One clock, clear axes and honest graphs: connect straight-line motion, projectiles and moving observers.
JEE Main · JEE Advanced · NEET UG75 min study guide9 indexed sections
BY THE END, YOU CAN
Interpret position, velocity and acceleration graphs and calculus.
Solve signed constant-acceleration and piecewise motion without extending a model beyond its interval.
Resolve two-dimensional motion, projectile trajectories and curvature components.
Solve one-dimensional overtaking and two-dimensional river, rain and closest-approach problems.
BEFORE YOU START
Basic vector components and Pythagoras.
Elementary differentiation, integration and quadratic equations.
Original educational lesson. Work through the examples before revealing the practice answers. Check the current official syllabus separately; historical question frequency does not establish eligibility or predict future questions.
CONNECT THE CONCEPT TO THE EXAM
Selected exam questions & worked solutions
These selected examples are separate from the original practice below. Each identifies its exam, year and wording. They do not represent every paper or every selected year.
Available in this lesson: JEE Advanced 2017. This is a limited lesson sample, not a complete exam-question bank.
1 sourced sample question available across the listed exams and years.
Independently derived and checked solution. Archived key finality is unverified; this is not presented as an official final answer.
An expanding sphere has fixed mass and spatially uniform density ρ. The fractional density-change rate (1/ρ)dρ/dt is constant. How does the surface speed v depend on the instantaneous radius R?
Reveal answer & worked solution
Answer: v is proportional to R (independently derived; archived option A).
Let the constant fractional density-change rate be c. Then v=dR/dt=−cR/3.
Expansion makes c negative, so −c/3 is a fixed positive factor. Hence v∝R; constant fractional density change does not mean constant speed.
Source: JEE (Advanced) 2017, Paper 2, Physics Q1; independently derived solution; official-key finality unverified. Solution written for this lesson.
01 / UNDERSTAND & APPLY
1. Describe the motion before choosing an equation
Kinematics describes where an object is and how its motion changes without yet asking what force causes the change. Start by choosing an origin, positive directions, a clock zero and a reference frame. Two observers may assign different positions and velocities to the same object without disagreeing about the physical event. A clearly labelled coordinate choice makes signs meaningful rather than decorative.
Position is a location relative to the origin. Displacement is final position minus initial position, so it depends only on endpoints. Distance is the length of the actual path and is never negative. A runner completing one lap has zero displacement and positive distance. Dividing these two quantities by the same elapsed time gives average velocity and average speed, respectively.
Instantaneous speed is the magnitude of instantaneous velocity. In one dimension, velocity can be positive or negative while speed cannot. Positive acceleration means the velocity component is increasing, not necessarily that the object is speeding up. If velocity is negative and acceleration positive, its speed initially decreases. The sign of va decides whether speed increases or decreases away from a turning point.
Use a point-particle approximation when the object’s size and rotation are irrelevant to the question. For a train passing a platform, size is relevant to the required travel distance even if each endpoint has the same translational velocity. Define the event precisely: first contact, centre alignment and complete clearance occur at different times.
v̄ = Δx/Δt; average speed = total distance/Δt
The elapsed time includes pauses unless the problem explicitly excludes them.
ORIGINAL WORKED EXAMPLE
Average is weighted by time
A cyclist covers equal 300 m stretches at 3 m s⁻¹ and 6 m s⁻¹ in the same direction. Find the average speed.
Times are 300/3 = 100 s and 300/6 = 50 s.
Total distance is 600 m and total time is 150 s, so average speed is 4 m s⁻¹.
The arithmetic mean 4.5 m s⁻¹ would apply to equal times, not equal distances.
Answer 4 m s⁻¹.
02 / UNDERSTAND & APPLY
2. Graph slopes and areas carry the physics
On a position–time graph, the secant slope is average velocity and the tangent slope is instantaneous velocity. The graph height is position, not speed. A horizontal segment means rest; a downward slope means negative velocity. Two curves crossing means equal position at that instant, not equal velocity. Equal velocities require equal tangent slopes.
On a velocity–time graph, slope is acceleration and signed area is displacement. Area below the time axis subtracts from displacement but contributes positively to distance. To calculate total distance, divide the interval wherever velocity changes sign and add the magnitudes of the signed areas. On an acceleration–time graph, area gives change in velocity, not velocity unless an initial value is supplied.
These graphical statements are calculus in geometric form. Velocity is dx/dt and acceleration is dv/dt. Integrating acceleration requires an integration constant fixed by initial velocity; integrating velocity requires an initial position. Differentiating a proposed trajectory is often easier and safer than forcing it into a constant-acceleration formula.
A sharp corner in an idealised velocity graph represents an abrupt change in acceleration, which can be a useful approximation. A discontinuity in velocity would require an impulsive change, not ordinary finite constant acceleration. A vertical jump in position is not a physically admissible continuous trajectory in this model. Read the axes and units before interpreting the shape.
v = dx/dt; a = dv/dt; Δx = ∫v dt; Δv = ∫a dt
These relations apply beyond constant acceleration. Limits and initial conditions are essential.
a = v dv/dx
Valid where velocity can be treated as a differentiable function of position along a one-dimensional trajectory.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
Quantitative axes. Constant slope −2 m/s². Positive and negative shaded regions carry explicit signed labels.
ORIGINAL WORKED EXAMPLE
A reversal hidden in a polynomial
For 0 ≤ t ≤ 4 s, x(t) = t² − 4t metres. Find displacement, distance and average speed.
Differentiate: v = 2t − 4 m s⁻¹. The object reverses at t = 2 s.
Positions are x(0) = 0, x(2) = −4 m and x(4) = 0.
Displacement is 0 m. Distance is |−4 − 0| + |0 − (−4)| = 8 m.
Average speed is 8/4 = 2 m s⁻¹, while average velocity is zero.
Answer Displacement 0; distance 8 m; average speed 2 m s⁻¹.
03 / UNDERSTAND & APPLY
3. Derive the constant-acceleration equations and respect their limits
If acceleration is constant, the velocity–time graph is a straight line. Its slope gives v = u + at. The area under that line is a trapezium, so displacement is (u + v)t/2. Substituting for v gives s = ut + at²/2. Eliminating t gives v² = u² + 2as. These are related descriptions of the same trajectory, not four unrelated tricks.
Keep a single signed axis throughout. If up is positive during free fall near Earth, a = −g on both ascent and descent. At the highest point velocity is zero momentarily but acceleration is still downward. Changing the sign of g because the object starts falling confuses direction of motion with direction of acceleration.
For piecewise motion, finish one interval and use its final position and velocity as the next interval’s initial conditions. A braking equation is valid only until the vehicle stops unless the physical mechanism really accelerates it backwards afterward. A negative root of a quadratic may describe an event before the chosen clock zero; retain roots only if they lie within the stated physical interval.
When acceleration varies with time, integrate it. When acceleration is specified as a function of position, the chain-rule form a = v dv/dx may be useful. The formula v² = u² + 2as cannot simply accept the final acceleration in place of a varying one. The average velocity (u + v)/2 also relies on a linear velocity–time relation.
ORIGINAL WORKED EXAMPLE
Reaction time plus braking
A car moves at 20 m s⁻¹. The driver reacts for 0.75 s, then brakes with constant acceleration −5 m s⁻². Find stopping distance.
During reaction, acceleration is taken as zero, so distance is 20 × 0.75 = 15 m.
During braking use v² = u² + 2as with v = 0: 0 = 400 − 10s, so s = 40 m.
Total stopping distance is 15 + 40 = 55 m. Braking time is 4 s; the full process lasts 4.75 s.
Answer 55 m.
04 / UNDERSTAND & APPLY
4. Components share one clock
In a plane, position is r = xi + yj. Differentiate each component to obtain v = vₓi + vᵧj and a = aₓi + aᵧj. The components obey their own equations, but they describe one event at one common time. You may choose axes to simplify acceleration; that does not make horizontal and vertical arrival times independent choices.
The magnitude of velocity is √(vₓ² + vᵧ²), and its direction follows from the components with attention to the quadrant. Acceleration is not generally parallel to velocity. A perpendicular acceleration changes direction without changing speed at that instant, while the component along the velocity changes speed. This explains why uniform circular motion has nonzero acceleration even though its speed is constant.
For a smooth curved path, resolve acceleration into tangential and normal components. The tangential component is dv/dt; the inward normal component is v²/R, where R is the local radius of curvature. These are perpendicular components, so their magnitudes combine by Pythagoras. In uniform circular motion the tangential component vanishes, but in nonuniform circular motion both may be present.
A useful sketch has a tangent arrow for velocity and an inward arrow for normal acceleration. At the top of a projectile’s path, velocity is horizontal and acceleration is vertical. At a general point, the acceleration can have both tangential and normal parts even though its Cartesian vector remains the same downward g.
r = xi + yj; v = (dx/dt)i + (dy/dt)j
Differentiate components in a fixed Cartesian basis.
aₜ = dv/dt; aₙ = v²/R
The normal component points toward the local centre of curvature; v is speed.
ORIGINAL WORKED EXAMPLE
Speed is not acceleration magnitude
A particle has r(t) = 3ti + t²j metres. Find velocity, acceleration and speed at t = 2 s.
Differentiate: v = 3i + 2tj m s⁻¹ and a = 2j m s⁻².
At 2 s, velocity is 3i + 4j, so speed is √(9 + 16) = 5 m s⁻¹.
Acceleration magnitude is 2 m s⁻². Its tangential component is (v·a)/|v| = 8/5 = 1.6 m s⁻²; the remaining normal component is 1.2 m s⁻².
Answer v = 3i + 4j m s⁻¹; a = 2j m s⁻²; speed 5 m s⁻¹.
05 / UNDERSTAND & APPLY
5. Projectile motion is simultaneous free fall and horizontal drift
For an ideal projectile near Earth, neglect air resistance and treat gravity as uniform. Choose x horizontal and y upward. Then aₓ = 0 and aᵧ = −g. If launch speed is u at angle θ above horizontal, initial components are u cosθ and u sinθ. Horizontal speed stays constant while vertical velocity changes linearly.
With the launch point as origin, x = u cosθ t and y = u sinθ t − gt²/2. Eliminating time gives a parabola when the horizontal component is nonzero. The top occurs when vᵧ = 0, but the speed there is |u cosθ|, not generally zero. A vertical launch is a special case in which the entire velocity vanishes at the top.
For landing at the launch height, the nonzero root of y = 0 gives flight time 2u sinθ/g. Substituting into x gives range u² sin2θ/g, with maximum at 45° under these assumptions. Complementary launch angles produce equal range but different flight times and peak heights. None of these equal-height formulas automatically applies to a launch from a cliff.
For a different landing height, solve the vertical equation using that height, select the future physical root, and then calculate horizontal displacement. A target collision requires both coordinates to match at the same time. If both freely moving objects have identical gravitational acceleration, their relative acceleration is zero until a collision or a constraint changes the motion.
x = u cosθ t; y = u sinθ t − ½gt²
Origin at launch, horizontal x, upward y, constant g and negligible drag.
H = u² sin²θ/(2g); R = u² sin2θ/g
H is rise above launch; the stated range R requires equal launch and landing heights.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
Position coordinates are quantitative. Velocity arrows are symbolic and not on the position scale; g=10 m/s² and air resistance is neglected.
ORIGINAL WORKED EXAMPLE
Launch from a height
A ball leaves a ledge 45 m above level ground horizontally at 12 m s⁻¹. Use g = 10 m s⁻². Find flight time, range and impact speed.
Vertical displacement relative to launch is −45 m: −45 = −5t², so t = 3 s.
Horizontal distance is 12 × 3 = 36 m.
Impact components are vₓ = 12 and vᵧ = −30 m s⁻¹. Speed is √(12² + 30²) = √1044 ≈ 32.3 m s⁻¹.
A horizontal launch does not have zero flight time; the equal-height range formula is inapplicable.
Answer 3 s; 36 m; approximately 32.3 m s⁻¹.
06 / UNDERSTAND & APPLY
6. Relative velocity works in one and two dimensions
The position of A relative to B is rₐᵦ = rₐ − rᵦ. Differentiating gives vₐᵦ = vₐ − vᵦ and aₐᵦ = aₐ − aᵦ. The order matters: reversing the observer reverses the relative vector. In one dimension, use signed velocities; “add speeds for opposite directions” is a consequence of subtraction, not a separate universal law.
For two objects with constant relative velocity, relative position is r₀ + vᵣt. A collision requires this entire vector to be zero at one time. The time of closest approach follows by minimizing its squared magnitude: t* = −(r₀·vᵣ)/|vᵣ|² for vᵣ ≠ 0. If vᵣ = 0, separation is constant and every allowed time is equally close. If this time is outside the allowed interval, the closest allowed position occurs at a boundary. Parallel tracks can yield zero longitudinal separation without collision.
In a river problem, boat velocity relative to ground equals boat velocity relative to water plus water velocity relative to ground. Aiming straight across gives the shortest crossing time for a fixed boat speed relative to water, but produces downstream drift. To land directly opposite, the boat must aim upstream so that its upstream component cancels the current. This is possible only if boat speed is at least current speed; a positive across-river component requires it to be greater.
Rain, wind and moving walkways use the same vector triangle. The angle at which rain appears to fall depends on rain velocity relative to the observer. Do not add magnitudes before resolving directions. These are nonrelativistic transformations with translating, nonrotating axes; a rotating reference frame requires additional care.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
Vector lengths use a common scale: the 3–4–5 triangle is quantitative. The current arrow is repeated at the boat-vector tip for tip-to-tail addition. Arrows are velocities, not spatial paths.
ORIGINAL WORKED EXAMPLE
Complete clearance in one dimension
Two trains 120 m and 180 m long approach on parallel tracks at 15 and 10 m s⁻¹. How long passes between their front ends meeting and complete clearance?
Choose the first train’s direction positive. Velocities are +15 and −10 m s⁻¹.
Relative speed is 15 − (−10) = 25 m s⁻¹. For complete clearance, one train must travel the sum of lengths relative to the other.
Time = (120 + 180)/25 = 12 s.
Answer 12 s.
ORIGINAL WORKED EXAMPLE
Two river objectives
A river is 120 m wide and flows at 3 m s⁻¹. A boat moves at 5 m s⁻¹ relative to water. Compare straight-across aiming with landing directly opposite.
Straight-across aiming gives across speed 5 m s⁻¹, so time is 24 s and downstream drift is 3 × 24 = 72 m.
For opposite-bank landing directly across, give the boat an upstream component of 3 m s⁻¹. Its remaining across component is √(5² − 3²) = 4 m s⁻¹.
Time is then 120/4 = 30 s. Aim at an angle arcsin(3/5) ≈ 36.9° upstream from the straight-across direction.
Answer Shortest crossing: 24 s with 72 m drift. Directly opposite: 30 s at 36.9° upstream.
07 / UNDERSTAND & APPLY
7. Turn verbal events into simultaneous equations
Questions about meeting, overtaking or catching up are questions about equal positions at a common time. Write each position using the same origin and clock, including any delayed start. Do not set velocities equal unless the question asks for equal velocities. A faster object can be behind, and equal velocity does not imply equal position.
For a chase that begins after a delay, either keep one global clock and use a shifted time for the delayed object or reset the clock and include the head start as an initial separation. Both approaches work; mixing them double-counts or removes the delay. Verify the resulting time by substituting into both position expressions.
Graph sketches make event checks quick. At a crossing of position graphs, compare slopes to see which object passes the other. At a minimum separation, relative velocity is perpendicular to relative position in two dimensions; in a one-dimensional interior minimum without collision, relative velocity is zero. Boundary cases must still be considered.
ORIGINAL WORKED EXAMPLE
Closest approach without collision
Initially B is 30 m east and 40 m north of A. Its velocity relative to A is 3 m s⁻¹ westward. Find the closest separation for t ≥ 0.
Relative position is (30 − 3t)i + 40j metres.
Squared separation is (30 − 3t)² + 1600, minimized when 30 − 3t = 0.
At t = 10 s the east–west separation vanishes but the north–south separation remains 40 m.
Answer 40 m at 10 s; no collision.
08 / UNDERSTAND & APPLY
8. Original practice with full reasoning
Use g = 10 m s⁻² where specified. Each problem is original practice rather than a claimed past-paper question.
ORIGINAL PRACTICE · 1
An object moves +8 m, then −3 m in 5 s. Find average velocity and average speed.
Reveal answer & reasoning
1 m s⁻¹ and 2.2 m s⁻¹.
Displacement is 5 m; distance is 11 m. Divide each by 5 s.
ORIGINAL PRACTICE · 2
Velocity changes uniformly from +6 to −2 m s⁻¹ over 4 s. Find displacement and distance.
Reveal answer & reasoning
8 m displacement; 10 m distance.
Acceleration is −2 m s⁻². Velocity reaches zero at 3 s. Signed areas are +9 m and −1 m, while distance adds their magnitudes.
ORIGINAL PRACTICE · 3
A ball is thrown vertically upward at 20 m s⁻¹. With g = 10, find velocity and acceleration at its highest point.
Reveal answer & reasoning
Velocity zero; acceleration −10 m s⁻² for upward positive.
The top occurs at 2 s. Gravity does not vanish when velocity does.
ORIGINAL PRACTICE · 4
For x = 2t³ metres, find velocity and acceleration at t = 2 s.
Reveal answer & reasoning
24 m s⁻¹ and 24 m s⁻².
Differentiate: v = 6t² and a = 12t. The numerical equality is coincidental and the units differ.
ORIGINAL PRACTICE · 5
A projectile is launched at 20 m s⁻¹ and 30° on level ground. Use g = 10. Find flight time and range.
Reveal answer & reasoning
2 s and 20√3 m ≈ 34.6 m.
Vertical launch component is 10 m s⁻¹, giving flight time 2 s. Horizontal component is 10√3 m s⁻¹.
ORIGINAL PRACTICE · 6
A particle has velocity 6i + 8j m s⁻¹ and acceleration −4i + 3j m s⁻². Is its speed changing at that instant?
Reveal answer & reasoning
No; its direction is changing.
v·a = −24 + 24 = 0, so tangential acceleration is zero. The nonzero acceleration is perpendicular to velocity.
ORIGINAL PRACTICE · 7
Two cars move east at 24 and 18 m s⁻¹. The faster starts 90 m behind. When does it catch up at constant speeds?
Reveal answer & reasoning
15 s.
Closing speed is 24 − 18 = 6 m s⁻¹. Divide separation by closing speed.
ORIGINAL PRACTICE · 8
Rain falls vertically at 8 m s⁻¹ relative to ground. A runner moves east at 6 m s⁻¹. Find apparent rain speed and direction.
Reveal answer & reasoning
10 m s⁻¹, downward and westward, 36.9° from vertical.
Subtract runner velocity: relative rain velocity is −6i − 8j. Magnitude is 10 and tanθ = 6/8.
ORIGINAL PRACTICE · 9
A swimmer’s speed relative to water is 2 m s⁻¹ and current is 3 m s⁻¹. Can the swimmer land directly opposite?
Reveal answer & reasoning
No under these steady-current assumptions.
The swimmer cannot supply a 3 m s⁻¹ upstream component with total relative speed only 2 m s⁻¹.
ORIGINAL PRACTICE · 10
A body is dropped from rest. With g = 10, how far does it fall during its third second?
Reveal answer & reasoning
25 m.
Distance by 3 s is 45 m and by 2 s is 20 m. Their difference is the distance during the interval 2 to 3 s.
09 / UNDERSTAND & APPLY
9. A reliable route through a motion problem
Draw the relevant positions and velocity directions, choose signed axes, and write the event condition before selecting an equation. Ask whether acceleration is actually constant. In multiple dimensions, resolve components and share the clock. In relative motion, state the observer and subtract vectors in the corresponding order.
After solving, check the result against the event and physical interval. Does a stopping-time calculation continue beyond rest? Does a range formula assume equal heights? Does a closest-approach time lie in the future? Do the two objects really occupy the same point, or merely the same x coordinate? These checks are often more informative than recomputing arithmetic.
Slope of x–t gives velocity; signed area of v–t gives displacement.
Distance requires absolute path contributions, split at reversals.
Zero velocity does not require zero acceleration.
Constant-acceleration formulas need constant acceleration over that interval.
A projectile has one flight time shared by both components.
Relative velocity is a vector difference, including in one dimension.
KEEP THE BIG IDEAS
Quick revision
Position, displacement and distance are distinct quantities.
Signs describe chosen axes; acceleration need not point along motion.
Component equations describe the same event at the same time.
The observer determines relative velocity; vector subtraction keeps the signs consistent.
The teaching notes, worked explanations and original practice are written for this lesson. Separately labeled selected exam questions identify their exam, year, source and whether wording is paraphrased or an exact excerpt. Educational references do not certify current exam coverage.