Every object in the universe is in motion — from subatomic electrons orbiting a nucleus to massive galaxies moving through space. In physics, Kinematics is the branch of mechanics that describes the motion of objects without considering the forces causing the motion. Motion in a Straight Line (One-Dimensional Motion) forms Chapter 3 of the CBSE Class 11 Physics syllabus and is a foundational scoring topic for JEE Main, JEE Advanced, and NEET.
This comprehensive guide covers frame of reference, distance vs displacement, average vs instantaneous velocity, uniform acceleration, graphical & calculus derivations of the three equations of motion, distance in the nth second, free fall under gravity, kinematic graphs (x-t, v-t, a-t), relative velocity in 1D, and five step-by-step solved entrance exam numericals.
- 1. Frame of Reference, Distance & Displacement
- 2. Speed, Velocity & Acceleration (Average vs Instantaneous)
- 3. Derivation of 3 Equations of Motion (Calculus & Graphical Methods)
- 4. Distance Traveled in the n-th Second Formula
- 5. Motion Under Gravity (Free Fall Formulas & Max Height)
- 6. Kinematic Graphs Analysis (Position-Time, Velocity-Time, Acceleration-Time)
- 7. Relative Velocity in One Dimension
- 8. Solved Entrance Exam Numerical Problems (JEE / NEET)
- 9. Frequently Asked Questions (FAQ)
1. Frame of Reference, Distance & Displacement
Motion is relative. An object is said to be in motion if its position changes with respect to a chosen origin and coordinate system called a Frame of Reference.
| Physical Parameter | Distance (s or d) | Displacement (Δx or &vec;s) |
|---|---|---|
| Definition | The actual length of the path traversed by an object during a given time interval. | The shortest straight-line vector distance from the initial position to the final position. |
| Vector / Scalar Type | Scalar Quantity (has magnitude only, no direction). | Vector Quantity (has both magnitude and direction). |
| Sign (Positive/Negative/Zero) | Always positive (> 0) for a moving object. Never zero for moving body. | Can be positive, negative, or zero (if object returns to starting point). |
| Magnitude Comparison | Distance ≥ |Displacement| (Distance equals displacement only in 1D motion without direction reversal). | |Displacement| ≤ Distance. |
2. Speed, Velocity & Acceleration (Average vs Instantaneous)
v_avg = Total Displacement / Total Time = Δx / Δt
• Instantaneous Velocity: Velocity at a specific instant of time:
v = lim (Δt → 0) [Δx / Δt] = dx / dt (Derivative of position w.r.t time).
a_avg = Change in Velocity / Time = (v - u) / Δt
• Instantaneous Acceleration:
a = dv / dt = d²x / dt² = v (dv / dx)
3. Derivation of 3 Equations of Motion (Calculus & Graphical Methods)
For an object moving with constant (uniform) acceleration a, initial velocity u, final velocity v, displacement s, and time t:
By definition, a = dv/dt ⇒ dv = a dt.
Integrating both sides with limits (t=0 to t, v=u to v):
Integral dv = a Integral dt ⇒ v - u = at ⇒ v = u + at.
2. Second Equation (s = ut + ½at²):
By definition, v = ds/dt ⇒ ds = v dt = (u + at)dt.
Integrating with limits (t=0 to t, s=0 to s):
Integral ds = u Integral dt + a Integral t dt ⇒ s = ut + ½at².
3. Third Equation (v² - u² = 2as):
By definition, a = v (dv/ds) ⇒ a ds = v dv.
Integrating with limits (s=0 to s, v=u to v):
a Integral ds = Integral v dv ⇒ a s = (v² - u²)/2 ⇒ v² - u² = 2as.
4. Distance Traveled in the n-th Second Formula
The distance traveled during the n-th second (s_n) is the displacement between time t = n-1 and t = n:
s_n = u + (a / 2)(2n - 1)
5. Motion Under Gravity (Free Fall Formulas & Max Height)
When an object moves under the sole influence of Earth's gravity (neglecting air resistance), it experiences a constant downward acceleration a = -g = -9.8 m/s² (or -10 m/s²).
| Motion Parameter | Object Dropped from Rest (u = 0, downward) | Object Thrown Upward with Initial Velocity u |
|---|---|---|
| Velocity at time t | v = gt | v = u - gt |
| Displacement / Height at time t | h = ½gt² | h = ut - ½gt² |
| Velocity-Height Relation | v² = 2gh ⇒ v = √(2gh) | v² = u² - 2gh |
| Maximum Height Reached (H_max) | N/A (moves downward) | H_max = u² / (2g) (occurs when v = 0) |
| Time of Ascent / Descent | Time to fall height h: t = √(2h/g) | t_ascent = t_descent = u / g | Total Time of Flight T = 2u / g |
6. Kinematic Graphs Analysis (Position-Time, Velocity-Time, Acceleration-Time)
• Horizontal straight line ⇒ Object at rest (v = 0).
• Slanted straight line ⇒ Uniform velocity (zero acceleration).
• Parabolic curve opening upward ⇒ Uniform positive acceleration.
• Parabolic curve opening downward ⇒ Deceleration (retardation).
• Area under v-t graph = Total Displacement.
• Total Area magnitude (above + below t-axis) = Total Distance.
• Slanted straight line ⇒ Constant acceleration.
• Area under Acceleration-Time (a-t) graph = Change in velocity (Δv).
7. Relative Velocity in One Dimension
The Relative Velocity of an object A with respect to object B is the velocity at which A appears to move when observed from the frame of reference of B.
v_BA = v_B - v_A = -v_AB (Relative velocity of B w.r.t A)
• Case 1: Both objects moving in the SAME direction (vA > 0, vB > 0):
v_AB = v_A - v_B (Magnitude decreases).
• Case 2: Objects moving in OPPOSITE directions (vA > 0, vB < 0):
v_AB = v_A - (-v_B) = v_A + v_B (Magnitude increases!).
8. Solved Entrance Exam Numerical Problems (JEE / NEET)
Final velocity v = 0, Stopping distance s = 200 m.
Using 3rd equation: v² - u² = 2as ⇒ 0² - (35)² = 2 a 200 ⇒ -1225 = 400a ⇒ a = -3.06 m/s².
Retardation = 3.06 m/s².
Using 1st equation: v = u + at ⇒ 0 = 35 - 3.06t ⇒ t = 35 / 3.06 = 11.44 s.
(a) Max height from top of building H = u² / (2g) = (20)² / (2 × 10) = 400 / 20 = 20 m.
Total height from ground = 25 m + 20 m = 45 m.
(b) Net displacement to reach ground s = -25 m.
Using s = ut + ½at² ⇒ -25 = 20t - ½(10)t² ⇒ -25 = 20t - 5t² ⇒ 5t² - 20t - 25 = 0.
Divide by 5: t² - 4t - 5 = 0 ⇒ (t - 5)(t + 1) = 0 ⇒ t = 5 s (since time t > 0).
Using formula: s_n = u + (a / 2)(2n - 1).
s_5 = 0 + (4 / 2)(2(5) - 1) = 2 × (10 - 1) = 2 × 9 = 18 m.
Relative initial velocity u_BA = vB - vA = 20 - 20 = 0 m/s.
Relative acceleration a_BA = aB - aA = 1 - 0 = 1 m/s².
Time t = 50 s.
Relative displacement s_BA = u_BA t + ½ a_BA t² = 0 + ½(1)(50)² = 2500 / 2 = 1250 m.
Original distance between driver of A and guard of B = 1250 m.
(a) Velocity v(t) = dx/dt = d/dt(3t² - 2t + 5) = 6t - 2.
At t = 2 s: v(2) = 6(2) - 2 = 12 - 2 = 10 m/s.
(b) Acceleration a(t) = dv/dt = d/dt(6t - 2) = 6 m/s² (Constant acceleration!).
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9. Frequently Asked Questions (FAQ)
Distance is the total scalar path length traveled by a moving body (always positive). Displacement is the shortest vector distance from the initial position to the final position (can be positive, negative, or zero).
(1) v = u + at, (2) s = ut + ½at², (3) v² - u² = 2as. These apply strictly when acceleration a is constant.
The slope of a v-t graph represents acceleration (a = dv/dt). The area under a v-t graph represents total displacement.
During free fall, all objects experience a constant downward acceleration due to gravity a = -g = -9.8 m/s² (or -10 m/s²), regardless of mass.
When two objects move in opposite directions, their relative velocity magnitude adds up: v_AB = v_A - (-v_B) = v_A + v_B.
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