Kinematics

Kinematics is the grammar of mechanics: before you can talk about why things move (forces), you must be able to describe how they move. Nearly every year NEET asks two to four questions from this area — graph reading, equations of uniformly accelerated motion, relative velocity, and projectiles — and the same tools reappear inside problems on dynamics, circular motion and even electromagnetism.

Locating a Particle: Frame of Reference, Path Length and Displacement

Motion has no absolute meaning. A passenger asleep in a moving train is at rest with respect to the train and moving at 90 km/h with respect to the platform. So every statement about motion is silently attached to a frame of reference — an origin plus a set of axes plus a clock. When a question says "a body is dropped from a balloon", the phrase quietly means dropped with zero velocity relative to the balloon, which is not zero relative to the ground.

For a particle moving along a line we specify its position $x$ (a signed number, with sign fixed by the chosen positive direction). Two different quantities describe how far it has gone between times $t_1$ and $t_2$:

• Path length (distance): the total length of the actual track traced out. It is a scalar, never negative, never decreasing with time.
• Displacement: $\Delta x = x_2 - x_1$, the straight-line vector from the starting point to the finishing point. It is independent of the route taken and can be positive, negative or zero.

A useful pair of consequences: displacement magnitude $\le$ path length, with equality only when the motion is along a straight line without reversing direction. Hence a runner completing one lap of a 400 m track has path length 400 m but zero displacement. When the motion is in two or three dimensions, position becomes a vector $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$ and displacement is $\Delta \vec{r} = \vec{r}_2 - \vec{r}_1$, whose magnitude is the length of the chord joining the two points on the trajectory.

Speed, Velocity and Acceleration

Rates of change come in two flavours — averaged over an interval, and instantaneous at a point.

• Average speed $= \dfrac{\text{total path length}}{\text{total time}}$ (scalar, always $\ge 0$).
• Average velocity $\vec{v}_{av} = \dfrac{\Delta \vec{r}}{\Delta t}$ (vector, points along the displacement).
• Instantaneous velocity $\vec{v} = \dfrac{d\vec{r}}{dt} = \lim_{\Delta t \to 0}\dfrac{\Delta \vec{r}}{\Delta t}$. Over a vanishingly small interval the chord merges with the curve, so instantaneous velocity is always tangential to the path, and its magnitude equals the instantaneous speed.
• Average acceleration $\vec{a}_{av} = \dfrac{\Delta \vec{v}}{\Delta t}$; instantaneous acceleration $\vec{a} = \dfrac{d\vec{v}}{dt} = \dfrac{d^2\vec{r}}{dt^2}$.

Note that average speed and the magnitude of average velocity are generally different numbers. A standard result: if a car covers the first half of a distance at $v_1$ and the second half at $v_2$, the average speed is the harmonic mean $\dfrac{2v_1v_2}{v_1+v_2}$; if instead it travels for equal times at $v_1$ and $v_2$, the average speed is the arithmetic mean $\dfrac{v_1+v_2}{2}$.

Acceleration is not "speeding up". It is any change in the velocity vector. Three separate situations all count as accelerated motion: increasing speed, decreasing speed, and constant speed with changing direction (uniform circular motion). Signs matter in one dimension: if $v$ and $a$ have the same sign the particle speeds up; opposite signs mean it slows down. A body thrown vertically upward has $a = -g$ throughout, including at the highest point where $v = 0$ momentarily — zero velocity with non-zero acceleration is perfectly consistent, because acceleration depends on how velocity is changing, not on its present value.

Reading Motion Graphs

Graph questions are cheap marks if you internalise two rules: slope gives the derivative, area gives the integral.

1. Position–time graph: slope at a point $=$ instantaneous velocity. A straight line means uniform velocity; a curve bending upward means increasing velocity (positive acceleration). The graph can never be a vertical line (that would mean infinite velocity), and for a single particle it cannot double back over the same time value.
2. Velocity–time graph: slope $=$ acceleration; area between the curve and the time axis $=$ displacement (areas below the axis counted negative). Adding the magnitudes of all areas without sign gives the path length instead.
3. Acceleration–time graph: area under it $=$ change in velocity.

Practical readings to remember: a $v$–$t$ graph that is a straight line through the origin is uniformly accelerated motion from rest; a horizontal $v$–$t$ line is uniform velocity; a $v$–$t$ line crossing the time axis means the body reverses direction at that instant. For a body thrown up and returning, the $v$–$t$ graph is a single straight line of slope $-g$ crossing the axis at the top of the flight, while the $x$–$t$ graph is a downward parabola.

Uniformly Accelerated Motion in a Straight Line

When $a$ is constant, calculus gives the three standard relations (with $u$ = initial velocity, $v$ = velocity at time $t$, $s$ = displacement in time $t$):

$$v = u + at \qquad s = ut + \tfrac{1}{2}at^2 \qquad v^2 = u^2 + 2as$$

Two more that save time:

• Displacement using average velocity: $s = \left(\dfrac{u+v}{2}\right)t$ — valid only for constant acceleration.
• Distance in the $n^{th}$ second: $s_n = u + \dfrac{a}{2}(2n-1)$. Note this is a distance travelled during a one-second interval, not the total distance in $n$ seconds.

All symbols are signed components along the chosen axis, so fix a positive direction first and stick to it. Substituting a "deceleration of 5 m/s²" as $a = +5$ is the single most common arithmetic disaster in this chapter.

Free fall is just this motion with $a = g \approx 9.8$ m/s² directed downward, air resistance ignored, and independent of the mass of the body. Handy results for a body projected vertically upward with speed $u$: time of ascent $= u/g$, maximum height $= u^2/2g$, total time of flight $= 2u/g$, and it returns to the launch point with speed $u$. For a body simply released from rest, distances covered in successive seconds are in the ratio $1:3:5:7\ldots$

Also worth remembering: stopping distance for a vehicle with initial speed $u$ and retardation $a$ is $u^2/2a$ — it scales as the square of the speed, so doubling the speed quadruples the stopping distance.

Motion in a Plane: Projectiles and Uniform Circular Motion

In two dimensions the great simplification is independence of components: motion along $x$ and along $y$ proceed as separate one-dimensional problems sharing a common clock.

For a projectile launched with speed $u$ at angle $\theta$ to the horizontal (neglecting air resistance):

• Horizontal: $a_x = 0$, so $v_x = u\cos\theta$ stays constant and $x = (u\cos\theta)t$.
• Vertical: $a_y = -g$, so $v_y = u\sin\theta - gt$ and $y = (u\sin\theta)t - \tfrac{1}{2}gt^2$.
• Eliminating $t$ gives $y = x\tan\theta - \dfrac{g x^2}{2u^2\cos^2\theta}$ — the trajectory is a parabola.
• Time of flight $T = \dfrac{2u\sin\theta}{g}$; maximum height $H = \dfrac{u^2\sin^2\theta}{2g}$; range $R = \dfrac{u^2\sin 2\theta}{g}$.
• $R$ is maximum at $\theta = 45°$, where $R_{max} = u^2/g$. Two complementary angles ($\theta$ and $90° - \theta$) give the same range but different heights and flight times. Also $R = 4H\cot\theta$.

At the topmost point the velocity is purely horizontal ($v_y = 0$) but the acceleration is still $g$ downward — the velocity and acceleration are perpendicular there. For a body projected horizontally from a height $h$ (e.g. from a moving aeroplane), the initial $u\sin\theta = 0$, so the time to fall is $\sqrt{2h/g}$, independent of the horizontal speed.

In uniform circular motion the speed is constant but the direction changes continuously, so there is an acceleration of magnitude $a_c = \dfrac{v^2}{r} = \omega^2 r$ directed always toward the centre (centripetal). Here $\omega = 2\pi/T = v/r$ is the angular speed. The centripetal acceleration is perpendicular to the velocity — which is exactly why it changes direction without changing magnitude.

Relative Velocity

The velocity of object A as seen from object B is $$\vec{v}_{AB} = \vec{v}A - \vec{v}B$$ with all the velocities on the right measured in the same (usually ground) frame. Consequences: $\vec{v}{AB} = -\vec{v}{BA}$, and for one-dimensional motion the vectors reduce to signed numbers, so two cars approaching head-on at 20 m/s and 15 m/s have relative speed 35 m/s, whereas moving in the same direction it is 5 m/s.

Two classic applications:

1. Rain–umbrella / river–boat: to a person walking with velocity $\vec{v}_m$, rain falling with $\vec{v}_r$ appears to come with velocity $\vec{v}_r - \vec{v}_m$; the umbrella must be tilted along that apparent direction.
2. River crossing: if a swimmer's speed in still water is $v$ and the river flows with speed $u$ across a width $d$, then heading straight across gives minimum crossing time $d/v$ with a downstream drift $ud/v$; to land exactly opposite, the swimmer must aim upstream at angle $\sin^{-1}(u/v)$ and the crossing time becomes $d/\sqrt{v^2-u^2}$ (possible only if $v > u$).

For two objects in free fall simultaneously, their relative acceleration is zero ($g - g = 0$), so relative velocity between the two stays constant throughout the fall — their separation grows or shrinks linearly with time even though each body individually has a non-zero acceleration in the ground frame. This is the standard trick behind "two balls dropped from different heights" or "one thrown up while the other is released" problems: shift to the frame of one falling body and the other appears to move at constant velocity, turning an accelerated-motion problem into a uniform-velocity one.

Common Mistakes and Exam Traps

• Wrong sign for deceleration. If a body decelerates at 5 m/s² while moving in the positive direction, $a$ must be substituted as $-5$, not $+5$, in every one of the three equations of motion — this single sign slip is the most frequent source of a wrong numerical answer in this chapter.
• Confusing distance with displacement in "average speed" questions. For a body that reverses direction — a ball thrown up and caught, a particle oscillating along a line — average speed uses the total path length, while average velocity uses only the net displacement; the two answers are numerically different even for the same time interval.
• Misreading the $n^{th}$-second formula. $s_n = u + \dfrac{a}{2}(2n-1)$ gives the distance covered during the $n^{th}$ second alone, not the total distance travelled in $n$ seconds; treating the two as the same quantity is a frequent slip under time pressure.
• Applying the flat-ground range formula to an elevated or inclined launch. $R = u^2\sin2\theta/g$ assumes the projectile lands at the same height from which it was launched. The moment a question introduces a cliff, a slope, or a target at a different level, the full $x(t)$ and $y(t)$ equations must be solved for the actual landing condition rather than quoting the standard range formula.