StudyBeacon
JEE Mains 2026 — Revision Capsule: Motion in a Straight Line
Crisp theory • High-yield formulas • Graph mastery • Micro problems • Quick traps
30-Second Core Idea
Position (x), Velocity (v), Acceleration (a) — track how they change with time, read graphs (slope & area), and distinguish constant vs variable acceleration. That's the job.
Core Definitions
- Position \(x(t)\): location on a line.
- Displacement \(\Delta x\): vector change in position.
- Distance: scalar path length (≥ displacement magnitude).
- Velocity \(v=\dfrac{dx}{dt}\). Instantaneous rate of change.
- Acceleration \(a=\dfrac{dv}{dt}\). Rate of change of velocity.
- Sign conveys direction in 1D — mind the + and −.
The Three Golden Lenses
Motion through time
Use \(x(t)\), differentiate/integrate to get \(v(t)\) and \(a(t)\).
Use \(x(t)\), differentiate/integrate to get \(v(t)\) and \(a(t)\).
Motion through velocity
Use \(a = v\,\dfrac{dv}{dx}\) when acceleration depends on \(x\) or \(v\).
Use \(a = v\,\dfrac{dv}{dx}\) when acceleration depends on \(x\) or \(v\).
Motion through graphs
Slope = velocity on x–t; area under v–t = displacement.
Slope = velocity on x–t; area under v–t = displacement.
Constant Acceleration (SUVAT)
Memorize these — they appear everywhere
v = u + ats = ut + ½ a t²v² = u² + 2ass = vt − ½ a t²s = (u+v)/2 · t⚠️ These are valid only for constant acceleration.
Motion Under Gravity
Take upward as +: \(a = -g \approx -9.8\ \text{m/s}^2\).
Time of flight
T = 2u/g
Max height
H = u² / (2g)
Time up = time down when landing at same level.
Graph Analysis — Read Like a Detective
- x–t graph: slope = velocity; curve = changing v (acceleration).
- v–t graph: slope = acceleration; area = displacement. (Area under curve may be negative.)
- a–t graph: area = change in velocity.
Piecewise v–t graphs are frequent — compute areas segment-wise (triangles + rectangles).
Variable Acceleration (Integration Approach)
Use \(a = v\,\dfrac{dv}{dx}\) when \(a\) depends on \(x\) or \(v\). Integrate carefully.
Example: If \(v = kx\), then \(a = v \frac{dv}{dx} = kx \cdot k = k^2 x\).
Relative Motion (1D)
Relative velocity: \(v_{AB} = v_A - v_B\).
If two objects approach each other, effective speed = sum of magnitudes; if moving same direction, subtract.
Common JEE Traps & Quick Tips
- Confusing distance with displacement — draw the path.
- Applying constant-acceleration formulas to variable a — check first.
- Wrong sign for \(g\) or velocity — choose and stick to a convention.
- For graphs, sign of area matters (negative area reduces displacement).
- Turning points: velocity = 0 (not acceleration).
Question Archetypes (High Frequency)
- Displacement from v–t graph (area calculation).
- Time of flight / max height under gravity.
- Relative motion meeting time / collision.
- Velocity as function of position: integrate using \(a = v\,dv/dx\).
- Piecewise constant acceleration — split into segments.
Micro Problem Set — Try First, Peek Later
- A body starts with \(u=5\ \text{m/s}\) and accelerates at \(3\ \text{m/s}^2\). Find displacement in \(t=4\ \text{s}\).
- Particle velocity \(v(t)=4t - t^2\). Find times when it stops.
- From v–t graph shaped as a triangle (base = 6 s, peak = 12 m/s), find displacement.
- Stone thrown up at \(u=15\ \text{m/s}\). Find time to reach \(10\ \text{m}\) height.
- Two bikes in opposite directions: speeds 20 m/s and 10 m/s. Relative velocity? Find time to meet if 300 m apart.
Show Answers & Solutions
- Use \(s = ut + \tfrac{1}{2}at^2\): \(s = 5\times4 + 0.5\times3\times16 = 20 + 24 = 44\ \text{m}.\)
- Set \(v=0\): \(4t - t^2 = 0 \Rightarrow t(4 - t)=0\). So \(t=0\) or \(t=4\ \text{s}\).
- Triangle area = \( \tfrac{1}{2}\times \text{base}\times \text{height} = 0.5\times6\times12 = 36\ \text{m}.\)
- Use \(s = ut - \tfrac{1}{2}gt^2 = 10\). With \(u=15\) and \(g\approx9.8\): solve \(15t - 4.9t^2 = 10\). Rearranged: \(4.9t^2 - 15t + 10 = 0\). Solve quadratic → \(t \approx 0.87\ \text{s}\) (ascending solution) and another larger time for descending if required.
- Relative speed = \(20 + 10 = 30\ \text{m/s}\). Time = distance / relative speed = \(300/30 = 10\ \text{s}.\)
One-Line Wrap-Up
Master x, v, a & graph reading — constant vs variable acceleration is the decisive skill for steady JEE scoring.
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