aa = 2 m s⁻²
T = 2(g + a) = 2 × 1224 N

Pro tip. The tension always lands between the two weights: 20 N < 24 N < 30 N. If yours falls outside that window a sign is wrong. Note also that the pulley's axle carries 2T = 48 N, not the 50 N total weight — an accelerating system presses less hard on its support.

In that 3 kg / 2 kg Atwood machine, the downward force the string exerts on the pulley is

50 N, the total weight of the blocks
48 N, which is 2T
10 N, the difference of the weights

The string pulls down on the pulley with T = 24 N on each side, so the axle carries 2T = 48 N. It falls short of 50 N because the system's centre of mass is accelerating downward at 0.4 m s⁻², and 5 × 0.4 = 2 N is exactly the shortfall.

5Friction, static and kinetic

Static friction is a response, not a value. It takes whatever magnitude is needed to prevent sliding, up to a ceiling of μ_s N — so writing f = μ_s N for a block that is sitting still is wrong unless the block is on the verge of moving. Once it does slip, kinetic friction locks to f_k = μ_k N, opposing the relative sliding, and μ_k < μ_s, which is why a stuck object lurches when it finally gives.

On an incline of angle θ the block is on the verge when tan θ = μ_s — the angle of repose. The mass cancels, so this is a direct measurement of μ_s with nothing but a protractor.

Figure. Drawn for a 10 kg block with μ_s = 0.5 and μ_k = 0.4, so N = 100 N. Below break-away friction simply matches the push and the graph is the line f = F. At F = 50 N it can do no more; the block slips and friction drops to a flat 40 N however hard you push after that.

How to decide what friction is doing

Test the verge firstCompare the driving force with μ_s N. If the drive is smaller, the body is static and friction equals the drive.
If it slips, switchFriction becomes μ_k N — a fixed number, independent of the speed and of the contact area.
Then write maOnly now is there an acceleration: on an incline, a = g(sin θ − μ_k cos θ), with no mass in it.

Released on a rough incline

A block is released from rest on a 37° incline with μ_s = 0.5 and μ_k = 0.25. Does it slide, and if so with what acceleration? Take g = 10 m s⁻², sin 37° = 0.6, cos 37° = 0.8.

Down-slope pull per unit mass: g sin 37° = 10 × 0.66 m s⁻²
Maximum static grip: μ_s g cos 37° = 0.5 × 10 × 0.84 m s⁻²
6 > 4, so it slips; kinetic grip = 0.25 × 82 m s⁻²
a = 6 − 24 m s⁻²

Pro tip. The mass cancels at every line: the acceleration depends only on g, θ and μ_k, never on how heavy the block is. Had μ_s been 0.8 the grip would have been 6.4 m s⁻² against a 6 m s⁻² pull and the block would not have moved at all — with friction sitting at exactly 6m newtons, not μ_s N.

A 10 kg block rests on a floor with μ_s = 0.5 (g = 10 m s⁻²). A horizontal push of 30 N is applied and the block still does not move. The friction force on it is

50 N
30 N
20 N

Static friction is self-adjusting: with no acceleration the horizontal forces must sum to zero, so friction is 30 N. μ_s N = 50 N is only its ceiling, reached at the verge of slipping and not before.

6Projectile motion

A projectile is two motions that never speak to each other. Horizontally there is no force, so a_x = 0 and x = (u cos θ)t. Vertically a_y = −g throughout, including at the top. The only thing the two share is the clock, and that is how every projectile question is solved: get the time from the vertical motion, then hand it to the horizontal one.

Doing that yields T = 2u sin θ/g, H = u² sin²θ/2g and R = u² sin 2θ/g, with a parabolic path y = x tan θ (1 − x/R). At the apex the vertical component has been spent but u cos θ survives untouched, so the speed there is u cos θ, never zero.

Figure. Plotted from y = 80(x/R)(1 − x/R) with R = 138.6 m, the real path for u = 40 m s⁻¹ at 30°: the apex is 20 m up at x = 69.3 m, exactly half the range, and the ball lands 4 s after launch. The curve is symmetric because the ascent and descent take the same time.

The standard route

Resolveu_x = u cos θ stays constant for the whole flight; u_y = u sin θ is the only thing gravity touches.
Time from the verticalThe projectile returns to launch height when the vertical displacement is zero, at T = 2u sin θ/g.
Range from the horizontalR = u_x T. Multiplying out gives u² sin 2θ/g, maximal at θ = 45° where R_max = u²/g.

Thrown at 30°

A ball is thrown from ground level at u = 40 m s⁻¹ and θ = 30° above the horizontal. With g = 10 m s⁻², find the time of flight, the maximum height and the range.

u_x = 40 cos 30° = 20√3, u_y = 40 sin 30°34.6 and 20 m s⁻¹
T = 2u_y/g = 40/104 s
H = u_y²/2g = 400/2020 m
R = u_x T = 20√3 × 480√3 ≈ 138.6 m

Pro tip. Cross-check in one line with R = 4H cot θ = 4 × 20 × √3 = 80√3 m. Throwing at the complementary angle 60° gives the identical range with H = 60 m instead, and the two heights obey H₁H₂ = 20 × 60 = 1200 = R²/16.

At the highest point of its flight, a projectile launched at 60° to the horizontal with speed u has speed

Zero
u/2
u√3/2

Only the vertical component vanishes at the apex. The horizontal component u cos 60° = u/2 is never acted on, so that is the speed there. u√3/2 is u sin 60° — the component that has just been used up.

7Relative velocity

The velocity of A as measured by B is the vector difference v_AB = v_A − v_B. Subtracting a vector means adding its reverse, so the construction is always the same: hang −v_B off the tip of v_A and read the closing arrow.

Differentiating gives a_AB = a_A − a_B, and that is where the leverage is. Two bodies in free fall share the acceleration −g ĵ, so their relative acceleration is exactly zero — sit on one of them and the other travels a straight line at constant speed. Collision and minimum-distance problems that look like calculus become geometry.

Figure. Subtracting is adding the reverse: hang −v_man off the tip of v_rain and the closing arrow from the start point is what the man measures. It is both faster than the rain really falls (20 against 17.3 m s⁻¹) and leaning 30° off vertical, arriving from in front of him — so the umbrella goes forward.

How to switch frames

Subtract, do not addv_AB = v_A − v_B. Getting the order backwards reverses the answer, which in a rain-or-river question means tilting the wrong way.
Check the relative accelerationIf a_A = a_B the relative motion is uniform, and the new frame is as good as an inertial one for the problem.
Solve in the new frameClosest approach becomes the perpendicular distance from the fixed body to the straight relative path.

Which way to tilt the umbrella

Rain falls vertically at 10√3 ≈ 17.3 m s⁻¹. A man walks east at 10 m s⁻¹. Find the speed and direction of the rain as he sees it.

v_rain − v_man = (0, −17.3) − (10, 0)(−10, −17.3) m s⁻¹
Apparent speed = √(10² + 17.3²) = √40020 m s⁻¹
tan θ from the vertical = 10 / 17.31/√3
θ, and the rain arrives from ahead, so he tilts forward by30°

Pro tip. Walking faster always tilts the umbrella further forward, never back — the relative velocity picks up a component opposite to your own motion, whichever way you go. Double his speed to 20 m s⁻¹ and tan θ = 20/17.3 = 1.155, a tilt of 49.1°.

Two stones are thrown from the same point at the same instant with different velocities. Ignoring air resistance, one stone sees the other move

Along a parabola
In a straight line at constant speed
In a circle about it

Both have acceleration −g ĵ, so a_AB = 0. With zero relative acceleration the relative velocity never changes, and a constant velocity traces a straight line at constant speed — this is why the separation between two projectiles grows linearly with time.

8Projectile on an incline

Firing up a slope looks like a new problem and is not. Rotate the axes: take x′ along the incline and y′ perpendicular to it. Gravity then splits into −g sin α along the slope and −g cos α across it, and the flight ends when the perpendicular displacement returns to zero — exactly the same condition as flat ground, with g cos α in place of g.

With β the launch angle measured from the incline, T = 2u sin β/(g cos α) and the range along the slope is R = u²[sin(2β + α) − sin α]/(g cos²α). Note that the slope-parallel acceleration is not zero, so unlike flat ground the along-slope motion is itself decelerating.

Figure. The same trajectory in ordinary horizontal-and-vertical coordinates, computed for u = 20 m s⁻¹ at 60°: the apex is 15 m up at x = 17.3 m and the particle meets the dashed 30° slope at (23.1, 13.3) after 2.31 s. The slope distance from origin to landing point is √(23.1² + 13.3²) = 26.7 m.

The rotated-axes recipe

Rotatex′ up the slope, y′ perpendicular. Resolve both u and g into these axes before writing anything else.
Time from y′Perpendicular motion is a flat-ground projectile with gravity g cos α, so T = 2u sin β/(g cos α).
Range from x′R = u cos β · T − ½(g sin α)T². Both terms matter; dropping the second is the usual slip.

Fired up a 30° slope

A particle is projected at u = 20 m s⁻¹ at 60° to the horizontal, up an incline of α = 30°. Take g = 10 m s⁻². Find the time of flight and the range measured along the slope.

Angle from the incline: β = 60° − 30°30°
T = 2u sin β/(g cos α) = 20 / 8.662.31 s
R = u²[sin(2β + α) − sin α]/(g cos²α) = 400(1 − 0.5)/7.526.7 m
Here 2β + α = 90°, so this is the maximum: u²/[g(1 + sin α)] = 400/1526.7 m ✓

Pro tip. Maximum range up a slope needs 2β + α = 90°, which means aiming along the bisector of the angle between the incline and the vertical. Fire down the same slope and every α flips sign: R_max = u²/[g(1 − sin α)] = 80 m here, three times as far for the identical throw.

A projectile is thrown up a 30° incline. Its range along the slope is greatest when the launch angle measured from the incline is

30°
45°
60°

R peaks when sin(2β + α) = 1, i.e. 2β + α = 90°, giving β = (90° − 30°)/2 = 30°. Equivalently the launch bisects the angle between the incline and the vertical. 45° is the flat-ground answer and is only correct when α = 0.

Notes

Equations of uniformly accelerated motion: For constant acceleration a these follow from integrating a=dv/dt: v=u+at, s=ut+\tfrac{1}{2}at^2 and v^2=u^2+2as. The displacement in the n-th second is s_n=u+\tfrac{a}{2}(2n-1), which is the average velocity over that one-second interval.
Newton's three laws: A body continues in rest or uniform motion unless acted on by a net external force (inertia); the net force equals the rate of change of momentum, \vec{F}=\tfrac{d\vec{p}}{dt}=m\vec{a} for constant mass; and every action has an equal and opposite reaction acting on a different body.
Projectile motion: With launch speed u at angle \theta, the horizontal motion is uniform (a_x=0) while the vertical motion has a_y=-g. This gives time of flight T=\tfrac{2u\sin\theta}{g}, maximum height H=\tfrac{u^2\sin^2\theta}{2g}, range R=\tfrac{u^2\sin 2\theta}{g}, and a parabolic path y=x\tan\theta-\tfrac{gx^2}{2u^2\cos^2\theta}.
Friction: Static friction is self-adjusting up to a maximum f_s^{max}=\mu_s N; once sliding begins kinetic friction f_k=\mu_k N opposes relative motion, with \mu_k<\mu_s. On an incline of angle \theta, a block just begins to slide when \tan\theta=\mu_s (angle of repose).
Relative motion: Velocity of A relative to B is \vec{v}_{AB}=\vec{v}_A-\vec{v}_B. Two projectiles in free fall share the same acceleration -g\hat{j}, so their relative acceleration is zero and one appears to move in a straight line at constant velocity relative to the other.

Formulas

Kinematic equations: v=u+at,\quad s=ut+\tfrac{1}{2}at^2,\quad v^2=u^2+2as,\quad s_n=u+\tfrac{a}{2}(2n-1)
Projectile parameters: T=\dfrac{2u\sin\theta}{g},\quad H=\dfrac{u^2\sin^2\theta}{2g},\quad R=\dfrac{u^2\sin 2\theta}{g}
Trajectory equation: y=x\tan\theta\left(1-\dfrac{x}{R}\right), with R_{max}=\dfrac{u^2}{g} at \theta=45^\circ
Newton's second law: \vec{F}_{net}=m\vec{a}=\dfrac{d\vec{p}}{dt}
Friction limits: f_s\le\mu_s N,\quad f_k=\mu_k N,\quad \text{angle of repose } \tan\theta=\mu_s
Projectile up an incline \alpha: R=\dfrac{u^2}{g\cos^2\alpha}\big[\sin(2\theta+\alpha)-\sin\alpha\big]

Exam traps & shortcuts

At \theta=45^\circ the range is maximum and R=4H; more generally R=4H\cot\theta, so you can get the range instantly from the peak height and launch angle.
Complementary angles \theta and 90^\circ-\theta give the same range, and their maximum heights satisfy H_1H_2=\tfrac{R^2}{16}.
Since two objects in free fall have zero relative acceleration, treat one as fixed and the other as moving in a straight line at constant relative velocity to solve collision/minimum-distance problems fast.

Reference tables

Every line should be reconstructible from the concept above it. If one is not, that concept is the one to reread.

Formula sheet
Quantity	Relation	Watch for
Uniform acceleration	v = u + at, s = ut + ½at², v² = u² + 2as	Constant a only
Distance in the n-th second	s_n = u + (a/2)(2n − 1)	A distance, not a speed
Displacement from a graph	signed area under the v–t curve	Unsigned area is distance
Second law	F_net = ma = dp/dt	ma only if m is constant
Friction	f_s ≤ μ_s N, f_k = μ_k N	f_s is what equilibrium needs
Angle of repose	tan θ = μ_s	Mass-free, so it measures μ_s
On a rough incline	a = g(sin θ − μ_k cos θ)	Only after it has slipped
Projectile	T = 2u sin θ/g, H = u² sin²θ/2g, R = u² sin 2θ/g	R = 4H cot θ
Trajectory	y = x tan θ (1 − x/R)	Apex at x = R/2
Relative velocity	v_AB = v_A − v_B	a_AB = 0 in free fall
Projectile up an incline	R = u²[sin(2β + α) − sin α]/(g cos²α)	β measured from the slope
Maximum range up a slope	R_max = u²/[g(1 + sin α)] at 2β + α = 90°	Use 1 − sin α downhill

The three equations of motion each omit one variable. Pick by what the question does not mention.

Which equation to reach for
You are given	You want	Use
u, a, t	v	v = u + at
u, a, t	s	s = ut + ½at²
u, a, s — no t anywhere	v	v² = u² + 2as
u, v, t — no a anywhere	s	s = ½(u + v)t
A v–t graph, a not constant	s	Area under the curve — the equations do not apply

Recap

Read only this the night before.

Equations: Three equations, each missing one variable: v = u + at has no s, s = ut + ½at² has no v, v² = u² + 2as has no t. Pick by what is absent, and only when a is constant.
Graphs: Slopes go down the chain x → v → a, areas go back up it. Signed area is displacement, unsigned area is distance.
Laws: F = dp/dt, with ma only for fixed mass. Third-law partners always sit on two different bodies, so N and mg are never a pair.
Friction: Static friction is whatever equilibrium needs, capped at μ_s N; kinetic friction is a flat μ_k N. Angle of repose: tan θ = μ_s.
Projectile: The two motions share only the clock. T = 2u sin θ/g, H = u² sin²θ/2g, R = 4H cot θ. Complementary angles share a range; the apex speed is u cos θ, not zero.
Relative: Subtract the vectors, then check the relative acceleration. In free fall it is zero, so one body sees the other run in a straight line at constant speed.
Slopes: Rotate the axes, use g cos α across the slope and g sin α along it. Maximum range up the slope when 2β + α = 90°.

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