Chapter 3: Motion in a Straight Line

Kinematics is the geometry of motion — we describe how an object moves without yet asking why. Restricting motion to one dimension (a straight line) lets us focus on the key ideas — position, displacement, velocity, acceleration — using just signed numbers, before vectors enter the picture in Chapter 4. Every kinematic formula you will use throughout Class XI and XII (and the JEE syllabus) is built on the ideas of this chapter.

A subtle point right at the start: we model a moving body as a point particle. This is justified when the size of the body is much smaller than the distances involved. A cricket ball traveling 100 m is a point. A spinning fan blade is not.

Key Concepts

3.1 Rest and Motion are Relative

An object is at rest with respect to a frame of reference if its position in that frame does not change with time, and in motion otherwise. The same object can be at rest in one frame and in motion in another. A passenger sitting in a moving train is at rest relative to the train but in motion relative to the platform. There is no absolute rest — every description of motion is relative to a chosen frame of reference.

A frame of reference is essentially a coordinate system attached to an observer, together with a clock. In one-dimensional kinematics we pick a single axis (call it $x$), an origin $O$, and a positive direction.

3.2 Position, Path Length, Displacement

Position $x(t)$ — the signed coordinate of the particle on the chosen axis at time $t$.

Path length (also called distance traveled) — the total length of the path actually followed; a scalar, always positive, never decreases.

Displacement $\Delta x = x_2 - x_1$ — the change in position; a signed scalar in 1D (a vector in 2D / 3D). Can be positive, negative, or zero.

Important: $\lvert\text{displacement}\rvert \le \text{path length}$ in every case. Equality holds iff the particle moves in one direction without ever turning back.

3.3 Average Speed and Average Velocity

Average speed over an interval $\Delta t$: $\bar{v}_{\text{speed}} = \frac{\text{total path length}}{\text{total time}}.$

Average velocity over an interval $\Delta t$: $\bar{v} = \frac{\Delta x}{\Delta t} = \frac{x_2 - x_1}{t_2 - t_1}.$

Speed is always $\ge 0$; velocity carries a sign (direction). Average speed and average velocity have the same magnitude only when motion is in one direction without reversing.

A famous trap. A particle moves $100$ m east in $20$ s, then $100$ m west in $30$ s. Average speed $= 200/50 = 4$ m/s. Average velocity $= 0$ (net displacement is zero). Never confuse the two.

3.4 Instantaneous Velocity and Speed

The instantaneous velocity at time $t$ is the limit of the average velocity as the interval shrinks: $v(t) = \lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t} = \frac{dx}{dt}.$

Instantaneous speed = $\vert v(t)\vert$. Speedometer reads instantaneous speed; velocity additionally encodes direction.

Geometrically, on an $x$–$t$ graph:

• Average velocity over an interval $=$ slope of the chord between the two endpoints.
• Instantaneous velocity $=$ slope of the tangent at the point.

3.5 Acceleration

Average acceleration over an interval: $\bar{a} = \frac{\Delta v}{\Delta t}.$

Instantaneous acceleration: $a(t) = \frac{dv}{dt} = \frac{d^2 x}{dt^2}.$

A useful chain-rule form (eliminates time): $a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = v\frac{dv}{dx}.$

Sign conventions:

• $v > 0$ — particle moves in $+x$ direction.
• $a > 0$ — velocity is increasing algebraically.
• $a v > 0$ — speed is increasing (accelerating).
• $a v < 0$ — speed is decreasing (decelerating / retarding).
• $a = 0$ means velocity is constant — not that the body is at rest!

3.6 Uniform vs Non-uniform Motion

• Uniform motion (uniform velocity): $v$ = const, so $a = 0$. On $x$–$t$ graph: straight line. On $v$–$t$ graph: horizontal line.
• Uniformly accelerated motion: $a$ = const. On $x$–$t$ graph: parabola. On $v$–$t$ graph: straight (slanted) line. On $a$–$t$ graph: horizontal line.
• Non-uniformly accelerated motion: $a$ varies with time. Graphs become curves.

3.7 Kinematic Equations for Uniformly Accelerated Motion

If $u$ is initial velocity, $v$ final velocity, $a$ constant acceleration, $s$ displacement during time $t$, then:

$\boxed{\;v = u + at\;}$ $\boxed{\;s = ut + \tfrac{1}{2} a t^2\;}$ $\boxed{\;v^2 = u^2 + 2 a s\;}$ $\boxed{\;s = \tfrac{1}{2}(u + v)\, t \;}$ $\boxed{\;s_n = u + \tfrac{a}{2}(2n - 1)\;}$

The last gives the displacement during the $n$-th second (not "in $n$ s") — note that its dimensions are length, not length/time, even though it looks otherwise.

Calculus derivation (constant $a$). Starting from $a = dv/dt$: $\int_u^v dv' = \int_0^t a\, dt' \;\Longrightarrow\; v - u = at \;\Longrightarrow\; v = u + at.$ Then $v = dx/dt = u + at$, so $\int_0^s dx = \int_0^t (u + at)\, dt \;\Longrightarrow\; s = ut + \tfrac{1}{2} a t^2.$ For the third, use $a = v\, dv/dx$: $\int_u^v v\, dv = \int_0^s a\, dx \;\Longrightarrow\; \tfrac{1}{2}(v^2 - u^2) = a s \;\Longrightarrow\; v^2 = u^2 + 2 a s.$

Graphical derivation. On a $v$–$t$ graph for constant acceleration, the line is straight with slope $a$, $v$-intercept $u$, value $v$ at time $t$. The area under the line from $0$ to $t$ is the displacement: $s = \tfrac{1}{2}(u + v) t = u t + \tfrac{1}{2}(v - u) t = u t + \tfrac{1}{2}(at) t = u t + \tfrac{1}{2} a t^2.$ Also $v - u = at$ (slope $\times$ run) gives the first equation. Combining $v = u + at$ with $s = \tfrac{1}{2}(u+v)t$ and eliminating $t$ gives $v^2 = u^2 + 2as$.

Derivation of $s_n$. Total displacement in $n$ s: $S_n = u n + \tfrac{1}{2} a n^2$. In $(n-1)$ s: $S_{n-1} = u(n-1) + \tfrac{1}{2} a (n-1)^2$. The displacement during the $n$-th second is the difference: $s_n = S_n - S_{n-1} = u + \tfrac{a}{2}(2n - 1).$

Sign convention. When applying these equations, fix the positive direction first (usually the direction of initial motion) and assign signs to $u$, $a$, $s$, $v$ consistently. A common pitfall is plugging in $-g$ for downward acceleration but then forgetting to flip the sign of an upward initial velocity.

3.8 Motion under Gravity (1D Free Fall)

Near Earth's surface (and in vacuum), all bodies experience the same downward acceleration of magnitude $g \approx 9.8\,\text{m/s}^2$ (often taken as $10$ for quick numerics). With upward as positive:

$v = u - g t, \quad y = u t - \tfrac{1}{2} g t^2, \quad v^2 = u^2 - 2 g y.$

Useful consequences for an object thrown straight up with speed $u$ from ground:

• Time to reach the highest point: $t_{\text{up}} = u / g$.
• Maximum height: $H = u^2 / (2g)$.
• Total time of flight (up + down): $T = 2u / g$.
• Speed on returning to launch point: $u$ (symmetry, by energy conservation).
• The time to reach height $h < H$ during ascent equals the time taken during descent from that height to the launch point.

3.9 Relative Velocity in 1D

If $v_A$ is the velocity of $A$ and $v_B$ that of $B$ in a common frame, the velocity of $A$ relative to $B$ is $v_{AB} = v_A - v_B.$ Similarly $v_{BA} = v_B - v_A = -v_{AB}$. Useful for problems involving two trains, two cars, river-and-swimmer (in 1D), etc.

Approach / separation. If $v_{AB} > 0$ and $A$ is behind $B$, they are approaching; if $A$ is ahead, they are separating. The time to meet/overtake is found by dividing the initial separation by $\vert v_{AB}\vert$ (when both move at constant velocity).

3.10 Position-Time, Velocity-Time, Acceleration-Time Graphs

Sample shapes (constant $a$):

• $x$–$t$ : parabola opening up if $a > 0$.
• $v$–$t$ : straight line, slope $a$, intercept $u$.
• $a$–$t$ : horizontal line.

Rules for reading graphs:

• Steepness = magnitude of derivative.
• Slope sign = sign of derivative.
• Curvature on $x$–$t$ tells the sign of acceleration: concave up $\Leftrightarrow a > 0$, concave down $\Leftrightarrow a < 0$.
• A horizontal $x$–$t$ segment means $v = 0$ (momentary rest).
• A $v$–$t$ curve crossing zero indicates a reversal of direction.

Special: $v$–$t$ graph and displacement vs distance. Area between $v$–$t$ curve and the $t$-axis, counting sign, is displacement. Counting all area as positive (folding negative parts up) gives distance. So if a $v$–$t$ graph crosses below the time axis, displacement < distance.

3.11 Some Common Motion Scenarios

(a) Stopping distance. A vehicle decelerates uniformly from $u$ to rest. Using $v^2 = u^2 - 2 a s$ with $v = 0$: $s_{\text{stop}} = \frac{u^2}{2a}.$ Note the quadratic dependence on initial speed — doubling speed quadruples stopping distance. A road-safety classic.

(b) Reaction time and braking distance. Real-life braking distance = reaction-time distance ($u \cdot t_R$) + braking distance ($u^2 / 2a$).

(c) Two-vehicle overtaking (uniform velocities). Time to overtake = (initial separation) / (relative velocity).

(d) Dropped ball + thrown ball. Two balls released from the same point at different times under gravity have a constant difference in their speeds (both accelerate at $g$) but the gap between them grows linearly with the time since the first was released.

Worked Examples

Example 3.1 — Distance vs displacement. A car drives $30$ km north and then $40$ km east. What are the distance traveled and the magnitude of displacement?

Solution. Distance $= 30 + 40 = 70$ km. Displacement $= \sqrt{30^2 + 40^2} = 50$ km, directed at $\tan^{-1}(40/30) = 53^\circ$ east of north. (Strictly this is 2D, but the path-length-vs-displacement contrast is the point.)

Example 3.2 — Average velocity vs average speed. A jogger runs at $10$ m/s for $30$ s, then walks at $1$ m/s for the next $300$ s, all in the same direction. Find average speed and average velocity.

Solution. Both quantities coincide here because direction is unchanged. Distance $= 10 \cdot 30 + 1 \cdot 300 = 300 + 300 = 600$ m. Time $= 330$ s. Average $= 600/330 \approx 1.82$ m/s.

Example 3.3 — Equations of motion. A train starts from rest and accelerates uniformly at $2$ m/s$^2$ for $10$ s, then moves at constant velocity for $30$ s, then decelerates uniformly to rest in $20$ s. Find total distance.

Solution. Phase 1: $u = 0$, $a = 2$, $t = 10$. $v_1 = u + at = 20$ m/s. $s_1 = \tfrac{1}{2} a t^2 = 100$ m. Phase 2: constant velocity $20$ m/s for $30$ s. $s_2 = 600$ m. Phase 3: $u = 20$, $v = 0$, $t = 20$. $a = (v - u)/t = -1$ m/s$^2$. $s_3 = \tfrac{1}{2}(u + v) t = \tfrac{1}{2}(20)(20) = 200$ m. Total $= 100 + 600 + 200 = 900$ m.

Example 3.4 — Stopping distance. A car traveling at $20$ m/s applies brakes that produce a deceleration of $5$ m/s$^2$. How far does it travel before stopping? How long does it take?

Solution. $v^2 = u^2 - 2 a s \Rightarrow 0 = 400 - 10 s \Rightarrow s = 40$ m. $v = u - a t \Rightarrow 0 = 20 - 5 t \Rightarrow t = 4$ s.

Example 3.5 — Free fall. A ball is thrown vertically upward with $u = 20$ m/s from a $25$ m tall tower. Take $g = 10$ m/s$^2$, upward positive. (a) Find the maximum height above ground. (b) Find the total time of flight to reach ground. (c) Find the speed on hitting the ground.

Solution. Choose origin at the top of the tower. Ground is at $y = -25$ m. (a) At max height $v = 0$. $v^2 = u^2 - 2 g y_{\max} \Rightarrow 0 = 400 - 20 y_{\max} \Rightarrow y_{\max} = 20$ m above tower top, i.e., $45$ m above ground. (b) For ground: $y = -25 = ut - \tfrac{1}{2}gt^2 = 20t - 5t^2$. Rearranging: $5t^2 - 20t - 25 = 0 \Rightarrow t^2 - 4t - 5 = 0 \Rightarrow (t-5)(t+1) = 0$. Take positive root: $t = 5$ s. (c) $v = u - g t = 20 - 10 \cdot 5 = -30$ m/s. Magnitude $30$ m/s, directed downward. (Cross-check: $v^2 = u^2 - 2g(-25) = 400 + 500 = 900 \Rightarrow \vert v\vert = 30$ m/s. $\checkmark$)

Example 3.6 — Relative velocity. A passenger walks north along the corridor of a train at $2$ m/s while the train moves north at $20$ m/s relative to the ground. What is the passenger's velocity (a) relative to the train and (b) relative to the ground?

Solution. (a) $2$ m/s north. (b) $20 + 2 = 22$ m/s north.

If instead the passenger walked south (against the train), the velocity relative to the ground would be $20 - 2 = 18$ m/s north — the passenger still moves north overall, despite walking south within the train. This is a classic JEE puzzle.

Example 3.7 — Use $s_n$. A particle starts from rest with constant acceleration $2$ m/s$^2$. Find the distance covered during the 5-th second.

Solution. $s_n = u + \tfrac{a}{2}(2n - 1) = 0 + 1 \cdot (2 \cdot 5 - 1) = 9$ m.

Example 3.8 — Graph reading. A particle's $v$–$t$ graph is a triangle: $v$ rises linearly from $0$ to $20$ m/s during $0 \le t \le 5$ s, then falls linearly from $20$ to $0$ during $5 \le t \le 10$ s. Find: (a) maximum acceleration; (b) displacement in 10 s; (c) average speed over 10 s.

Solution. (a) On the rising leg, $a = 20/5 = 4$ m/s$^2$; on the falling leg, $a = -4$ m/s$^2$. Maximum magnitude $= 4$ m/s$^2$. (b) Area under the $v$–$t$ graph (entirely above $t$-axis here) $= \tfrac{1}{2} \cdot 10 \cdot 20 = 100$ m. (c) $100 / 10 = 10$ m/s.

Example 3.9 — Ball dropped from height. A ball is dropped from a height $h$. Find (a) time to fall, (b) speed on impact, (c) ratio of distances covered in successive seconds.

Solution. Take down positive, $u = 0$. (a) $h = \tfrac{1}{2} g t^2 \Rightarrow t = \sqrt{2h/g}$. (b) $v = g t = \sqrt{2 g h}$. (c) Distance in $n$-th second: $s_n = \tfrac{g}{2}(2n - 1)$. So $s_1 : s_2 : s_3 : \ldots = 1 : 3 : 5 : 7 : \ldots$. Galileo's odd-number rule — discovered experimentally by him in the 1600s, decades before the calculus that derives it.

Common Traps

• Confusing average speed with average velocity. Equal only if motion is unidirectional.
• "Negative velocity means slowing down." No. Negative velocity means moving in $-x$ direction. Slowing down means $\vert v\vert$ decreases — equivalently, $a$ and $v$ have opposite signs.
• "At max height, acceleration is zero." No. Velocity is zero, but acceleration is still $g$ downward (otherwise the ball would stay up forever).
• Forgetting to choose a sign convention before plugging numbers. Decide once which direction is positive; stick to it for $u, v, a, s$ alike.
• Using $v^2 = u^2 + 2as$ to a problem with reversal of direction. The formula is valid only between two states with constant acceleration; if the particle reverses, split into segments.
• Misreading $s_n$. $s_n$ is the distance during the $n$-th second (between $t = n-1$ and $t = n$), not the total distance in $n$ seconds. Watch the question.
• In free-fall problems, dropping the negative sign of displacement. A ball that ends below its launch point has negative $y$ (with up = positive).
• Mistaking the $v$–$t$ slope for displacement. Slope is acceleration; area is displacement.
• Splitting motion incorrectly. If acceleration changes (e.g., from free fall to landing on a spring), each phase needs its own set of equations.
• Two-stage problems: time-to-meet errors. When two bodies start at different times, write their positions as functions of a common clock, then set positions equal.
• Galileo's odd-number rule: applies only when starting from rest. With $u \ne 0$, the ratio differs.
• Wrongly applying $v_{\text{avg}} = \tfrac{1}{2}(u+v)$. This is only valid for uniformly accelerated motion in a single phase.
• Reaction time in braking problems. During reaction time the vehicle moves at constant $u$ (zero deceleration). Braking distance is additional.

Quick Recap

• Position is a signed coordinate; displacement is the change in position; path length is the total distance traveled along the path.
• $\vert \Delta x\vert \le \text{path length}$, equality only when motion is unidirectional.
• $v = dx/dt$, $a = dv/dt = v \, dv/dx$.
• For constant $a$: $v = u + at$, $s = ut + \tfrac{1}{2} a t^2$, $v^2 = u^2 + 2as$, $s = \tfrac{1}{2}(u+v) t$, $s_n = u + \tfrac{a}{2}(2n-1)$.
• Free fall: replace $a$ by $-g$ (upward positive) or $+g$ (downward positive). Be consistent.
• For a body thrown straight up with speed $u$: $H = u^2/(2g)$, $T = 2u/g$, return speed $= u$.
• Galileo's odd-number rule: distances in successive seconds from rest are in $1:3:5:7:\ldots$.
• Relative velocity in 1D: $v_{AB} = v_A - v_B$.
• Graphs:
  • $x$–$t$ slope = velocity, curvature gives sign of $a$.
  • $v$–$t$ slope = $a$; area under = displacement.
  • $a$–$t$ area = change in velocity.
• Sign of $av$ tells whether the body is speeding up ($av > 0$) or slowing down ($av < 0$).

Formula Summary

The next chapter generalises every idea here to two dimensions — and brings in vectors, projectile motion, circular motion, and 2D relative velocity.