Motion in a Straight Line

When an object moves, we can describe how far it has travelled in two distinct ways. Distance is the total path length covered — it is a scalar quantity, always positive, and it accumulates regardless of direction. Displacement is the straight-line vector from the initial position to the final position — it has both magnitude and direction, and it can be zero, positive, or negative.

A classic example: if you walk 4 m east and then 4 m west, your distance is 8 m but your displacement is 0 m. The two quantities are only equal when the object moves in a single straight line without reversing direction.

This inequality always holds. Equality holds only when the path is straight and the object never reverses. This is one of the most tested relationships in NEET.

Speed is the rate of change of distance — a scalar. Velocity is the rate of change of displacement — a vector. Both have SI units of m/s, but they carry very different physical information.

Average speed is never negative. Average velocity can be zero (circular trip returning to start), negative (net motion in the negative direction), or positive.

Instantaneous speed is the magnitude of the instantaneous velocity vector. For one-dimensional motion: $\text{speed} = |v|$.

When are speed and velocity equal? When the object moves in a single direction without reversing — so that total distance equals magnitude of displacement. In that case, average speed = |average velocity|.

Acceleration is the rate of change of velocity with respect to time. Like velocity, it is a vector quantity — it has both magnitude and direction.

Uniform acceleration: When acceleration is constant — both magnitude and direction unchanged. The equations of motion (next section) apply strictly here.

Non-uniform acceleration: When acceleration changes with time. The equations of motion no longer apply; we must use calculus ($v = \int a\, dt$, $s = \int v\, dt$).

Deceleration (Retardation): When the acceleration vector is opposite to the velocity vector, the object slows down. This is called retardation. Retardation is not a separately defined quantity — it is simply a negative acceleration (when the chosen positive direction is the direction of motion).

Three kinematic equations relate the five quantities: initial velocity $u$, final velocity $v$, acceleration $a$, time $t$, and displacement $s$. They are valid only when acceleration is constant (uniform).

Equation 1 — velocity–time relation (derived from definition of acceleration):

Equation 2 — displacement–time relation (using average velocity = (u+v)/2 for uniform acceleration):

Equation 3 — velocity–displacement relation (eliminate $t$ from equations 1 and 2):

Displacement in the $n$-th second (distance covered specifically in the $n$-th second, not in $n$ seconds):

This formula gives the displacement in exactly the $n$-th second. Note the unit: it is displacement, so it can be negative. Also, $n$ must be a positive integer ≥ 1.

Any object moving only under Earth's gravitational pull (ignoring air resistance) is in free fall. The acceleration is $g = 9.8 \approx 10$ m/s² directed vertically downward.

Sign convention (standard for NEET): Take upward as positive (+ve), downward as negative (−ve). Then $a = -g$.

Object thrown straight upward with speed $u$:

Object dropped from rest from height $h$:

Key insight at the highest point: At maximum height, velocity $v = 0$ but acceleration $= g$ downward. The object is momentarily at rest but is not in equilibrium — it is still accelerating downward.

A displacement–time (s–t) graph plots displacement on the y-axis against time on the x-axis. The slope of the s–t graph at any point equals the instantaneous velocity at that moment.

Slope of tangent at any point on the curve = instantaneous velocity at that instant.

Slope of secant (chord) between two points = average velocity over that interval.

Area under the s–t graph has no standard physical meaning — do not confuse it with the area under the v–t graph.

A velocity–time (v–t) graph is the most information-rich kinematic graph. Two things to remember always:

• Slope of v–t graph = acceleration
• Area under v–t graph = displacement

SVG diagram — common v–t graph shapes:

Area above the time-axis represents positive displacement (motion in +ve direction). Area below the time-axis represents negative displacement. The net area gives net displacement; total (unsigned) area gives total distance.

An acceleration–time (a–t) graph plots acceleration on the y-axis against time. Its key property:

This is simply the integral form of $a = dv/dt$, so $\int a\, dt = \Delta v$.

For variable acceleration, the a–t graph approach is the cleanest method. Find the area under the a–t graph to get the change in velocity over any interval, then integrate the resulting v–t relation to get displacement.

Chain of integration:

When two objects move along the same line, their motion relative to each other is described by relative velocity.

Note: $v_{AB} = -v_{BA}$.

Two objects moving toward each other (approaching): if A moves at $+v_A$ and B moves at $-v_B$ (both magnitudes positive), the rate of decrease of separation = $v_A + v_B$.

Two objects moving in the same direction: rate of change of separation = $|v_A - v_B|$.

Time to meet (or close a gap $d$):

Typical NEET problem — two trains: Train A (length $L_A$, speed $v_A$) and Train B (length $L_B$, speed $v_B$) moving in the same direction. Time for A to completely pass B:

If they move in opposite directions, the relative speed is $v_A + v_B$.

This chapter produces a large number of NEET questions, and certain traps appear year after year. Knowing them is as important as knowing the theory.

Trap 1 — Sign convention for $g$

If you take upward as positive, then $a = -g = -10$ m/s². Forgetting the negative sign is the single most common error in this chapter. Always write the sign convention at the top of your solution.

Trap 2 — "At the highest point, speed = 0 but acceleration ≠ 0"

At the topmost point of a vertically thrown ball, $v = 0$ but $a = g$ downward. The ball is not in equilibrium. A question may ask: "What is the acceleration when velocity is zero?" — the answer is $g$, not zero.

Trap 3 — Uniform retardation: stopping distance and time

If speed is doubled, stopping distance becomes four times (since $s \propto u^2$). NEET often asks for this ratio.

Trap 4 — Average velocity ≠ (u+v)/2 for non-uniform acceleration

The formula $v_{\text{avg}} = (u+v)/2$ is derived from the assumption of constant acceleration. For variable acceleration, you must use $v_{\text{avg}} = \Delta s / \Delta t$.

Trap 5 — $n$-th second formula gives displacement, not distance

$s_n = u + a(n - \frac{1}{2})$ gives displacement in the $n$-th second, which can be negative (if the object is decelerating and reverses direction before the $n$-th second ends). Never equate it to distance without checking the sign.

Trap 6 — Free fall from a moving vehicle

An object dropped from a horizontally moving vehicle has the same horizontal velocity as the vehicle at the moment of release. In the ground frame it follows a parabola; in the vehicle's frame it falls straight down. NEET sometimes asks what the passenger sees — always specify the reference frame.