Motion in a Straight Line — NEET notes

Position, Path Length & Displacement

Every kinematics problem starts with defining WHERE the object is. Position needs a reference point (origin), path length is the total distance travelled (always positive), and displacement is the shortest straight-line change in position (can be negative in 1D). NEET loves the distance vs displacement distinction.

Displacement = x_f - x_i | Distance |Displacement|

Position & Reference Point

Position is always measured relative to a chosen origin on a coordinate axis. Different origins give different position values for the same object.

To describe motion in a straight line, we lay down an x-axis along the line of motion and pick an origin O. The position of an object is its coordinate x on this axis — positive on one side, negative on the other.

• Position is a signed quantity — it can be positive or negative depending on which side of the origin the object is
• The choice of origin is arbitrary — physics doesn't change if you shift the origin. But once chosen, be consistent
• In 1D, position is fully described by a single number x (with sign)
• NCERT uses the convention: rightward = positive, leftward = negative (unless stated otherwise)

Distance vs Displacement

Distance is the total length of the path travelled (scalar, always non-negative). Displacement is the change in position from start to end (vector in general, signed in 1D).

Path length (distance) = total length of the actual route taken. It's always 0 and depends on the path. Displacement = final position minus initial position: x = x_f - x_i . It depends only on the starting and ending points — not the route.

x = x_f - x_i (displacement in 1D)

Quick recall

An object moves from x = 2 m to x = 8 m, then back to x = 5 m. Find distance and displacement.

Distance = |8 - 2| + |5 - 8| = 6 + 3 = 9 m. Displacement = x_f - x_i = 5 - 2 = +3 m (positive direction).

Sign Convention in 1D

In 1D kinematics, every quantity (position, velocity, acceleration) is signed. The sign tells you direction. Getting the sign wrong is the #1 source of errors in kinematics.

In straight-line motion, direction is encoded in the sign. Once you choose a positive direction (say, rightward), then: a positive velocity means moving right, a negative velocity means moving left. A positive acceleration means speeding up (if moving right) or slowing down (if moving left).

• Pick one direction as positive before solving — don't change mid-problem
• For vertical motion: upward = positive is the most common convention. Then a = -g = -9.8 m/s ^2
• Displacement can be negative — it just means the object ended up on the negative side of the starting point
• If velocity and acceleration have the same sign → object is speeding up. Opposite signs → slowing down

Speed & Velocity

Speed tells you how fast, velocity tells you how fast AND in which direction. Average speed = total distance / total time. Average velocity = displacement / time. Instantaneous velocity is the limit as the time interval shrinks to zero — it's the slope of the x-t graph. NEET tests the numerical difference between average speed and average velocity regularly.

v_ avg = x t | speed_ avg = total distance t

Average Speed & Average Velocity

Average speed uses total distance; average velocity uses displacement. They can give very different answers for the same journey.

Average speed is the total path length divided by total time. It is a scalar — always positive. Average velocity is displacement divided by time. It is a vector (signed in 1D) — can be positive, negative, or zero.

Average speed = total distance t v = x t = x_f - x_i t_f - t_i

• For a round trip (return to start): average velocity = 0, but average speed 0
• Average speed = |Average velocity| ONLY for straight-line, one-directional motion
• SI unit for both: m/s. CGS: cm/s. Common: km/h ( 1 m/s = 3.6 km/h)

Quick recall

A car travels 100 km at 60 km/h, then 100 km at 40 km/h. Find average speed for the entire journey.

Time _1 = 100/60 = 5/3 h, Time _2 = 100/40 = 5/2 h. Total time = 5/3 + 5/2 = 25/6 h. Average speed = 200 ÷ (25/6) = 48 km/h. Note: it's NOT 50 km/h (the simple average). For equal distances, use the harmonic mean: 2 60 40 60 + 40 = 48 km/h.

Instantaneous Velocity

Instantaneous velocity is the velocity at a single instant — defined as the limit of average velocity as the time interval approaches zero. Graphically, it's the slope of the x-t curve.

As you shrink the time interval t to zero, the average velocity approaches a definite value — the instantaneous velocity. Mathematically, it's the derivative of position with respect to time.

v = _ t 0 x t = dx dt

• Instantaneous velocity = slope of the x - t (position-time) graph at that point
• Instantaneous speed = |instantaneous velocity| = magnitude without the sign
• For uniform motion (constant velocity), instantaneous velocity = average velocity at all times
• The sign of instantaneous velocity tells the direction of motion at that instant

Acceleration

Acceleration is the rate of change of velocity — how quickly the velocity is changing. Like velocity, it has a sign in 1D. Positive acceleration doesn't always mean 'speeding up' — it depends on the direction of motion. Uniform acceleration (constant a ) is the foundation for all kinematic equations.

a = dv dt = d^2x dt^2

Average & Instantaneous Acceleration

Average acceleration = change in velocity / time taken. Instantaneous acceleration = derivative of velocity = slope of v-t graph.

Average acceleration over a time interval: a = v t = v_f - v_i t_f - t_i . Instantaneous acceleration at an instant: a = dv dt . Just as velocity is the derivative of position, acceleration is the derivative of velocity (and the second derivative of position).

a = v_f - v_i t a = dv dt = d^2 x dt^2

• SI unit of acceleration: m/s ^2 (metres per second per second)
• Acceleration is a vector — in 1D, it has a sign (+ or -)
• Instantaneous acceleration = slope of v - t graph at that point
• For uniform acceleration, a is constant → instantaneous = average at all times
• Acceleration CAN be zero even if velocity is non-zero (constant velocity motion)

Speeding Up vs Slowing Down

An object speeds up when velocity and acceleration point the same way (same sign). It slows down when they point opposite ways (opposite signs). 'Retardation' = deceleration = slowing down.

Speeding up: v and a have the same sign (both positive or both negative). The speed (magnitude of velocity) increases. Slowing down (retardation/deceleration): v and a have opposite signs. The speed decreases.

Quick recall

A ball is thrown upward with v = +25 m/s. Acceleration = -9.8 m/s ^2 throughout. Is it speeding up or slowing down (a) while going up, (b) while coming down?

(a) Going up: v > 0 , a < 0 → opposite signs → slowing down. (b) Coming down: v < 0 , a < 0 → same signs → speeding up.

Kinematic Equations for Uniform Acceleration

The three (plus one bonus) kinematic equations are the workhorse of NEET mechanics. They ONLY work when acceleration is constant. Every free-fall problem, every braking problem, every projectile component uses these equations. Know when to use which — that's the real skill.

v = u + at | s = ut + 1 2 at^2 | v^2 = u^2 + 2as

The Big Three

Three equations, five variables ( u, v, a, s, t ). Each equation connects four of them — so if you know any three, you can find the rest. The trick is picking the right equation.

These equations are valid ONLY when acceleration a is constant (uniform acceleration). The five variables are: u (initial velocity), v (final velocity), a (acceleration), s (displacement), t (time).

v = u + at ...(1)

s = ut + 1 2 at^2 ...(2)

v^2 = u^2 + 2as ...(3)

• All three equations assume constant acceleration — do NOT use them if a is changing
• s in these equations is displacement, NOT distance. If the object reverses direction, s can be negative
• All quantities are signed — use sign convention consistently
• Equation 3 ( v^2 = u^2 + 2as ) is the most useful when time is not given or needed

Displacement in the n th Second

The displacement in the nth second (not in n seconds) has its own formula. It's a linear function of n, which makes it useful for ratio-based problems.

The displacement in the n -th second means the displacement from t = (n-1) to t = n . It's NOT the total displacement in n seconds — it's the displacement during that one specific second.

s_n = u + a 2 (2n - 1)

• s_n = displacement in the n -th second (a one-second interval), NOT in n seconds
• For u = 0 : s_n = a 2 (2n-1) , giving the famous 1:3:5:7:... ratio
• The ratio holds regardless of the value of a — the odd-number pattern is universal for uniformly accelerated motion from rest
• This equation is derived by subtracting: s_n = s(n) - s(n-1) where s(t) = ut + 1 2 at^2

Quick recall

A body starting from rest has uniform acceleration. If it covers 5 m in the 3rd second, what is the acceleration?

s_3 = u + a 2 (2 3 - 1) = 0 + a 2 (5) = 5a 2 . Given s_3 = 5 m: 5a 2 = 5 → a = 2 m/s ^2 .

Quick recall

A body starting from rest covers distances d_1, d_2, d_3 in the 1st, 2nd, and 3rd seconds. Find d_1 : d_2 : d_3 .

d_1 : d_2 : d_3 = 1 : 3 : 5 (odd number ratio for uniform acceleration from rest). This extends: d_4 = 7, d_5 = 9 , etc.

Motion Graphs (x-t, v-t, a-t)

Graphs are the visual language of kinematics. The slope of an x-t graph gives velocity. The slope of a v-t graph gives acceleration. The area under a v-t graph gives displacement. NEET asks 3-5 graph-based questions every year — learn to READ graphs, not just draw them.

Slope of x-t = v | Slope of v-t = a | Area under v-t = s

Position-Time Graphs

The x-t graph shows how position changes with time. A straight line means constant velocity. A curve means changing velocity (acceleration). The slope at any point gives the instantaneous velocity.

In a position-time ( x - t ) graph, the slope at any point equals the instantaneous velocity at that time. A positive slope → moving in the positive direction. A negative slope → moving in the negative direction. Zero slope → momentarily at rest.

• Slope of x - t graph = dx dt = instantaneous velocity
• Steeper slope → greater speed. Flat line → at rest
• A curve that gets steeper → object is accelerating. A curve that flattens → object is decelerating
• The x - t graph CANNOT have a vertical segment — that would mean the object is at two places at the same time (impossible)

Velocity-Time Graphs

The v-t graph is the MOST information-rich graph in kinematics. Its slope gives acceleration, its area gives displacement, and the sign of v tells the direction of motion.

In a velocity-time ( v - t ) graph: slope = instantaneous acceleration, area under the curve = displacement. If the v - t graph goes below the time axis ( v < 0 ), the area below the axis is negative displacement (motion in the negative direction).

• Area above t -axis → positive displacement. Area below t -axis → negative displacement
• Displacement = algebraic sum of areas (with signs). Distance = sum of absolute areas
• When the v - t graph touches the t -axis, v = 0 → object is momentarily at rest (may be a turning point)
• For uniform acceleration: v - t graph is a straight line → area is a triangle or trapezium

Quick recall

A v-t graph is a straight line from (0, 10) to (5, 0) then from (5, 0) to (8, -6) . Find displacement and distance from t = 0 to t = 8 s.

Area 1 (triangle above axis): 1 2 5 10 = 25 m. Area 2 (triangle below axis): 1 2 3 6 = 9 m (negative). Displacement = 25 + (-9) = 16 m. Distance = 25 + 9 = 34 m.

Converting Between Graphs

Given any one graph, you can construct the others. From x-t: differentiate (slope) to get v-t. From v-t: differentiate to get a-t, integrate (area) to get x-t. Master these conversions.

The three motion graphs are connected by calculus: x slope v slope a (differentiation) and a area v area x (integration). NEET often gives one graph and asks you to identify the corresponding other graph.

• x - t → v - t : find the slope of x - t at each point → that's the value of v at that time
• v - t → a - t : find the slope of v - t at each point → that's the value of a at that time
• v - t → x - t : find the area under v - t up to each time → that gives x (add to initial position)
• a - t → v - t : find the area under a - t up to each time → that gives v (add to initial velocity)

Free Fall & Motion Under Gravity

Free fall is motion under gravity alone — no air resistance. Near Earth's surface, g 9.8 m/s ^2 (downward, constant). All kinematic equations apply with a = g . NEET tests time of flight, maximum height, and the symmetry of upward-downward motion extensively.

a = -g = -9.8 m/s^2 (taking upward as positive)

Free Fall Basics

In free fall, the ONLY force acting is gravity. The acceleration is g = 9.8 m/s ^2 downward, regardless of the object's mass, shape, or size. A feather and a hammer hit the ground simultaneously in vacuum.

Free fall = motion under gravity alone, with no air resistance, no push, no pull — just gravity. Galileo proved (and Apollo 15 astronauts demonstrated on the Moon) that all objects in free fall have the same acceleration g , regardless of mass.

• g = 9.8 m/s ^2 (often rounded to 10 m/s ^2 in NEET for quick calculation)
• g is the same for all objects — a 1 kg ball and a 100 kg ball fall equally fast in vacuum
• In real life, air resistance makes lighter/larger objects fall slower — but NEET problems usually say 'neglect air resistance'
• The direction of g is always downward — toward the center of the Earth

Kinematic Equations for Free Fall

All kinematic equations work for free fall. The key is the sign convention: if upward is positive, then a = -g . If downward is positive, then a = +g .

Taking upward as positive (NCERT convention): a = -g = -9.8 m/s ^2 . The kinematic equations become:

v = u - gt

h = ut - 1 2 gt^2

v^2 = u^2 - 2gh

• For a body dropped from rest: u = 0 , so v = gt , h = 1 2 gt^2 , v^2 = 2gh (taking downward as positive)
• For a body thrown upward with speed u : it rises until v = 0 , then falls back
• Maximum height: Set v = 0 in v^2 = u^2 - 2gh → h_ = u^2 2g
• Time to reach max height: Set v = 0 in v = u - gt → t_ up = u g
• Total time of flight (return to same level): T = 2u g (by symmetry, t_ up = t_ down )

Quick recall

A ball is thrown upward with u = 40 m/s. Take g = 10 m/s ^2 . Find (a) max height, (b) time of flight, (c) speed on return to the ground.

(a) h_ = u^2 2g = 1600 20 = 80 m. (b) T = 2u g = 80 10 = 8 s. (c) Speed on return = u = 40 m/s (by symmetry — same speed, opposite direction).

Relative Velocity in 1D

Velocity is always measured relative to something — usually the ground. When you need the velocity of A as seen by B, use v_ AB = v_A - v_B . This simple subtraction solves train-passing problems, rain-man problems, and 'two objects approaching each other' problems. NEET tests this concept every year.

v_ AB = v_A - v_B

Relative Velocity

The velocity of A relative to B is what an observer sitting on B would measure for A's motion. In 1D, it's a simple subtraction with signs.

If object A moves with velocity v_A and object B moves with velocity v_B (both measured w.r.t. the ground), then the velocity of A relative to B (what B 'sees') is:

v_ AB = v_A - v_B

• v_ AB = v_A - v_B → velocity of A as seen by B
• v_ BA = v_B - v_A = -v_ AB → velocity of B as seen by A (equal magnitude, opposite direction)
• Same direction: relative velocity = difference of speeds → objects approach/separate slowly
• Opposite directions: relative velocity = sum of speeds → objects approach/separate quickly

Rain-Man & River Problems (1D Concepts)

The rain-umbrella problem and river-boat problem are standard relative velocity applications. The 1D version involves finding how rain appears to a moving person.

When rain falls vertically and you walk forward, the rain appears to come at an angle — hitting your face instead of just your head. This is because the rain's velocity relative to you has both a vertical component (rain's actual velocity) and a horizontal component (negative of your velocity).

• Rain velocity relative to man: v _ rain, man = v _ rain - v _ man
• If rain is vertical and man moves horizontally, the relative velocity vector is tilted — the angle of tilt increases with man's speed
• For 1D (same line): if two cars approach each other at 60 km/h and 40 km/h, relative approach speed = 100 km/h
• The concept extends to 2D (river-boat problems), but NEET often keeps it simple with 1D or basic angle calculations

Quick recall

Two trains of length 100 m each approach from opposite directions at 36 km/h and 54 km/h. How long do they take to completely pass each other?

Relative speed = 36 + 54 = 90 km/h = 25 m/s. Total distance to cover = 100 + 100 = 200 m. Time = 200/25 = 8 seconds.

Quick recall

A 200 m long train overtakes a 100 m long train moving in the same direction at 30 km/h. The faster train moves at 50 km/h. Find the time to completely overtake.

Relative speed = 50 - 30 = 20 km/h = 50 9 m/s. Total distance = 200 + 100 = 300 m. Time = 300 50 9 = 300 9 50 = 54 seconds.

Stopping Distance & Reaction Time

When you see a hazard and hit the brakes, the car doesn't stop instantly. There's a reaction time delay (during which the car keeps moving at full speed), then a braking phase (during which the car decelerates). Total stopping distance = reaction distance + braking distance. This is a real-life application of kinematics that NEET loves.

d_ braking = v^2 2a | d_ total = vt_r + v^2 2a

Reaction Time

Reaction time is the time between seeing a danger and actually pressing the brake. During this time, the car continues at its original speed. Typical human reaction time: 0.2 to 0.5 seconds.

When a driver sees an obstacle, the brain needs time to process the information, decide to brake, and send signals to the foot. This delay is the reaction time t_r . During t_r , the car moves at its original speed v without any deceleration.

d_ reaction = v t_r

• Typical reaction time: 0.2 s (alert driver) to 1+ s (drowsy/distracted driver)
• Reaction distance v — doubles if speed doubles
• NCERT mentions a simple experiment: have someone drop a ruler and catch it — the distance fallen gives your reaction time via t_r = 2d/g
• Alcohol, fatigue, and phone use increase reaction time significantly

Braking Distance

After brakes are applied, the car decelerates at rate a . The braking distance is proportional to v^2 — double the speed, quadruple the braking distance. This is why speed limits exist.

Once the brakes engage, the car decelerates uniformly from speed v to 0 . Using v^2 = u^2 + 2as with final velocity = 0:

0 = v^2 - 2a d_ brake d_ brake = v^2 2a

d_ total = v t_r _ reaction distance + v^2 2a _ braking distance

• Braking distance v^2 — the most dangerous term at high speeds
• Reaction distance v — significant but grows slower than braking distance
• Total stopping distance = reaction distance + braking distance
• Wet/icy roads reduce a (lower friction) → braking distance increases dramatically
• Braking distance depends on a , which depends on friction: a = g for flat ground

Quick recall

A car travelling at 20 m/s has a reaction time of 0.5 s and braking deceleration of 5 m/s ^2 . Find the total stopping distance.

Reaction distance = 20 0.5 = 10 m. Braking distance = 20^2 2 5 = 400 10 = 40 m. Total = 10 + 40 = 50 m.

Quick recall

If the car in the previous problem was going at 40 m/s instead, what is the new total stopping distance?

Reaction distance = 40 0.5 = 20 m. Braking distance = 40^2 10 = 160 m. Total = 20 + 160 = 180 m. Speed doubled → braking distance quadrupled (40 → 160), total distance more than tripled (50 → 180).