JEE Main chapters · Class 11 · Physics

Kinematics: JEE Main Physics formulas and notes

Kinematics is the base for Laws of Motion, Work, Energy and Rotation. JEE Mains usually gives 2 to 3 questions from it. The favourites are motion graphs, variable acceleration, relative velocity (river and rain), projectile motion and vectors. Aim for 90 seconds per question. About 10 hours in total is enough.

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Quick revision

• v = u + at, s = ut + ½at², v² = u² + 2as
• Slope of x–t = v, slope of v–t = a, area under v–t = displacement
• Equal distances: v(av) = (2v₁v₂)/(v₁+v₂), equal times: (v₁+v₂)/(2)
• From rest, successive seconds: 1 : 3 : 5 : 7
• Stopping distance ∝ v²
• Up and down: same speed at the same height
• v = k√x ⇒ a = (k²)/(2)
• R = √(A² + B² + 2AB cosθ)
• A⃗·B⃗ = AB cosθ and |A⃗ × B⃗| = AB sinθ
• Relative velocity: subtract the vectors
• River: least time t = (d)/(v(b)), head straight across
• T = (2u sinθ)/(g), H = (u² sin²θ)/(2g), R = (u² sin2θ)/(g)
• tanθ = (4H)/(R) and R(max) = (u²)/(g)
• a(c) = (v²)/(r) = ω²r and |Δv| = 2v sin(θ)/(2)

Formulas with solved examples

Basics

Distance, displacement, speed and velocity

• average speed = total distance ÷ total time
• average velocity = total displacement ÷ total time
• instantaneous velocity v = dx/dt
• average acceleration = Δv ÷ Δt

How to apply. Distance adds up the whole path and is never negative. Displacement is the straight-line change of position and can be zero. Speed is a scalar; velocity is a vector.

Example. A runner goes once round a circle of radius r: distance 2πr, displacement 0, average velocity 0. Half a round: distance πr, displacement 2r.

Average speed over two parts

• equal distances: v(av) = (2v₁v₂)/(v₁+v₂)
• equal times: v(av) = (v₁+v₂)/(2)
• in general: total distance ÷ total time

How to apply. Never average the speeds directly unless the times are equal. For three equal distances use 3v₁v₂v₃ ÷ (v₁v₂ + v₂v₃ + v₃v₁). If one part is made of two equal times, find its own average first.

Example. First half of the distance at 6 m/s. The second half in two equal times at 9 and 15 m/s, so that half has speed 12 m/s. v(av) = 2 × 6 × 12 ÷ (6 + 12) = 8 m/s.

Velocity and acceleration from x(t) or v(x)

• v = dx/dt
• a = dv/dt = (d²x)/(dt²)
• a = v dv/dx
• x(t) given: differentiate
• a(t) given: integrate

How to apply. Differentiate the position to get the velocity, and again for the acceleration. When the acceleration is a function of position, use a = v dv/dx and integrate. Put the time into the answer only at the end.

Example. x = t³ − 6t² + 20t + 15. v = 3t² − 12t + 20 and a = 6t − 12. a = 0 at t = 2 s, so v = 12 − 24 + 20 = 8 m/s.

Integration and distance

• Δv = ∫a dt
• Δx = ∫v dt
• distance = ∫|v| dt
• if v changes sign, split the integral at v = 0

How to apply. Displacement is the plain integral of the velocity. Distance needs the speed |v|, so split the integral wherever the velocity changes sign.

Example. v = αt + βt². Distance from t = 1 s to 2 s = ∫(αt + βt²)dt = (3α)/(2) + (7β)/(3).

Standard special relations

• v = k√x ⇒ a = (k²)/(2) (constant)
• v = kx ⇒ a = k²x
• t = αx² + βx ⇒ a = −2αv³
• x² = At² + 2Bt + C ⇒ acceleration = (AC − B²)/(x³)

How to apply. These patterns appear again and again. Each comes from a = v dv/dx or from differentiating the given relation twice.

Example. v = 4√x gives a = 16 ÷ 2 = 8 m/s², a constant. For t = αx² + βx: v = 1 ÷ (2αx + β), so the retardation is 2αv³.

Graphs

Reading x–t, v–t and a–t graphs

• slope of x–t = velocity
• slope of v–t = acceleration
• area under v–t = displacement
• area under |v|–t = distance
• area under a–t = change in velocity

How to apply. Read the axes and the units first. Count areas as triangles, rectangles and trapeziums. A v–t line below the axis is motion in the negative direction: its area is negative displacement but positive distance.

Example. v rises uniformly from 0 to 10 m/s in 4 s: slope = 2.5 m/s², area = ½ × 10 × 4 = 20 m.

Shapes of the graphs

• x–t straight line ⇒ constant velocity
• x–t parabola ⇒ constant acceleration
• v–t straight line ⇒ constant acceleration
• a–t flat line ⇒ constant acceleration
• x–t flat ⇒ at rest

How to apply. Match the shape of the graph with the order of the motion. If the x–t curve bends upward, its slope (the velocity) is increasing. Parts of the graph with a sharp kink mean a sudden change of velocity.

Example. x = 5t² gives a parabola in x–t, a straight line v = 10t in v–t and a flat line a = 10 m/s² in a–t.

v–x and a–x graphs

• a = v (dv)/(dx) = v × (slope of v–x)
• area under a–x = ½(v² − u²)
• straight v–x line ⇒ a = k v with a sign set by the slope

How to apply. From a v–x graph, read the slope at the point and multiply it by v. The a–x graph is not the same shape as the v–x graph. For a straight v–x line, a = (slope) × v is again a straight line in x, but the acceleration is not constant (for the example below it rises from −40 to 0 m/s² as x goes from 0 to 10 m).

Example. v = 20 − 2x: slope −2, so a = −2v = −40 + 4x, a straight line in a–x with value −40 m/s² at x = 0.

Equations of motion

Equations of motion (constant acceleration)

• v = u + at
• s = ut + ½at²
• v² = u² + 2as
• s = (u+v)/(2) t

How to apply. They hold only when the acceleration is constant. Write u, v, a, s and t, pick the equation with the unknown and without the quantity you do not need, and fix the sign of the direction first.

Example. u = 4 m/s, a = 2 m/s², t = 5 s: s = 4(5) + ½(2)(25) = 45 m and v = 4 + 10 = 14 m/s.

Distance in the nth second and in equal intervals

• s(nth) = u + (a(2n − 1))/(2)
• from rest, s ∝ t²
• distances in successive equal seconds from rest are 1 : 3 : 5 : 7

How to apply. The nth second means the interval from t = n − 1 to t = n. For a ratio question, form s(nth) for the two values of n and divide.

Example. From rest with a = 10 m/s²: distance in the 5th second = 10 × 9 ÷ 2 = 45 m. The first three seconds cover 5, 15 and 25 m.

Stopping distance and retardation

• v² = 2ad ⇒ d ∝ v² for the same braking
• stopping time t = (v)/(a) ∝ v
• distance in the last second before rest = (a)/(2)

How to apply. Put the final velocity as zero. If the speed doubles, the stopping distance becomes four times and the time doubles. For a bullet losing equal fractions of speed in each plank, use v² = u² + 2as plank by plank.

Example. A car at 40 km/h stops in 40 m. At 80 km/h the same braking needs 4 × 40 = 160 m.

Catching up and meeting

• put both bodies on one axis and equate x₁ = x₂
• or use relative values u(rel), a(rel), s(rel) in s = ut + ½at²
• two bodies from the same point meet when their displacements are equal

How to apply. Choose one common starting time and one origin. Add the gap to one of the displacements. A quadratic often gives two roots, so keep the one that makes sense.

Example. A car 200 m behind a bus; both start from rest with a = 4 and 2 m/s². The relative acceleration is 2 m/s²: ½ × 2 × t² = 200, t = 10√2 s.

Speed up, then slow down

• rest to rest in time t with a = α then −β: v(max) = (αβt)/(α+β)
• t₁ : t₂ = β : α
• total distance = ½ v(max) t = (αβt²)/(2(α+β))
• average speed = ½ v(max)

How to apply. The v–t graph is a triangle. Its area is the total distance and its height is the maximum velocity. The times are inversely proportional to the accelerations.

Example. α = 2 m/s², β = 3 m/s², t = 10 s: v(max) = 2 × 3 × 10 ÷ 5 = 12 m/s and distance = ½ × 12 × 10 = 60 m.

Free fall

Free fall and vertical throw

• dropped: h = ½gt²
• v = gt
• v² = 2gh
• thrown up: time up = (u)/(g)
• H = (u²)/(2g)
• total time in the air = (2u)/(g)

How to apply. Take g = 10 m/s² unless told otherwise. Take upward as positive, so a = −g for a body thrown up. For a throw from a height, use s = ut − ½gt² with s = −h at the ground.

Example. Dropped from 80 m: t = 4 s, v = 40 m/s. Thrown up at 30 m/s: time up 3 s, H = 45 m, back at the start after 6 s.

Symmetry of the up and down journey

• same speed at the same height on the way up and down
• time up = time down
• seen at the same height at t₁ (up) and t₂ (down): u = (g(t₁+t₂))/(2), h = ½gt₁t₂

How to apply. The velocity at a height has the same size up and down, but opposite direction. If two times for the same height are given, add them to find u and multiply them for h.

Example. Seen at the same height at t = 1 s and 3 s: u = 20 m/s and h = ½ × 10 × 1 × 3 = 15 m.

Dropping from a moving carrier

• a body released from a balloon or helicopter keeps the velocity of the carrier
• balloon rising at u: −h = ut − ½gt² (up positive)
• from a horizontal plane: the horizontal velocity stays, the vertical one starts at 0

How to apply. The first thing to write is the initial velocity of the object: it is the velocity of the carrier at the moment of release. Then use the ground as the point where s = −h.

Example. A balloon rises at 10 m/s and drops an object at 75 m: −75 = 10t − 5t², so t = 5 s. The balloon is then at 75 + 10 × 5 = 125 m.

Ball bouncing from the floor

• rebound speed = e × impact speed
• heights h, e²h, e⁴h …
• total time = (1+e)/(1−e)√(2h)/(g)
• total distance = h(1+e²)/(1−e²)

How to apply. The ball loses a fixed fraction of its speed at every bounce. The heights form a geometric series, so sum them. If a ball rises to a fraction of its height, that fraction is e².

Example. h = 5 m and the ball rises to 81% of its height: e = 0.9. Average speed = 5 × 1.81 ÷ (1.9² × 1) = 2.5 m/s.

Drops falling at equal intervals

• drops leave the tap at equal gaps T
• distances fallen at one instant are in the ratio 1 : 4 : 9 …
• gaps between neighbouring drops are in the ratio 1 : 3 : 5 …

How to apply. Take the instant the first drop lands. If the third drop is just starting, the fall time is 2T. The second drop has fallen for T only, so it has covered a quarter of the height.

Example. Shower height 9.8 m. Fall time = √2 s, so T = √2 ÷ 2 s. The second drop has fallen ½ × 9.8 × ½ = 2.45 m and is 7.35 m above the floor.

Vectors

Adding and subtracting vectors

• R = √(A² + B² + 2AB cosθ)
• tanβ = (B sinθ)/(A + B cosθ)
• |A⃗ − B⃗| = √(A² + B² − 2AB cosθ)
• R(max) = A + B at 0°, R(min) = |A − B| at 180°

How to apply. θ is the angle between the two vectors placed tail to tail. For a difference, flip B⃗ (add 180°) and use the same rule. β is the angle that the resultant makes with A⃗.

Example. Forces 3 N and 4 N: at 90° R = 5 N, at 0° R = 7 N, at 180° R = 1 N. Two equal vectors A at 120° give a resultant A.

Components and unit vectors

• A⃗ = Aₓî + A_yĵ + A_zk̂
• |A⃗| = √(Aₓ² + A_y² + A_z²)
• Aₓ = A cosθ, A_y = A sinθ (θ from the x-axis)
• unit vector  = (A⃗)/(|A⃗|)

How to apply. Resolve along the axes, add component by component, then combine with Pythagoras. The angle with the y-axis swaps sin and cos. To make a vector parallel to another, multiply the unit vector by the length.

Example. A⃗ = 3î + 4ĵ: |A⃗| = 5 and unit vector = 0.6î + 0.8ĵ. A vector of length 10 at 30° to the y-axis has x-part 5 and y-part 8.66.

Dot and cross product

• A⃗·B⃗ = AB cosθ = AₓBₓ + A_yB_y + A_zB_z
• |A⃗ × B⃗| = AB sinθ
• A⃗ × B⃗ = −B⃗ × A⃗
• perpendicular ⇒ A⃗·B⃗ = 0
• parallel ⇒ A⃗ × B⃗ = 0
• component of A⃗ along B⃗ = (A⃗·B⃗)/(|B⃗|)

How to apply. The dot product is a number, the cross product is a vector (use the determinant of î, ĵ, k̂). Use the dot product to test for perpendicular vectors and to find the angle or a projection.

Example. A⃗ = 2î + 3ĵ + k̂ and B⃗ = î − ĵ + k̂: A⃗·B⃗ = 2 − 3 + 1 = 0, so they are perpendicular.

Vectors of equal size

• |A⃗ + B⃗| = 2A cos(θ)/(2)
• |A⃗ − B⃗| = 2A sin(θ)/(2)
• |A⃗ + B⃗| = n|A⃗ − B⃗| ⇒ cosθ = ((n²−1)(A²+B²))/(2AB(n²+1)), which is (n²−1)/(n²+1) when A = B
• three equal vectors 120° apart add to zero
• vectors forming a closed polygon add to zero

How to apply. Use the half-angle forms for equal magnitudes. For a closed figure drawn in order, the vector sum is zero, and for the sides of a polygon the last side is minus the sum of the others.

Example. For two vectors of equal magnitude, if |A⃗ + B⃗| = 2|A⃗ − B⃗| then n = 2 and cosθ = 3/5, so θ ≈ 53°.

Relative motion

Relative velocity

• v⃗(A wrt B) = v⃗(A) − v⃗(B)
• same direction: v₁ − v₂
• opposite directions: v₁ + v₂
• at angle θ: √(v₁² + v₂² − 2v₁v₂ cosθ)
• time to meet = gap ÷ relative speed

How to apply. Relative velocity is the velocity of one body as seen from another. Subtract vectors, do not add speeds. After you find the relative speed you can treat the other body as if it is at rest.

Example. A at 20 m/s east and B at 30 m/s east: v(A wrt B) = −10 m/s. Moving towards each other, the relative speed is 50 m/s. At right angles with 3 and 4 m/s it is 5 m/s.

River and boat

• least time: t = (d)/(v(b)), drift = v(r) × t
• shortest path (v(b) > v(r)): sinθ = (v(r))/(v(b)) upstream, t = (d)/(√(v(b)² − v(r)²))
• ground velocity = boat + river

How to apply. The river speed does not change the time to cross when the boat points straight across. To land directly opposite, the boat must point upstream so that its upstream part cancels the river speed.

Example. River 4 km/h, boat 5 km/h, width 1 km: least time = 12 min, drift = 0.8 km. Straight across: sinθ = 4/5, across speed 3 km/h, time = 20 min. A swimmer at 4 km/h heading straight across a 1 km river lands 750 m downstream: t = 0.25 h, river speed = 3 km/h.

Rain and umbrella

• v⃗(rain, man) = v⃗(rain) − v⃗(man)
• rain falls vertically at v and the man walks at u: tanθ = (u)/(v) from the vertical, speed √(u² + v²)
• tilt the umbrella forward by θ

How to apply. Subtract the man’s velocity from the rain’s velocity. The man walking faster makes the rain appear to come more from the front, so the umbrella is tilted more.

Example. Rain falls vertically at 8 m/s and a man walks at 6 m/s: the rain seems to move at 10 m/s, tanθ = 6/8, θ ≈ 37° from the vertical.

Trains, crossing and meeting

• time to cross = (L₁ + L₂)/(relative speed)
• same direction: v₁ − v₂
• opposite directions: v₁ + v₂
• passing a pole or a signal: only the length of the train

How to apply. Add the lengths of both trains when they pass each other, and use the relative speed. The ratio of the two times is the inverse ratio of the relative speeds.

Example. Trains of 60 m (80 km/h) and 120 m (30 km/h): same direction relative speed 50, opposite 110, so t(same direction) : t(opposite) = 110 : 50 = 11 : 5.

River faster than the boat

• v(r) > v(b): the boat cannot land opposite and always drifts downstream
• least drift: head upstream at angle θ to the river direction reversed with cosθ = (v(b))/(v(r))
• least drift = d (√(v(r)² − v(b)²))/(v(b))

How to apply. Draw the river velocity, then a circle of radius v(b) around its tip; the ground velocity is the tangent from the origin. The boat heading is perpendicular to that ground velocity.

Example. River 5 km/h, boat 3 km/h, width 1 km: least drift = 1 × 4 ÷ 3 = 1.33 km.

Closest approach

• two bodies at constant velocity: take v(rel) = v₁ − v₂ and the position vector r⃗ of one from the other
• least distance = r sinφ, φ the angle between r⃗ and v(rel)
• time = (r cosφ)/(v(rel))

How to apply. Make one body stationary by working with the relative velocity. The least distance is the perpendicular from that body to the line of the relative velocity.

Example. A is 100 m west of B. A moves east at 10 m/s, B moves north at 10 m/s. v(rel of B wrt A) = (−10, 10), so |v(rel)| = 10√2, φ = 45°; least distance = 100 sin45° = 70.7 m, at t = 100 cos45° ÷ (10√2) = 5 s.

Projectile

Projectile on level ground

• T = (2u sinθ)/(g)
• H = (u² sin²θ)/(2g)
• R = (u² sin2θ)/(g)
• R(max) = (u²)/(g) at θ = 45°

How to apply. Split the velocity into u cosθ (horizontal, constant) and u sinθ (vertical, changes with g). Use the vertical part for the time and the height, and the horizontal part for the range.

Example. u = 20 m/s, θ = 30°: T = 2 s, H = 5 m, R = 34.6 m. At 45° the range is 40 m, the largest possible.

Path, speed and angle at any time

• y = x tanθ − (g x²)/(2u² cos²θ)
• v = √((u cosθ)² + (u sinθ − gt)²)
• tanφ = (u sinθ − gt)/(u cosθ)
• at the top: speed = u cosθ and KE = KE₀ cos²θ

How to apply. The path is a parabola. The horizontal part stays u cosθ and only the vertical part changes. The angle φ of the velocity with the horizontal falls to zero at the top and becomes negative afterwards.

Example. u = 25 m/s at 53° (cos 0.6, sin 0.8), t = 1 s: vₓ = 15 m/s, v_y = 20 − 10 = 10 m/s, speed = √325 ≈ 18 m/s.

Ratios and special angles

• tanθ = (4H)/(R)
• R(max) = (u²)/(g) and at 45° R = 4H
• θ and (90° − θ) give the same range
• for these two angles T₁T₂ = (2R)/(g) and R = 4√(H₁H₂)

How to apply. These ratio results save a lot of time. Divide the formulas for H and R to remove u. For the same range, the two angles add up to 90°.

Example. R = H when tanθ = 4. If the maximum range of a thrower is 100 m, the greatest height he can reach is u² ÷ 2g = 50 m.

Horizontal projection from a height

• t = √(2h)/(g)
• x = u t = u√(2h)/(g)
• v = √(u² + 2gh)
• y = (g x²)/(2u²)

How to apply. The initial vertical velocity is zero, so the time is that of a free fall from the height h. The speed on landing uses both velocity parts.

Example. A ball is thrown horizontally at 10 m/s from a 20 m cliff: t = 2 s, range = 20 m, landing speed = √(100 + 400) ≈ 22.4 m/s.

Projectile on an inclined plane

• along the slope a = −g sinα, perpendicular to it −g cosα
• T = (2u sinθ)/(g cosα)
• range up the slope R = (2u² sinθ cos(θ + α))/(g cos²α)
• R(max) = (u²)/(g(1 + sinα))

How to apply. Take axes along and perpendicular to the plane, with θ measured from the plane. The same formulas with α replaced by −α hold for a plane that slopes down.

Example. On a 30° slope, u = 2 m/s at 15° above the slope: R = 2 × 4 × sin15° cos45° ÷ (10 × cos²30°) ≈ 0.2 m = 20 cm.

Equal-range angles and other cases

• angles 45° + α and 45° − α: same range
• T ratio = (1 + tanα)/(1 − tanα)
• H ratio (45° − α : 45° + α) = (1 − sin2α)/(1 + sin2α)
• from a moving vehicle: add the vehicle’s velocity
• radius of curvature at the top = (u² cos²θ)/(g)

How to apply. For a moving vehicle, the horizontal velocity of the object is the vehicle’s velocity plus the throw. Two bodies in the air have zero relative acceleration, so one moves in a straight line as seen from the other.

Example. α = 15°: T ratio = 1.268 ÷ 0.732 = √3 ≈ 1.73, and the ranges are equal.

Circular motion

Angular quantities

• ω = (2π)/(T) = 2πf
• v = ωr
• α = (dω)/(dt)
• a(t) = αr
• for uniform motion θ = ωt

How to apply. Convert revolutions to radians (1 rev = 2π). Revolutions per minute become ω = 2π × rpm ÷ 60. The speed of every point of a rotating body is ωr, so it grows with r.

Example. A wheel of radius 9 m makes 120 revolutions in 3 min: f = 2/3 Hz, ω = 4π/3 rad/s, a = ω²r = 16π² m/s².

Centripetal acceleration

• a(c) = (v²)/(r) = ω²r = (4π²r)/(T²) = 4π²f²r
• directed to the centre
• same ω: a ∝ r
• same speed: a ∝ (1)/(r)

How to apply. In uniform circular motion the speed is constant but the velocity changes, so there is an acceleration towards the centre. The acceleration vector is not tangent to the circle.

Example. v = 10 m/s, r = 5 m: a = 100 ÷ 5 = 20 m/s². Two cars finishing a circle in the same time have a ratio of acceleration equal to the ratio of radii.

Change in velocity

• |Δv⃗| = 2v sin(Δθ)/(2)
• quarter turn: v√2
• half turn: 2v
• full turn: 0
• average acceleration = (|Δv⃗|)/(Δt)

How to apply. The speed stays the same, so the initial and final velocity form an isosceles triangle. Use the formula for any angle. The average acceleration is not the same as the centripetal acceleration.

Example. v = 10 m/s, turn through 60°: |Δv| = 2 × 10 × sin30° = 10 m/s.

Non-uniform circular motion

• a = √(a(t)² + a(c)²)
• a(t) = (dv)/(dt) = αr
• a(c) = (v²)/(r)
• if a(t) = a(c): (dv)/(dt) = (v²)/(r)

How to apply. Split the acceleration into the part along the tangent (changes the speed) and the part towards the centre (changes the direction). The two parts are perpendicular, so use Pythagoras.

Example. a(t) = 3 m/s² and a(c) = 4 m/s²: total acceleration = 5 m/s², at tanφ = 3/4 from the radius.

Rotational kinematics (constant α)

• ω = ω₀ + αt
• θ = ω₀t + ½αt²
• ω² = ω₀² + 2αθ
• number of turns n = θ ÷ 2π

How to apply. Same as the straight-line equations with s → θ, u → ω₀, v → ω, a → α. They need constant angular acceleration. Use the same sign rules.

Example. A wheel starts from rest with α = 4 rad/s²: after 5 s, ω = 20 rad/s, θ = ½ × 4 × 25 = 50 rad, turns = 50 ÷ 2π ≈ 8.

Most asked question types

How do I solve “Equations of motion with given data” questions in JEE Main?

A body moves with constant acceleration. Three of u, v, a, s, t are given; find another. 1. List u, v, a, s, t and mark the unknown. 2. Choose the equation that has the unknown and only the given quantities. 3. Solve and check the sign and the units. Example: u = 0, a = 5 m/s², s = 90 m. v² = 0 + 2(5)(90) = 900, so v = 30 m/s.

How do I solve “Distance in the nth second and ratios” questions in JEE Main?

Find the distance in one particular second, or the ratio of distances in successive intervals. 1. Write s(nth) = u + a(2n − 1)/2. 2. Put in u, a and n (or divide the two expressions for a ratio). 3. Remember: from rest the successive seconds give 1 : 3 : 5. Example: u = 2 m/s, a = 4 m/s², n = 3: s = 2 + 4(5)/2 = 12 m.

How do I solve “Stopping distance and braking” questions in JEE Main?

A vehicle is brought to rest, or a bullet is slowed by planks. Find the distance, the time or the speed after part of the distance. 1. Set the final velocity to zero. 2. Use v² = u² + 2as. The distance is proportional to u². 3. Remember that doubling the speed gives four times the distance. Example: 40 km/h needs 40 m, so 80 km/h needs 4 × 40 = 160 m.

How do I solve “Average speed and average velocity” questions in JEE Main?

A body covers parts of the journey with different speeds. Find the average speed or the average velocity. 1. Find the total distance (or displacement) and the total time. 2. For equal distances use 2v₁v₂ ÷ (v₁ + v₂). 3. Do not average the speeds unless the times are equal. Example: 40 km/h for the first half of the distance and 60 km/h for the rest: v(av) = 2 × 40 × 60 ÷ 100 = 48 km/h.

How do I solve “Motion graphs” questions in JEE Main?

Given an x–t, v–t, a–t or v–x graph, find the velocity, acceleration, displacement or distance, or choose the matching graph. 1. Read the axes and units first. 2. Use the slope for the rate and the area for the total. 3. Note the sign: below the axis means negative. Example: A v–t graph is a triangle from 0 to 6 s with a peak of 12 m/s. Displacement = ½ × 6 × 12 = 36 m.

How do I solve “Calculus and special relations” questions in JEE Main?

The position, velocity or time is given as a formula. Find the velocity, acceleration, distance or the relation between a and v. 1. Differentiate for velocity and acceleration; integrate for displacement. 2. Use a = v dv/dx when a depends on x. 3. Use the standard results: v = k√x gives a = k²/2, and t = αx² + βx gives a = −2αv³. Example: x = 5t² − 4t + 5: v = 10t − 4, at t = 2 s v = 16 m/s.

How do I solve “Free fall and vertical throw” questions in JEE Main?

A body is dropped, or thrown up or down. Find the time, the velocity or the height. 1. Choose the positive direction and fix the sign of g. 2. Use v = u + at and s = ut + ½at² with s = −h at the ground when it is below the start. 3. For symmetric motion use time up = time down. Example: Thrown up at 20 m/s: time to the top 2 s, H = 20 m, total time in the air 4 s.

How do I solve “Relative velocity, rain and boats” questions in JEE Main?

Find the velocity of one body as seen from another, the shortest time to cross a river or the angle of the umbrella. 1. Write both velocities as vectors. 2. Subtract the one you are looking from. 3. For a river use width ÷ v(boat) for the least time and, when v(boat) > v(river), sinθ = v(river) ÷ v(boat) for the shortest path. Example: Wind blows east at 10 m/s and a plane flies north at 40 m/s relative to the air: its ground velocity is √(10² + 40²) = 41.2 m/s.

How do I solve “Vectors: resultant, dot and cross” questions in JEE Main?

Find the resultant of two vectors or forces, the angle between vectors, or a component. 1. Write the vectors in î, ĵ, k̂ form or use A, B and θ. 2. Use R = √(A² + B² + 2AB cosθ) or the dot product for the angle. 3. Check the special cases: 0°, 90° and 180°. Example: A⃗ = 2î + 3ĵ + k̂ and B⃗ = î − ĵ + k̂: A⃗·B⃗ = 0, so the angle is 90°.

How do I solve “Projectile: T, H, R and ratios” questions in JEE Main?

A ball or a bullet is projected at an angle. Find the time, the maximum height, the range or a ratio of these. 1. Split u into u cosθ and u sinθ. 2. Use the vertical part for T and H, the horizontal part for R. 3. For ratios use tanθ = 4H/R and the fact that θ and 90° − θ give the same range. Example: u = 40 m/s, θ = 60°: T = 2(40)(0.866)/10 = 6.9 s, R = 1600 × 0.866/10 = 139 m.

How do I solve “Horizontal projection and slopes” questions in JEE Main?

A body is thrown horizontally from a height, or a projectile moves on an inclined plane. 1. For a horizontal throw, the vertical start velocity is zero: t = √(2h/g). 2. Find the range from x = ut and the landing speed from √(u² + 2gh). 3. For a slope, take axes along and perpendicular to it. Example: Thrown horizontally at 10 m/s from 20 m: t = 2 s, range 20 m, landing speed 22.4 m/s.

How do I solve “Circular motion” questions in JEE Main?

A body moves in a circle at constant speed. Find the acceleration, the change in velocity or the speed from revolutions. 1. Convert the given rotation to ω in rad/s. 2. Use a = v²/r = ω²r for the acceleration. 3. Use |Δv| = 2v sin(θ/2) for the change in velocity. Example: v = 10 m/s, turn through 60°: |Δv| = 2 × 10 × sin30° = 10 m/s.

Common myths and the truth

• Myth: Average speed equals the magnitude of the average velocity. Truth: They are equal only when the motion is in one direction. Average speed uses distance, average velocity uses displacement.
• Myth: A projectile has zero acceleration at its highest point. Truth: Vertical velocity is zero there, but the acceleration is still g downward.
• Myth: The horizontal range is maximum at 90°. Truth: The range is maximum at 45°. At 90° the projectile goes straight up and the range is zero.
• Myth: The area under the a–t graph gives displacement. Truth: The area under a–t gives the change in velocity. The area under v–t gives displacement.
• Myth: A curved x–t graph means a curved v–t graph. Truth: A parabola in x–t means constant acceleration, which is a straight line in v–t.
• Myth: A v–t line below the time axis means the object is slowing down. Truth: It means the velocity is negative. Slowing down depends on the sign of the acceleration relative to the velocity.
• Myth: Relative velocity is found by adding speeds. Truth: Subtract vectors. Two equal speeds in the same direction have zero relative speed, in opposite directions they have 2v.
• Myth: A body in uniform circular motion has no acceleration because its speed is constant. Truth: The velocity changes direction, so there is a centripetal acceleration v²/r towards the centre.
• Myth: Average speed over two equal distances is the simple mean of the speeds. Truth: It is 2v₁v₂ ÷ (v₁ + v₂). The simple mean is only for equal times.
• Myth: For a boat crossing a river, a faster river takes the boat longer to cross. Truth: If the boat points straight across, the crossing time is width ÷ boat speed. The river only changes the drift.

Mistakes to avoid

• Sign: take one direction as positive and keep it for the whole problem.
• At the top of a projectile’s path the vertical velocity is zero, but the acceleration is still g downward. The horizontal acceleration is always zero.
• Angle with the horizontal, not the vertical, unless told. The angle with the y-axis swaps sin and cos.
• Average speed is distance ÷ time; average velocity is displacement ÷ time.
• In relative velocity the order matters: v(A wrt B) = v(A) − v(B).
• When an object is thrown up, the time up is u/g and the total time in the air is 2u/g.
• The equations of motion do not work when the acceleration is not constant.
• For graphs with kinks, work piece by piece.

Simulations for Kinematics

• Average vs instantaneous velocity — Shrink the time interval and watch the secant slope approach the tangent slope on the x–t graph.
• Reaction and braking distance — Follow a car through reaction time, braking, and a complete stop.
• The catch-up problem — Release a faster cart after a delay and find where it catches a moving target.
• Motion in one dimension — Follow a cart and watch its position graph curve as its velocity changes.
• Rain and the tilted umbrella — Walk through vertical rain and find the umbrella tilt that keeps you dry using relative velocity.
• Relative velocity — Compare two carts moving along parallel tracks.
• Vertical throw and the turning point — Throw a ball upward and follow height, velocity, and acceleration through its highest point.
• Area under a v–t graph — Read displacement as the area beneath a velocity–time graph.
• Projectile with air resistance — Add linear air drag and compare the path with the ideal vacuum parabola.
• Uniform circular motion — Watch speed stay constant while velocity changes direction.
• Conical pendulum — Whirl a bob in a horizontal circle and see how the cone angle sets the speed and period.
• Horizontal projectile — Launch a ball horizontally from a cliff and watch its parabolic drop.
• Projectile motion — Launch an object and see how horizontal and vertical motion combine to form its path.
• Projectile landing on a slope — Launch above a hillside and locate the point where the projectile meets the rising ground.
• Six particles chasing on a hexagon (HC Verma) — Six particles at the corners of a regular hexagon of side a each move at constant speed v toward the…
• Three particles chasing each other (HC Verma) — Three particles at the corners of an equilateral triangle each move at speed v straight toward the next…
• Boat crossing a river — Steer upstream and combine boat velocity with river flow.
• Vector addition: the parallelogram — Resolve two vectors into components and build their resultant, head to tail.

More: Class 11 Physics simulations · all JEE Main chapters.