In plain English
We turn a moving object into measurable questions.
- Where is it?Position, x
- How fast and in which direction?Velocity, v
- How is its velocity changing?Acceleration, a
H. C. Verma Chapter 3 aligned • self-study refresher
Rebuild kinematics from meaning.
Start with what motion means. Then see why differentiation is slope, why integration is area, and where the familiar motion equations actually come from.
observe
represent
calculate
motionx → v → aposition → differentiate → differentiate
By the end
Students can explain before they calculate
- Choose a reference frame and keep a consistent sign convention.
- Distinguish distance, displacement, speed, velocity, and acceleration.
- Explain Δx/Δt, dx/dt, slope, differentiation, and integration in plain language.
- Translate among motion, tables, graphs, and equations.
- Use constant-acceleration equations only when their condition is satisfied.
- Resolve plane motion and compare observations from different frames.
Module 1 • begin here
What is kinematics?
Kinematics is the language used to describe motion. It tells us what an object is doing without yet asking what force caused it.
Position x(t)
Where the object is at time t, measured from a chosen origin.
Velocity v(t)
How quickly position changes, including the direction of that change.
Acceleration a(t)
How quickly velocity changes. It may change speed, direction, or both.
How to use this lesson
Do not just read—pause and test the picture in your mind.
- 1Read the idea in plain language.
- 2Predict what the graph or object will do.
- 3Move the slider and compare with your prediction.
- 4Open the check only after answering for yourself.
Quick check: which statement is kinematics?
A. “Gravity causes the ball to accelerate.”
B. “The ball’s velocity changes by 9.8 m/s every second.”
Answer: B. It describes how motion changes. A explains the cause, so it belongs to dynamics.
Lesson 1 • describing motion
A coordinate is the beginning of the story
“Where?” is answered by position. “How did it change?” is answered by displacement. “How much ground?” is answered by distance.
Reference frame
Motion is always described relative to an observer and a coordinate system. Choose an origin, a positive direction, and when t = 0 before assigning signs.
x = −4 m tells location, not direction of travel.
Distance
Distance is the total path length. It remembers every part of the route, is a scalar, and cannot be negative.
distance ≥ |displacement|
Displacement
Displacement compares only the final and initial positions. Its sign records direction along the chosen axis.
Δx = xf − xi
Distance: the whole route10 m|first leg| + |second leg|
Displacement: endpoint change+2 mxfinal − xinitial
One complete interval
Average quantities remember the whole journey.
Average speedtotal distance / Δt
Average velocityΔx / Δt
A round trip can have zero average velocity and non-zero average speed. The next module shows how a whole-interval average becomes a value at one instant.
Modules 3–5 • the mathematics bridge
Calculus in kinematics is two ideas: slope and area
Differentiation asks, “How fast is this changing now?” Integration asks, “How much change has accumulated?” The symbols look compact; the ideas are visual.
Delta means a finite change
Δx / Δt describes an interval
Observe at two separate times. Δx = x2 − x1 and Δt = t2 − t1. Their ratio is the average velocity over that entire interval.
Derivative means the limiting rate
dx / dt describes one instant
Keep shrinking the interval around the chosen time. If the average slopes approach one value, that limit is the instantaneous velocity.
Important: dx/dt is not simply “one tiny distance divided by one tiny time.” It is the value approached by Δx/Δt as the observation interval shrinks toward zero.
Interactive idea 1 • slope
Average velocity becomes instantaneous velocity
The blue secant joins two observations. Move the second observation closer to t = 2 s. Its slope approaches the orange tangent slope at that instant.
Finite intervalvavg = Δx / Δt
One instantv = limΔt→0 Δx / Δt = dx / dt
Secant slope = 6 m/s; tangent slope = 4 m/s.
Differentiation • follow the change
Position → velocity → acceleration
On an x–t graph, slope is velocity. On a v–t graph, slope is acceleration. Differentiating once changes position into velocity; differentiating again changes velocity into acceleration.
x(t)metresdifferentiate
d/dt →v(t) = dx/dtm/sdifferentiate
d/dt →a(t) = dv/dtm/s²
d/dt →v(t) = dx/dtm/sdifferentiate
d/dt →a(t) = dv/dtm/s²
Three patterns worth seeing
| Position x(t) | Velocity v(t) | Acceleration a(t) | Motion story |
|---|---|---|---|
| 5 + 2t | 2 | 0 | Constant velocity |
| 2t² | 4t | 4 | Constant positive acceleration |
| 10 + 8t − t² | 8 − 2t | −2 | Moves right, slows, then reverses |
Here x is in metres and t in seconds; the coefficients carry the units needed to make each term a position.
Check the slope idea before continuing
For x(t) = 5t² − 2t, what are v(t) and the velocity at t = 1.5 s?
Answer: v = dx/dt = 10t − 2, so v(1.5) = 13 m/s.
Interactive idea 2 • accumulation
Integration adds many small displacements
During one short slice, displacement is approximately v Δt. Add the rectangles, then make them thinner. Their total approaches the exact area under the velocity–time graph.
Small sliceΔx ≈ v Δt
Continuous totalΔx = ∫ v(t) dt
Rectangle estimate = 14 m; exact area = 16 m.
Area under v–t
Gives signed displacement
Δx = ∫t₁t₂ v(t) dt
- Above the time axis: positive displacement.
- Below the time axis: negative displacement.
- For total distance, add magnitudes: ∫ |v(t)| dt.
Area under a–t
Gives change in velocity
Δv = ∫t₁t₂ a(t) dt
The area gives change, not the final velocity by itself. Add the initial value: vfinal = vinitial + Δv.
Integration • rebuild the motion
Acceleration → velocity → position
Integration reverses differentiation, but an initial value is needed. The same acceleration can belong to many motions that begin with different velocities or positions.
Recover velocityv(t) = v(t0) + ∫t₀t a(τ) dτ
Recover positionx(t) = x(t0) + ∫t₀t v(τ) dτ
τ is only a placeholder inside the integral; the final answer is a function of t.Interactive motion lab
Make the motion, then read its graphs
Change one quantity at a time. The same time cursor links the number line, x–t, v–t, and a–t views.
Equal-time position dotsThe object is moving right and slowing down because velocity and acceleration have opposite signs.
t = 2.2 sTime2.2 s
Position4.76 m
Velocity3.6 m/s
Acceleration-2 m/s²
Read vertically at one time: the cursor gives the position, velocity, and acceleration of the same object at the same instant.
The sign logic
Acceleration describes change, not automatically “speeding up”
Compare the directions of v and a. Speed changes according to whether their signs agree.
→ →
→ ←
← ←
← →
Same sign: speed increases. Opposite signs: speed decreases.
Module 6 • one assumption
Build the motion equations one question at a time
The goal is not to memorise four lines. It is to see how one straight velocity–time graph can be read in four useful ways.
What are we trying to find?
After t seconds, what are the final velocity and the displacement?
We begin with an initial velocity u and a constant acceleration a. We want to predict v and Δx.
- u
- initial velocity
- v
- final velocity
- a
- constant acceleration
- t
- elapsed time
- Δx
- displacement = xfinal − xinitial
final = initial + change in velocity
motion from u + extra motion from a
Use the slope to find final velocity
Constant acceleration means the v–t line has a constant slope.
a = (v − u)/t
at = v − u
v = u + at
Start at u; acceleration changes velocity by at during t seconds.
Use the area to find displacement
The whole shaded area is the displacement. Split it into two simple pieces.
rectangle = u × t = ut
triangle = ½ × t × (v − u)
The rectangle is motion from the initial velocity. The triangle is the extra displacement produced by acceleration.
Replace the triangle height
From step 1, v − u = at. Put that into the triangle area.
Δx = ut + ½t(v − u)
Δx = ut + ½t(at)
Δx = ut + ½at²
Whole shaded area = displacement Δx. Unit check: (m/s) × s = m. This drawing shows positive a; signed quantities keep the equations valid when a is negative.
Pause before continuing: what do the two shaded pieces mean?
Rectangle ut: the displacement that would occur if the object simply kept its initial velocity u.
Triangle ½at²: the extra displacement created because the velocity changes steadily.
A third form • use the whole trapezium
When u and v are known
A straight-line v–t graph has average height (u + v)/2. Multiplying that average velocity by time gives the same shaded area.
vavg = (u + v)/2
Δx = ½(u + v)t
This endpoint average works because acceleration is constant and v(t) is linear.A fourth form • remove time
When t is unknown or unnecessary
No new physical law is introduced. Rewrite the same information so time disappears.
2aΔx = 2uat + a²t²
2aΔx = (u + at)² − u² = v² − u²
v² = u² + 2aΔx
See the same first two equations using integration
1. Accumulate acceleration:
v − u = ∫0t a dτ = at ⇒ v = u + at
2. Accumulate velocity:
Δx = ∫0t (u + aτ) dτ = ut + ½at²
The graph-area route and the integration route are two descriptions of the same accumulation.
Choose by what is absent
Confirm constant acceleration, then omit what you neither know nor need.
| Equation | Quantity absent | Useful when |
|---|---|---|
| v = u + at | Δx | Finding velocity change over time |
| Δx = ut + ½at² | v | Finding position after a known time |
| Δx = ½(u + v)t | a | Initial and final velocities are known |
| v² = u² + 2aΔx | t | Time is unknown or unnecessary |
A reliable six-step routine
Represent first, calculate second
- 01Sketch the motion and choose positive direction.
- 02List u, v, a, Δx, t with signs and units.
- 03Mark the required unknown.
- 04Choose the relation with the fewest extra unknowns.
- 05Solve, preferably symbolically first.
- 06Interpret the sign; check units and plausibility.
Interactive bridge • equation → shape
Change an equation and watch all three graphs respond
Build one constant-acceleration motion, then connect each coefficient to a visible feature: vertical intercept, starting slope, curvature, and area.
2. Read the equation family
Same motion, three functions of time
x(t) = x₀ + ut + ½at² — differentiate with respect to t → v(t) = u + at — differentiate again → a(t) = a
Position equationx(t) = 2t
Velocity equationv(t) = 2
Acceleration equationa(t) = 0
- x₀ = x(0)
- vertical intercept of the x–t graph
- u = v(0)
- initial x–t slope and vertical intercept of v–t
- a = dv/dt
- x–t curvature, v–t slope, and a–t value; the t² coefficient is ½a
What changed? This preset creates one complete motion story. Move one coefficient next and predict which graph will change.
3. Read one instant on every graph
Move the shared time cursor
Graph key: the slanted dashed line is the tangent (instantaneous slope), the vertical dashed line is the shared time cursor, and +/− inside shading marks signed area. Each vertical axis may rescale, so compare the axis numbers as well as visual steepness.
- Position x
- 6 m
- Velocity v
- 2 m/s
- Acceleration a
- 0 m/s²
At t = 3 seconds, x = 6 metres, v = 2 metres per second, and a = 0 metres per second squared. The object is moving right at constant velocity.
The object is moving right at constant velocity.
Velocity is constant, so there is no turning point in this interval.
Signed-area check: area under v–t = Δx = 6 m; area under a–t = Δv = 0 m/s. Distance requires adding area magnitudes if v changes sign.
Graph-reading caution: x–t is position versus time, not a drawing of the object’s path.
Sign caution: negative acceleration means acceleration points left; it does not automatically mean slowing down.
Check all three coefficients before you leave
1. If only x₀ changes: the v–t and a–t graphs remain identical. A constant position shift disappears when x(t) is differentiated.
2. If only u increases: the initial x–t slope and the v–t vertical intercept increase; the a–t graph is unchanged.
3. If a changes sign: x–t changes concavity and the v–t slope reverses. Remember the factor of two: if x(t) contains −t², then ½a = −1, so a = −2 m/s²—not −1 m/s².
Challenge: x(t) = 5 − 2t + 2t² has x₀ = 5 m, u = −2 m/s, and a = +4 m/s². It reverses when v = −2 + 4t = 0, at t = 0.5 s.
Worked examples
Use the result to tell a physical story
Open each solution only after students have drawn a motion sketch and predicted the sign.
01A round tripdistance, displacement, averages
A walker goes from x = 0 to x = 12 m in 6 s, then returns to x = 4 m in 4 s.
Distance = 12 + 8 = 20 m. Displacement = 4 − 0 = +4 m.
Average speed = 20/10 = 2.0 m/s. Average velocity = 4/10 = +0.4 m/s.
The route affects distance; only the endpoints affect displacement.
02A braking vehiclenegative acceleration with positive velocity
A car moving at 20 m/s brakes uniformly with a = −4 m/s².
From v = u + at: 0 = 20 − 4t, so t = 5 s.
From v² = u² + 2aΔx: 0 = 400 − 8Δx, so Δx = 50 m.
The negative sign tells direction of acceleration; the speed falls because v and a have opposite signs.
03A vertical throwvelocity zero, acceleration non-zero
A ball is thrown upward at 19.6 m/s. Choose upward positive, so a = −9.8 m/s².
At the top, v = 0: 0 = 19.6 − 9.8t, so ttop = 2.0 s.
0 = 19.6² + 2(−9.8)Δy, so maximum rise = 19.6 m. Return time = 4.0 s.
At the top v = 0 for an instant, but a remains −g.
04Reaction plus brakingone journey, two motion models
A car travels at 72 km/h (= 20 m/s). Reaction time is 0.75 s; then braking acceleration is −5 m/s².
Reaction distance = 20 × 0.75 = 15 m. Braking distance follows 0 = 20² + 2(−5)d, so d = 40 m.
Total stopping distance = 55 m.
Do not apply one acceleration to the whole event: reaction and braking are different intervals.
H. C. Verma 3.7–3.9
Extend the same ideas to a plane and a new frame
Two-dimensional motion becomes manageable when perpendicular components are analysed independently.
One vector, two component stories
Write position as r = x i + y j. Differentiate each component:
v = vx i + vy j
a = ax i + ay j
The x and y equations share the same time, but otherwise can be solved separately.
Horizontal calm, vertical fall
Ignoring air resistance and taking upward positive:
x = (u cos θ)t
y = (u sin θ)t − ½gt²
Horizontal velocity stays constant; vertical velocity changes by −g every second. Their combination traces a parabola.
Subtract the observer’s motion
The position and velocity of A as seen from B are:
rA/B = rA/G − rB/G
vA/B = vA/G − vB/G
Relative velocity answers how quickly separation changes, including its direction.
Original example • same launch and landing level
Launch at 20 m/s and 30°; take g = 10 m/s²
ux = 20 cos 30° = 10√3 m/s; uy = 20 sin 30° = 10 m/s.
Time of flightT = 2uy/g = 2 s
Maximum heightH = uy²/(2g) = 5 m
RangeR = uxT = 20√3 ≈ 34.6 m
Condition check: the compact time-of-flight and range results above assume launch and landing at the same height. The component equations remain the safer starting point in other cases.
Predict • choose • explain
Concept checks
Feedback explains the physics, not just the correct option.
Teacher sequence
Three lessons with a repeated rhythm
Predict → represent → calculate → interpret. Keep that rhythm visible in every activity.
Describe motion
- Reference frames and signs
- Distance vs displacement demo
- Average vs instantaneous quantities
- Exit check: round trip
Read graphs
- Velocity and acceleration signs
- Linked simulator predictions
- Slope and signed-area meanings
- Graph-story discussion
Build and apply equations
- Derive from definition and graph
- Free fall with sign convention
- Plane motion and relative frames
- Mixed practice and exit ticket
Practice and assessment
Student worksheet
Eight original problems move from definitions to graphs, free fall, and relative motion.
SishtaKinematics practice
Name ____________________ Class __________ Date __________
- A person walks from x = 0 to x = 30 m in 20 s, then returns to x = 10 m in 10 s. Find distance, displacement, average speed, and average velocity.
- A train has u = 5 m/s and constant a = 1.5 m/s² for 8 s. Find its final velocity and displacement.
- A car moving at 24 m/s stops uniformly in 6 s. Find its acceleration and stopping distance.
- A ball is thrown vertically upward at 14 m/s. Take g = 9.8 m/s². Find the time to the top, maximum rise, and total time back to launch height.
- A v–t graph rises uniformly from 0 to 8 m/s during 0–4 s, remains at 8 m/s during 4–7 s, and falls uniformly to zero during 7–9 s. Find the acceleration in each interval and the total displacement.
- An object moves at +4 m/s for 3 s, then at −2 m/s for 2 s. Find displacement, distance, average velocity, and average speed.
- Velocity changes linearly from −6 m/s to +6 m/s in 4 s. Find acceleration, turning time, displacement, and distance.
- Runner A passes a marker at 2 m/s. Runner B passes the same marker 5 s later at 3 m/s. Assuming constant velocities, when and where does B catch A?
Graph sketch: choose one problem and represent the motion below.
xt
vt
at
Exit ticket
Complete the sentence: An object can have zero velocity but non-zero acceleration when…
Teacher answer key
- 50 m; +10 m; 1.67 m/s; +0.333 m/s.
- 17 m/s; 88 m.
- −4 m/s²; 72 m.
- 1.43 s; 10.0 m; 2.86 s.
- +2, 0, and −4 m/s²; displacement 48 m.
- 8 m; 16 m; 1.6 m/s; 3.2 m/s.
- 3 m/s²; turns at 2 s; displacement 0; distance 12 m.
- 10 s after B starts (15 s after A); x = 30 m.
Exit ticket: for example, at the highest point of a vertical throw.