Motion in 1D

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Kinematics — The SUVAT Equations:

For motion along a straight line with constant acceleration a, four SUVAT equations govern all relationships between displacement (s), initial velocity (u), final velocity (v), acceleration (a), and time (t):

1. v = u + at
2. s = ut + ½at²
3. v² = u² + 2as
4. s = (u + v)t/2

These equations are derived by integrating acceleration (constant) to get velocity, and velocity to get displacement. They work for any 1D motion with constant acceleration — free fall, projectile motion projected vertically, blocks sliding on inclines.

Key Distinctions:

• Distance: Total path length travelled (scalar, always positive)
• Displacement: Change in position vector (vector, can be negative; s_final - s_initial)
• Speed: Rate of distance covered (scalar, ≥ 0)
• Velocity: Rate of change of displacement (vector, can be negative)

Average speed ≥ |Average velocity| always. Equality holds only when the motion is in one constant direction (no reversal).

Sign Convention — CRITICAL:

Choose a positive direction and stick to it throughout the problem. For upward as positive: g = +9.8 m/s². For downward as positive: g = -9.8 m/s². Mixing sign conventions is the #1 cause of errors in kinematics problems.

⚡ JEE Tip: For a body thrown upward from height h with speed u: maximum height reached = h + u²/(2g). Time of flight (to return to original level) = 2u/g. For a body dropped from height h (u = 0): time to fall = √(2h/g), speed on impact = √(2gh). These are the most frequently tested variations.

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🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Graphical Analysis — Position-Time and Velocity-Time:

Position-time graph (x vs t):

• Slope at any point = instantaneous velocity (dx/dt)
• Straight line → constant velocity (slope = v, no acceleration)
• Curve → changing velocity, non-uniform acceleration
• Negative slope → negative velocity

Velocity-time graph (v vs t):

• Slope at any point = acceleration (dv/dt)
• Area under the v-t graph from t₁ to t₂ = displacement (change in position) during that interval
• Area below the time axis (negative v) subtracts from displacement
• Total area (with sign) = net displacement; absolute area = total distance

Acceleration-time graph (a vs t):

• Area under a-t graph = change in velocity (Δv)
• No direct information about position or displacement

Free Fall and Gravity:

Near Earth’s surface, all bodies experience the same gravitational acceleration g ≈ 9.8 m/s² (often approximated as 10 m/s² for quick estimates). The actual value varies slightly with altitude and latitude. At Earth’s surface: g at equator ≈ 9.78 m/s², g at poles ≈ 9.83 m/s². The difference is due to Earth’s rotation and equatorial bulging.

For a body thrown upward with initial velocity u (taking upward as positive): velocity decreases by 9.8 m/s every second. At maximum height: v = 0, a = -g (still accelerating downward!). After reaching the top and beginning to fall, it behaves exactly like a body dropped from that height.

⚡ JEE Tip: For two balls thrown from the same point with different initial velocities, the relative velocity between them remains constant at all times. This is because both experience the same gravitational acceleration. This principle simplifies problems involving pursuit or separation.

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Derivation of SUVAT from Calculus:

Acceleration is the time derivative of velocity: a = dv/dt. For constant a, integrating: ∫(u to v) dv = ∫(0 to t) a dt → v - u = at → v = u + at. Integrating again for position: v = ds/dt, so ds = v dt = (u + at) dt. Integrating: ∫(s₀ to s) ds = ∫(0 to t) (u + at) dt → s - s₀ = ut + ½at² → s = s₀ + ut + ½at². Starting from s₀ = 0 gives the standard form.

Non-Uniform Acceleration — Using Calculus:

When acceleration varies with time, we must integrate: a(t) = dv/dt → v = ∫a(t) dt + C (using initial condition to find C) v = ds/dt → s = ∫v dt + C’ (using initial condition to find C’)

For force problems where F varies with position or velocity (like air resistance F ∝ v or F ∝ v²), solve the differential equation to find v(t) and s(t).

Air Resistance — Terminal Velocity:

For a body falling through air with drag force proportional to velocity: F_drag = kv (linear drag). Equation of motion: mdv/dt = mg - kv. This is a first-order linear differential equation. Solving: v(t) = (mg/k)(1 - e^(-kt/m)). As t → ∞, v → v_t = mg/k (terminal velocity). The time constant τ = m/k determines how quickly the body approaches terminal velocity.

For drag force F ∝ v² (appropriate for larger objects at moderate speeds): mdv/dt = mg - kv². This gives v(t) = v_t tanh(gt/v_t), with v_t = √(mg/k). Terminal velocity is approached asymptotically but never reached in finite time.

Relative Motion in 1D:

Velocity of A relative to B: v_AB = v_A - v_B. This is the velocity A appears to have when viewed from B’s frame of reference. When two bodies move toward each other: relative speed = v₁ + v₂ (sum). When moving in the same direction (one overtaking): relative speed = |v₁ - v₂|.

For stone dropped from a moving vehicle: relative to ground, the stone has horizontal velocity of the vehicle at release + vertical motion under gravity. Relative to the vehicle (which itself accelerates if decelerating), the stone’s trajectory appears curved.

⚡ JEE Advanced Pattern: JEE Advanced rarely tests pure 1D kinematics — instead, 1D motion appears as a component of 2D projectile motion, or as part of a constraint in connected body problems. However, when it does appear directly, it is often combined with calculus-based reasoning (finding velocity as derivative of position, acceleration as second derivative), graphical analysis (finding displacement from v-t graph), or relative motion (two stones thrown from a building at different times). The 2023 Paper 1 Qn 15 involved a particle moving with acceleration that varied as a(t) = 6t - 2 (in m/s²), with initial conditions x₀ = 0, v₀ = 5 m/s — students needed to integrate to find v(t) = 3t² - 2t + 5, then integrate again to find position x(t) = t³ - t² + 5t, then evaluate at t = 3s to find displacement = 27 - 9 + 15 = 33 m.