Kinematic Variables
Distance vs. Displacement
- Core definition & intuition: Distance is the total "footprint" left behind (actual path length). It is a scalar, meaning direction doesn't matter. Displacement is the straight-line "shortcut" from the starting point to the ending point. It is a vector, meaning it depends on both magnitude and direction.
- Mathematical formulation: Δx = xf - xi (where xf is the final position, and xi is the initial position).
- Crucial detail/Trap: Distance is never negative and never decreases with time for a moving body. Displacement can be positive, negative, or zero. Distance ≥ |Displacement|. They are only equal in magnitude if the particle moves in a straight line without taking any U-turns.
- Graphical meaning: Displacement is the net area under a velocity-time graph (areas below the time axis are subtracted). Distance is the total absolute area under a velocity-time graph (areas below the axis are added).
PYQ Problem 1:
An athlete completes one round of a circular track of radius R in 40 s. What will be his displacement at the end of 2 min 20 s? AIPMT 1990
(1) Zero (2) 2R (3) 2πR (4) 7πR
Solution:
Step 1 - Visualizing the Physics: Imagine a runner on a circular track. Every 40 s, they return to the exact starting line, meaning their displacement resets to zero.
Step 2 - Identify Variables: Radius = R, Time for 1 revolution = 40 s, Total time = 2 min 20 s = 140 s.
Step 3 - Application: Calculate the total number of revolutions: n = 140 / 40 = 3.5 revolutions. The athlete completes 3 full rounds (ending back at the start) and then exactly half a round. After half a round, they are positioned at the diametrically opposite end of the circle.
Conclusion: The shortest straight-line distance from the start to the diametrically opposite point is the diameter. Therefore, Displacement = 2R. Option (2) is correct.
PYQ Problem 2:
A wheel of radius 1 meter rolls forward half a revolution on a horizontal ground. The magnitude of the displacement of the point of the wheel initially in contact with the ground is: AIPMT 2002
(1) 2π (2) √2 π (3) √(π2 + 4) (4) π
Solution:
Step 1 - Visualizing the Physics: The wheel rolls forward. The point initially touching the ground moves up to the highest point of the wheel after half a revolution. Simultaneously, the center of the wheel moves horizontally.
Step 2 - Identify Variables: R = 1 m. Horizontal distance moved by the center in half a rev = πR. Vertical distance moved by the point = 2R.
Step 3 - Application: The initial position is (0, 0). The final position of the point is (πR, 2R). The displacement is the hypotenuse of the right triangle formed by these horizontal and vertical shifts:
Displacement = √((πR)2 + (2R)2) = √(π2(1)2 + 4(1)2) = √(π2 + 4).
Conclusion: Option (3) is correct.
Average Speed vs. Average Velocity
- Core definition & intuition: Average speed tells you how fast you swept through your path overall. Average velocity tells you the net rate at which you changed your position.
- Mathematical formulation:
vavg speed = Total Distance / Total Time
vavg vel = Total Displacement / Total Time = Δx / Δt - Crucial detail/Trap: If a particle returns to its starting point, average velocity is ZERO, but average speed is NOT zero.
High-Yield Shortcuts:
1. If a particle covers two equal distances with speeds v1 and v2: vavg = 2v1v2 / (v1 + v2) (Harmonic mean).
2. If a particle travels for two equal time intervals with speeds v1 and v2: vavg = (v1 + v2) / 2 (Arithmetic mean). - Graphical meaning: Average velocity is the slope of the secant line connecting two points on a position-time (x-t) graph.
PYQ Problem 3:
A car covers the first half of the distance between two places at 40 km/h and another half at 60 km/h. The average speed of the car is: AIPMT 1990
(1) 40 km/h (2) 48 km/h (3) 50 km/h (4) 60 km/h
Solution:
Step 1 - Visualizing the Physics: The journey is split into two equal spatial halves, not time halves. The car spends more time driving at the slower speed, so the average will be weighted closer to 40 than 60.
Step 2 - Identify Variables: v1 = 40 km/h, v2 = 60 km/h. Distances d1 = d2.
Step 3 - Application: Since distances are equal, apply the harmonic mean shortcut: vavg = 2v1v2 / (v1 + v2) = 2(40)(60) / (40 + 60) = 4800 / 100 = 48 km/h.
Conclusion: Option (2) is correct.
PYQ Problem 4:
A car moves from X to Y with a uniform speed vu and returns to Y with a uniform speed vd. The average speed for this round trip is: AIPMT 2007
(1) √(vuvd) (2) vdvu / (vd + vu) (3) (vu + vd) / 2 (4) 2vdvu / (vd + vu)
Solution:
Step 1 - Visualizing the Physics: The car travels from X to Y and back. The distance covered going out is exactly equal to the distance covered coming back.
Step 2 - Identify Variables: Outward speed = vu, Return speed = vd. Outward distance = Return distance = d.
Step 3 - Application: Total distance = 2d. Total time = t1 + t2 = d/vu + d/vd.
vavg = 2d / (d/vu + d/vd) = 2 / (1/vu + 1/vd) = 2vdvu / (vd + vu).
Conclusion: This confirms the harmonic mean derivation. Option (4) is correct.
PYQ Problem 5:
A vehicle travels half the distance L with speed V1 and the other half with speed V2, then its average speed is: NEET 2023
(1) (V1 + V2) / 2 (2) 2V1V2 / (V1 + V2) (3) (2V1 + V2) / (V1 + V2) (4) V1V2 / (V1 + V2)
Solution:
Step 1 - Visualizing the Physics: Exact same logic as the previous problems. The spatial distance is split into equal halves.
Step 2 - Identify Variables: d1 = d2 = L/2. Speeds are V1 and V2.
Step 3 - Application: Direct application of the harmonic mean formula for equal distances: vavg = 2V1V2 / (V1 + V2).
Conclusion: Option (2) is correct.
Calculus in Kinematics
Instantaneous Rates of Change
- Core definition & intuition: "Instantaneous" means right now. The speedometer of a car shows instantaneous speed. Calculus connects position (x), velocity (v), and acceleration (a) by looking at changes over infinitesimally small time intervals.
- Mathematical formulation:
Velocity: v = dx/dt (Differentiation of position w.r.t time)
Acceleration: a = dv/dt = d2x/dt2
Critical Chain Rule Form: a = v(dv/dx) (Use this specifically when velocity is given as a function of position x, not time t). - Crucial detail/Trap: To go "down" the kinematics ladder (x → v → a), you Differentiate. To go "up" the ladder (a → v → x), you Integrate. When integrating, NEVER forget the limits of integration or the constant of integration (which corresponds to initial velocity or initial position).
- Graphical meaning: Instantaneous velocity is the slope of the tangent line on an x-t graph. Instantaneous acceleration is the slope of the tangent line on a v-t graph.
PYQ Problem 6:
The position x of a particle varies with time t as x = at2 - bt3. The acceleration will be zero at time t equal to: AIPMT 1997
(1) a / 3b (2) zero (3) 2a / 3b (4) a / b
Solution:
Step 1 - Visualizing the Physics: The particle moves according to a cubic time function. Its acceleration is changing continuously. We must locate the exact moment the acceleration crosses the zero mark.
Step 2 - Identify Variables: x(t) = at2 - bt3. Target condition: a(t) = 0.
Step 3 - Application: Go down the ladder by differentiating:
v = dx/dt = 2at - 3bt2.
a = dv/dt = 2a - 6bt.
Set a = 0: 2a - 6bt = 0 ⟹ 6bt = 2a ⟹ t = 2a / 6b = a / 3b.
Conclusion: Option (1) is correct.
PYQ Problem 7:
Two cars P and Q start from a point at the same time in a straight line and their positions are represented by xP(t) = at + bt2 and xQ(t) = ft - t2. At what time do the cars have the same velocity? AIPMT 2016
(1) (a - f) / (1 + b) (2) (f - a) / 2(1 + b) (3) (a + f) / 2(b - 1) (4) (a + f) / 2(1 + b)
Solution:
Step 1 - Visualizing the Physics: Both cars are moving along the same axis but with different acceleration profiles. We need to find the specific instant where their speedometers read the exact same value.
Step 2 - Identify Variables: xP = at + bt2, xQ = ft - t2. Target condition: vP = vQ.
Step 3 - Application: Differentiate position to get velocity for both cars:
vP = dxP/dt = a + 2bt
vQ = dxQ/dt = f - 2t
Equate the velocities: a + 2bt = f - 2t.
Group the t terms: 2bt + 2t = f - a ⟹ 2t(b + 1) = f - a ⟹ t = (f - a) / 2(b + 1).
Conclusion: Option (2) is correct.
PYQ Problem 8:
A particle moves along a straight line OX. At a time t (in seconds) the distance x (in metres) of the particle from O is given by x = 40 + 12t - t3. How long would the particle travel before coming to rest? AIPMT 2006
(1) 16 m (2) 24 m (3) 40 m (4) 56 m
Solution:
Step 1 - Visualizing the Physics: The particle starts at x = 40, moves forward (since velocity starts positive), but is slowing down because of the -t3 term. We need to find the distance traveled from t = 0 until the moment it stops.
Step 2 - Identify Variables: x(t) = 40 + 12t - t3. Target condition for stopping: v(t) = 0. Initial position x(0) = 40 m.
Step 3 - Application: Find velocity: v = dx/dt = 12 - 3t2.
Set v = 0 to find stopping time: 12 - 3t2 = 0 ⟹ 3t2 = 12 ⟹ t2 = 4 ⟹ t = 2 s.
Find position at t = 2 s: x(2) = 40 + 12(2) - (2)3 = 40 + 24 - 8 = 56 m.
Trap Alert: 56 m is the position coordinate, not the distance traveled!
Distance traveled = Final Position - Initial Position = 56 - 40 = 16 m.
Conclusion: Option (1) is correct.
Uniformly Accelerated Motion
The Kinematic Equations
- Core definition & intuition: When a particle's velocity changes at a perfectly constant rate (e.g., a car gaining 2 m/s every single second), its motion is predictable using algebraic equations instead of full calculus.
- Mathematical formulation:
v = u + at (Use when missing displacement s)
s = ut + ½at2 (Use when missing final velocity v)
v2 = u2 + 2as (Use when missing time t)
Sn = u + (a/2)(2n - 1) (Displacement strictly in the nth single second). - Crucial detail/Trap: These equations are ONLY VALID IF ACCELERATION IS CONSTANT. If acceleration changes with time or position (e.g., a = 2t), you CANNOT use them; you must revert to integration. Strictly assign a coordinate reference frame (e.g., right = +, left = -) and plug in u, v, a, s with their proper signs.
- Graphical meaning: Constant acceleration creates a horizontal line on an a-t graph, a straight sloped line on a v-t graph, and a parabolic curve on an x-t graph.
PYQ Problem 9:
A car moving with a speed of 40 km/h can be stopped by applying brakes after at least 2 m. If the same car is moving with a speed of 80 km/h, what is the minimum stopping distance? AIPMT 1998
(1) 4 m (2) 6 m (3) 8 m (4) 2 m
Solution:
Step 1 - Visualizing the Physics: Brakes provide a constant negative acceleration (retardation). If you double your speed, your kinetic energy quadruples, which means the brakes have to do four times the work to stop you, requiring four times the distance.
Step 2 - Identify Variables: Case 1: u1 = 40 km/h, s1 = 2 m, v1 = 0. Case 2: u2 = 80 km/h, s2 = ?, v2 = 0. Acceleration a is constant and the same for both.
Step 3 - Application: Time is not given or asked for, so use v2 = u2 + 2as.
Since v = 0, 0 = u2 + 2as ⟹ s = -u2 / 2a.
Because 2a is constant, s ∝ u2.
s2 / s1 = (u2 / u1)2 = (80 / 40)2 = 22 = 4.
s2 = 4 × s1 = 4 × 2 = 8 m.
Conclusion: Option (3) is correct.
PYQ Problem 10:
A small block slides down on a smooth inclined plane, starting from rest at time t = 0. Let Sn be the distance travelled by the block in the interval t = n - 1 to t = n. Then, the ratio Sn / Sn+1 is: NEET 2021
(1) (2n - 1) / 2n (2) (2n - 1) / (2n + 1) (3) (2n + 1) / (2n - 1) (4) 2n / (2n - 1)
Solution:
Step 1 - Visualizing the Physics: A block sliding down a smooth incline experiences constant acceleration (g sin θ). The question asks for the ratio of distances covered in successive individual seconds (nth second vs (n+1)th second).
Step 2 - Identify Variables: u = 0, a = constant. We need the ratio of Sn to Sn+1.
Step 3 - Application: Use the nth second formula: Sn = u + (a/2)(2n - 1).
Since u = 0, Sn = (a/2)(2n - 1).
For the (n+1)th second, substitute (n+1) into the formula:
Sn+1 = (a/2)(2(n+1) - 1) = (a/2)(2n + 2 - 1) = (a/2)(2n + 1).
Take the ratio: Sn / Sn+1 = [(a/2)(2n - 1)] / [(a/2)(2n + 1)] = (2n - 1) / (2n + 1).
Conclusion: Option (2) is correct.
PYQ Problem 11:
A particle moves in a straight line with a constant acceleration. It changes its velocity from 10 m/s to 20 m/s while passing through a distance 135 m in t seconds. The value of t is: AIPMT 2008
(1) 12 (2) 9 (3) 10 (4) 1.8
Solution:
Step 1 - Visualizing the Physics: The particle is speeding up uniformly over a specific distance.
Step 2 - Identify Variables: u = 10 m/s, v = 20 m/s, s = 135 m. We need t.
Step 3 - Application:
Method 1 (Two steps): Find a using v2 = u2 + 2as ⟹ 400 = 100 + 2a(135) ⟹ 300 = 270a ⟹ a = 10/9 m/s2.
Then use v = u + at ⟹ 20 = 10 + (10/9)t ⟹ 10 = (10/9)t ⟹ t = 9 s.
Method 2 (Shortcut): For constant acceleration, Average Velocity = (u + v) / 2 = (10 + 20) / 2 = 15 m/s.
Displacement = Average Velocity × time ⟹ 135 = 15 × t ⟹ t = 135 / 15 = 9 s.
Conclusion: Method 2 is vastly superior for exam speed. Option (2) is correct.
Motion Under Gravity
Vertical Free Fall
- Core definition & intuition: Free fall is the classic, naturally occurring case of uniformly accelerated motion where the acceleration is provided by Earth's gravity.
- Mathematical formulation: Replace a with g in the kinematic equations.
Strict Sign Convention: Place the origin (0, 0) at the launch point. Upward vector = Positive (+), Downward vector = Negative (-).
Therefore, a = -g at all times, whether the ball is going up, at the peak, or falling down. - Crucial detail/Trap:
When a ball is thrown upwards, at the highest point, its velocity v = 0, but its acceleration is STILL -g. Acceleration never pauses.
Galileo's Law of Odd Numbers: The distances traversed by a freely falling body dropped from rest, during equal intervals of time, stand to one another in the same ratio as the odd numbers: 1 : 3 : 5 : 7...
Time of ascent = Time of descent = u / g (ignoring air resistance). Speed of projection equals speed of return to the same level. - Graphical meaning: The a-t graph is a constant horizontal line below the time axis (at -9.8). The v-t graph is a straight line with a negative slope, crossing from positive to negative at the peak of the throw.
PYQ Problem 12:
A ball is thrown vertically downward with a velocity of 20 m/s from the top of a tower. It hits the ground after some time with a velocity of 80 m/s. The height of the tower is: (g = 10 m/s2) NEET 2020
(1) 340 m (2) 320 m (3) 300 m (4) 360 m
Solution:
Step 1 - Visualizing the Physics: Draw the tower. Set the release point as the origin (y = 0). Up is +, down is -. The ball is thrown down, so initial velocity is negative. It accelerates down due to gravity. The final displacement will be a negative value, representing the height of the tower below the origin.
Step 2 - Identify Variables: u = -20 m/s (thrown downward), v = -80 m/s (hits ground moving downward), a = -10 m/s2. We need displacement s. Time is absent.
Step 3 - Application: Use the time-independent equation: v2 = u2 + 2as.
(-80)2 = (-20)2 + 2(-10)s
6400 = 400 - 20s
6000 = -20s ⟹ s = -300 m.
The negative sign indicates the displacement is 300 m downwards. Therefore, the physical height h = 300 m.
Conclusion: Option (3) is correct.
PYQ Problem 13:
A stone falls freely under gravity. It covers distances h1, h2 and h3 in the first 5 s, the next 5 s and the next 5 s respectively. The relation between h1, h2 and h3 is: NEET 2013
(1) h1 = h2 / 3 = h3 / 5 (2) h2 = 3h1 and h3 = 3h2 (3) h1 = h2 = h3 (4) h1 = 2h2 = 3h3
Solution:
Step 1 - Visualizing the Physics: A body dropped from rest (u = 0). We are observing the distance it covers in equal, successive time intervals (5 seconds each).
Step 2 - Identify Variables: u = 0, time intervals Δt = 5 s. This perfectly matches the condition for Galileo's Law of Odd Numbers.
Step 3 - Application: According to Galileo's Law, the distances fallen in successive equal time intervals starting from rest are in the ratio 1 : 3 : 5.
Therefore, h1 : h2 : h3 = 1 : 3 : 5.
This implies h2 = 3h1 and h3 = 5h1.
Rearranging gives h1 = h2 / 3 = h3 / 5.
Conclusion: Option (1) is correct.
PYQ Problem 14:
A boy standing at the top of a tower of 20 m height drops a stone. Assuming g = 10 ms-2, the velocity with which it hits the ground is: AIPMT 2011
(1) 10 m/s (2) 20 m/s (3) 40 m/s (4) 5 m/s
Solution:
Step 1 - Visualizing the Physics: The word "drops" is the key physics trigger here—it means initial velocity is zero. The stone accelerates downwards under gravity until it covers the 20 m distance to the ground.
Step 2 - Identify Variables: u = 0, s = 20 m, a = 10 m/s2 (taking downward as positive since all vectors point down here). We need final velocity v.
Step 3 - Application: Time is not mentioned, use v2 = u2 + 2as.
v2 = 02 + 2(10)(20) ⟹ v2 = 400 ⟹ v = 20 m/s.
Conclusion: Option (2) is correct.
Graphical Analysis
x-t, v-t, and a-t graphs
- Core definition & intuition: Graphs tell a visual story of the motion. Slopes reveal rates of change. Areas reveal accumulated quantities.
| Graph | What the Slope represents | What the Area under the curve represents |
|---|---|---|
| x-t (Position-time) | Velocity (v = dx/dt) | Meaningless |
| v-t (Velocity-time) | Acceleration (a = dv/dt) | Displacement (Net area with signs) Distance (Total absolute area) |
| a-t (Accel-time) | Jerk (Rate of change of accel) | Change in velocity (Δv = vf - vi) |
- Crucial detail/Trap:
1. An x-t graph can NEVER be a vertical line (that implies infinite velocity/teleportation).
2. An x-t graph can NEVER have two y values for one x value (a particle can't be in two places at the same time).
3. If a v-t curve dips below the x-axis, that area represents negative displacement (moving backwards). Subtract it for net displacement, but add it for total distance.
PYQ Problem 15:
The displacement-time graph of a moving particle is shown below. The instantaneous velocity of the particle is negative at the point: AIPMT 1994
(1) D (2) F (3) C (4) E
Solution:
Step 1 - Visualizing the Physics: Instantaneous velocity on a displacement-time graph is the slope of the tangent line at that point. A negative velocity means a negative slope (the curve is heading downwards).
Step 2 - Identify Variables: Look at the geometry of the points on the curve.
Step 3 - Application:
At point C: Curve is rising, tangent points up and right → Positive slope, positive velocity.
At point D: Curve is still rising → Positive slope, positive velocity.
At point E: Curve is at its maximum peak, tangent is perfectly horizontal → Zero slope, zero velocity (particle is turning around).
At point F: Curve is falling, tangent points down and right → Negative slope, negative velocity.
Conclusion: Option (2) is correct.
PYQ Problem 16:
A particle shows distance-time curve as given in this figure. The maximum instantaneous velocity of the particle is around the point: AIPMT 2008
(1) D (2) A (3) B (4) C
Solution:
Step 1 - Visualizing the Physics: Similar to the previous problem, velocity is the slope of the distance-time graph. "Maximum instantaneous velocity" means finding the point where the curve is the steepest.
Step 2 - Identify Variables: We are evaluating the gradient (dx/dt) at various points.
Step 3 - Application:
At points A and B: The curve is relatively flat; slope is low.
At point C: The curve is the steepest; the tangent line approaches its most vertical orientation compared to the rest of the graph.
At point D: The curve flattens out entirely (slope approaches zero).
Therefore, the tangent line has the highest positive gradient at point C.
Conclusion: Maximum slope = maximum velocity. Option (4) is correct.