Chapter 2 · Motion in a Straight Line

2.3 Kinematic Equations and Free Fall

For constant acceleration, four short equations connect $u$, $v$, $a$, $t$ and $x$. They predict braking distances, chases, and the flight of anything dropped or thrown straight up under gravity.

Highway engineers place "slow down" signs before a sharp bend at a distance worked out from one equation. A fielder running in for a high catch is solving the same problem in her head. Forensic investigators measure a car's skid marks and tell the court how fast it was going before the brakes locked. In each case the acceleration is (very nearly) constant, and then the whole motion follows from a handful of equations that you can derive in two lines — from the area under a v–t graph, or from the calculus of Sections 2.1 and 2.2.

The most important constant acceleration of all is the one we live with: near the Earth's surface, every freely falling body — a feather in a vacuum tube, a cricket ball, a falling raindrop before air drag builds up — gains $9.8\ \text{m/s}$ of downward velocity every second. Galileo discovered this by slowing the fall down on gentle ramps and timing it with his pulse and a water clock. In this section you will derive the kinematic equations, learn a strategy for choosing among them, and apply them to free fall and vertical throws.

The kinematic equations for constant acceleration

Suppose a body moves along the $x$-axis with a constant acceleration $a$. At $t=0$ it is at $x_0$ with velocity $u$; at a later time $t$ it is at $x$ with velocity $v$. Because $a$ is constant, the v–t graph is a straight line of slope $a$ starting at height $u$.

📐 Result 2.4 — Equations of motion for constant acceleration

If the acceleration $a$ is constant, then for all $t\ge0$:

$$v=u+at \qquad (1)$$ $$x-x_0=ut+\tfrac12at^2 \qquad (2)$$ $$v^2=u^2+2a\,(x-x_0) \qquad (3)$$ $$x-x_0=\tfrac12(u+v)\,t \qquad (4)$$

All quantities except $t$ carry signs relative to the chosen positive direction. These equations are false when the acceleration changes with time — then use $v=\dfrac{dx}{dt}$, $a=\dfrac{dv}{dt}$ and areas directly.

Graphical derivation. (1) The slope of the straight v–t line is $a=\dfrac{v-u}{t}$, so $v=u+at$. (4) The displacement is the area under the line, a trapezium with parallel sides $u$ and $v$ and width $t$: $x-x_0=\tfrac12(u+v)t$. (2) Split the same area into a rectangle $ut$ and a triangle $\tfrac12\,t\,(v-u)=\tfrac12\,t\,(at)$: $x-x_0=ut+\tfrac12at^2$. (3) From (1), $t=\dfrac{v-u}{a}$; put it into (4): $x-x_0=\dfrac{(u+v)(v-u)}{2a}=\dfrac{v^2-u^2}{2a}$.

Calculus derivation. We need a velocity whose rate of change is the constant $a$ and whose value at $t=0$ is $u$: $v=u+at$ works, since $\dfrac{d}{dt}(u+at)=a$ by Result 2.1 (and it is the only such function). Next we need a position with $\dfrac{dx}{dt}=u+at$ and value $x_0$ at $t=0$: $x=x_0+ut+\tfrac12at^2$ works, because $\dfrac{d}{dt}\left(x_0+ut+\tfrac12at^2\right)=u+at$. In the language of Result 2.3, $x-x_0=\displaystyle\int_0^t(u+at)\,dt=ut+\tfrac12at^2$. Equations (3) and (4) then follow by algebra.

🧭 Problem-solving strategy — using the kinematic equations
  1. Check that the acceleration is constant (or constant in each stage — then treat each stage separately, the final velocity of one being the initial velocity of the next).
  2. Draw a line, mark the origin, and choose the positive direction. Give every vector quantity ($u$, $v$, $a$, $x-x_0$) its sign.
  3. List the five quantities $u$, $v$, $a$, $t$, $x-x_0$: three are known, one is wanted, one is not needed.
  4. Pick the equation that does not contain the unneeded quantity: no $x$ → (1); no $v$ → (2); no $t$ → (3); no $a$ → (4).
  5. Solve, then check signs, units and whether the answer is reasonable (a quadratic in $t$ may give a root that is negative or belongs to a different part of the motion).
✏️ Example 1 — Braking from $60\ \text{km/h}$
A car moving at $60\ \text{km/h}$ brakes with a constant deceleration of $7.0\ \text{m/s}^2$. Find (a) its braking distance and (b) its braking time. (c) What is the braking distance from $120\ \text{km/h}$?
Set up Positive = direction of motion. $u=\dfrac{60}{3.6}=16.7\ \text{m/s}$, $v=0$, $a=-7.0\ \text{m/s}^2$.
(a) No $t$ Equation (3): $0=u^2+2a\,d$, so $d=\dfrac{u^2}{2|a|}=\dfrac{(16.67)^2}{14}=19.8\ \text{m}$.
(b) No $x$ Equation (1): $t=\dfrac{0-16.67}{-7.0}=2.4\ \text{s}$.
(c) $d\propto u^2$: doubling the speed gives four times the distance, $4\times19.8=79\ \text{m}$.
Check Equation (4): $\tfrac12(16.67+0)(2.38)=19.8\ \text{m}$ ✓. This is the braking distance in the chapter teaser.
✏️ Example 2 — Stopping distance and reaction time
A driver moving at $90\ \text{km/h}$ sees a hazard. Her reaction time is $0.75\ \text{s}$ and the car then decelerates at $6.25\ \text{m/s}^2$. Find the reaction distance, braking distance and total stopping distance.
Stage 1 During the reaction time $a=0$: $u=25\ \text{m/s}$, reaction distance $=ut_r=25\times0.75=18.75\ \text{m}$.
Stage 2 Braking distance $=\dfrac{u^2}{2|a|}=\dfrac{625}{12.5}=50\ \text{m}$.
Total $d_s=ut_r+\dfrac{u^2}{2|a|}=18.75+50=68.75\approx69\ \text{m}$.
Insight The reaction part grows like $u$, the braking part like $u^2$, so at high speed braking dominates. A phone call that adds $0.5\ \text{s}$ to the reaction time adds another $12.5\ \text{m}$ at this speed.
✏️ Example 3 — Distance covered in the $n$th second
(a) Show that a body with constant acceleration $a$ and initial velocity $u$ covers a displacement $s_n=u+\tfrac a2(2n-1)$ during the $n$th second (between $t=n-1$ and $t=n$, in seconds). (b) A body starts from rest with $a=4.0\ \text{m/s}^2$. Find the distances it covers in the 1st, 2nd and 3rd seconds.
(a) By (2), $s_n=x(n)-x(n-1)=\left[un+\tfrac12an^2\right]-\left[u(n-1)+\tfrac12a(n-1)^2\right]=u+\tfrac12a(2n-1)$. (Here "$u$" and "$a$" stand for their numerical values in m/s and m/s², and $s_n$ comes out in metres.)
(b) $u=0$: $s_n=2.0(2n-1)$: $s_1=2.0\ \text{m}$, $s_2=6.0\ \text{m}$, $s_3=10\ \text{m}$.
Pattern From rest, the distances in successive equal intervals are in the ratio $1:3:5:\ldots$ — Galileo's law of odd numbers.
Check Total in 3 s: $2+6+10=18\ \text{m}=\tfrac12(4.0)(3)^2$ ✓.
✏️ Example 4 — The chase
A car speeding at a constant $20\ \text{m/s}$ passes a stationary police motorcycle. At that instant the motorcycle starts from rest with a constant acceleration of $3.0\ \text{m/s}^2$. (a) When and where does it catch the car? (b) How fast is the motorcycle going then? (c) What is the largest gap between them before the catch?
Positions Origin at the motorcycle: car $x_c=20t$; motorcycle $x_m=\tfrac12(3.0)t^2=1.5t^2$.
(a) $1.5t^2=20t$ gives $t=0$ (the start) or $t=\dfrac{20}{1.5}=13.3\ \text{s}$, at $x=20\times13.33=267\ \text{m}$.
(b) $v=at=3.0\times13.33=40\ \text{m/s}$ — exactly twice the car's speed, as equation (4) predicts (equal displacements in equal times need equal average velocities: $\tfrac12(0+v)=20$).
(c) The gap $20t-1.5t^2$ grows while the car is faster and shrinks after; it is largest when the speeds are equal, $3.0t=20$, $t=6.67\ \text{s}$: gap $=133-66.7=67\ \text{m}$.
Check Interactive 2 runs this chase; with $v_c=20$, $a=3.0$ and no delay its readouts give $13.3\ \text{s}$, $267\ \text{m}$, $40\ \text{m/s}$ and $67\ \text{m}$.

Free fall

📘 Definition 2.6 — Free fall and the acceleration due to gravity

A body is in free fall when the only influence on its motion is gravity — air resistance is negligible. Near the Earth's surface every freely falling body, whatever its mass, size or material, has the same constant downward acceleration, the acceleration due to gravity,

$$g=9.8\ \text{m/s}^2\ \ \text{(downwards)} .$$

This holds while the body is going up, at the top, and coming down. ($g$ varies slightly from place to place; you will see why in Chapter 7.)

For vertical motion we use a $y$-axis. With upwards positive, the acceleration is $a=-g$, and Result 2.4 becomes $$v=u-gt,$$ $$y-y_0=ut-\tfrac12gt^2,$$ $$v^2=u^2-2g\,(y-y_0).$$ You may instead take downwards as positive (then $a=+g$); either choice works if you keep to it for the whole problem.

📐 Result 2.5 — Vertical motion under gravity (no air resistance)

(i) Dropped from rest through a height $h$: time $t=\sqrt{\dfrac{2h}{g}}$, final speed $v=\sqrt{2gh}$. Distances fallen in successive equal time intervals are in the ratio $1:3:5:7:\ldots$

(ii) Thrown straight up with speed $u$ and caught at the launch height: time to the top $\dfrac ug$, maximum height $H=\dfrac{u^2}{2g}$, time of flight $\dfrac{2u}g$. The time up equals the time down, and the body passes every height with the same speed going up and coming down.

✏️ Example 5 — A catch in the outfield
A fielder throws a ball straight up at $14.7\ \text{m/s}$ and catches it at the same height. Find (a) the time to the top, (b) the maximum height, (c) the time of flight, (d) the velocity and position $2.0\ \text{s}$ after the throw.
(a) Up positive, $a=-9.8\ \text{m/s}^2$. At the top $v=0$: $t=\dfrac{14.7}{9.8}=1.5\ \text{s}$.
(b) $H=\dfrac{u^2}{2g}=\dfrac{(14.7)^2}{19.6}=11.0\ \text{m}$.
(c) $2\times1.5=3.0\ \text{s}$.
(d) $v=14.7-9.8(2.0)=-4.9\ \text{m/s}$ (moving down); $y=14.7(2.0)-4.9(2.0)^2=29.4-19.6=9.8\ \text{m}$ above the hand.
Check $v^2=u^2-2gy$: $(4.9)^2=24.0$ and $(14.7)^2-19.6(9.8)=216.1-192.1=24.0$ ✓.
✏️ Example 6 — Thrown up from a rooftop
A stone is thrown straight up at $10\ \text{m/s}$ from the edge of a roof $25\ \text{m}$ above the ground and falls past the edge to the ground. Find (a) when it hits the ground, (b) its speed then. (c) Explain the second root of the quadratic.
Set up Origin at the roof edge, up positive: $u=+10\ \text{m/s}$, $a=-9.8\ \text{m/s}^2$, the ground is at $y=-25\ \text{m}$.
(a) $-25=10t-4.9t^2\Rightarrow4.9t^2-10t-25=0\Rightarrow t=\dfrac{10\pm\sqrt{100+490}}{9.8}=\dfrac{10\pm24.29}{9.8}$. The positive root is $t=3.5\ \text{s}$.
(b) $v^2=u^2-2g(y-y_0)=100-2(9.8)(-25)=590$, so the speed is $24\ \text{m/s}$ ($24.3$), moving down.
(c) The other root, $t=-1.5\ \text{s}$, is when a stone following the same parabola would have had to leave the ground to pass the roof at $10\ \text{m/s}$ upwards. It is outside our motion, so we reject it.
Check In stages: rises $\dfrac{100}{19.6}=5.10\ \text{m}$ in $1.02\ \text{s}$, then falls $30.1\ \text{m}$ from rest in $\sqrt{2(30.1)/9.8}=2.48\ \text{s}$; total $3.50\ \text{s}$ ✓.
Interactive 1 — Free-Fall and Vertical-Throw Lab

Launch a ball straight up (or down, with $u\lt0$) from a height $h_0$. The left panel shows strobe images at equal time intervals; the graphs show $y(t)$ and $v(t)$ computed from $a=-g$. Set $u=0$ to drop the ball and check Galileo's $1:3:5:7$ pattern. Upwards is positive.

14.7 m/s0 m
time $t$—
height $y$ / velocity $v$—
maximum height—
time to reach ground—
speed on landing—
⚠️ Common mistakes in free-fall problems

1. "At the top the acceleration is zero." No: $v=0$ there, but $a=-g$ throughout — otherwise the ball would hover.

2. Mixed signs. With up positive, a ball below the launch point has $y\lt0$, a falling ball has $v\lt0$, and $a=-9.8\ \text{m/s}^2$ — always.

3. Using the equations across a change of acceleration. Reaction time ($a=0$) then braking ($a\ne0$) are two stages; a ball that bounces starts a new stage at the bounce.

4. Forgetting air resistance is ignored. A feather or a parachutist is not in free fall; their acceleration falls towards zero as drag grows (Chapter 9).

Relative motion of two bodies

When two bodies move along the same line, each has its own position and velocity in the ground frame, so problems like Example 4 can always be solved by writing $x(t)$ for both and setting them equal. JEE and NEET often shorten such problems by describing one body as seen from the other.

🚀 Beyond the board syllabus (JEE Main · NEET · JEE Advanced) — relative velocity in one dimension

In JEE Main, NEET and JEE Advanced; not in the CBSE syllabus.

If bodies A and B move along the $x$-axis with positions $x_A$, $x_B$ and velocities $v_A$, $v_B$ (all measured from the ground), the position, velocity and acceleration of B relative to A — as A would measure them — are

$$x_{BA}=x_B-x_A,$$ $$v_{BA}=v_B-v_A,\qquad a_{BA}=a_B-a_A .$$

And $v_{AB}=-v_{BA}$. Two bodies moving the same way have a relative speed equal to the difference of their speeds; moving towards each other, the sum. In A's frame you can use the kinematic equations with relative quantities, provided the relative acceleration is constant.

Mini-example. Two trains run on parallel tracks in the same direction at $54\ \text{km/h}$ ($15\ \text{m/s}$) and $90\ \text{km/h}$ ($25\ \text{m/s}$). Relative to the slower one, the faster moves at $10\ \text{m/s}$. If the faster train is $200\ \text{m}$ long and the slower $150\ \text{m}$, overtaking (front of one to rear of the other clearing completely) needs a relative displacement of $350\ \text{m}$: $350/10=35\ \text{s}$. In opposite directions the relative speed is $40\ \text{m/s}$ and they pass in $8.75\ \text{s}$.

Free fall seen from free fall. Two balls in free fall have $a_{BA}=(-g)-(-g)=0$: each sees the other move in a straight line at constant velocity. This is why the two balls in Card IV, part 5 below meet at $t=\dfrac{40\ \text{m}}{20\ \text{m/s}}$.

This topic is not in the current board syllabus for Class 11; it is examined in JEE Main and NEET. It returns in two dimensions in Section 3.3.

Interactive 2 — The Chase

A car passes a stationary police motorcycle at constant speed $v_c$. After a delay $t_d$ the motorcycle sets off from rest with constant acceleration $a$. The x–t graphs show both; they cross at the catch. The green bar marks the largest gap, at the instant the two graphs are parallel (equal velocities). Press Play.

20 m/s3.0 m/s²0 s
catch time—
catch distance—
motorcycle speed at catch—
largest gap (at time)—

Exercises

I
Equations of motion (constant acceleration).
  • 1 A car speeds up from $5.0\ \text{m/s}$ to $25\ \text{m/s}$ at $2.5\ \text{m/s}^2$. Find the time taken and the distance covered.
  • 2 A body starts from rest with $a=3.0\ \text{m/s}^2$. Find its velocity and the distance covered after $6.0\ \text{s}$.
  • 3 A train moving at $72\ \text{km/h}$ is brought uniformly to rest in $400\ \text{m}$. Find its acceleration and the time taken.
  • 4 A bullet enters a wooden block at $200\ \text{m/s}$ and stops after penetrating $5.0\ \text{cm}$. Assuming uniform deceleration, find the deceleration and the time it takes to stop.
  • 5 A body has $u=2.0\ \text{m/s}$ and $a=4.0\ \text{m/s}^2$. How far does it go in the 5th second?
  • 6 A uniformly accelerated body covers $20\ \text{m}$ in the 2nd second and $40\ \text{m}$ in the 4th second. Find $u$ and $a$.
  • 7 Derive $v^2=u^2+2a(x-x_0)$ from equations (1) and (2) alone.
  • 8 A car's speed halves while it brakes uniformly over $60\ \text{m}$. How much further does it travel before stopping?
1 $t=\dfrac{25-5.0}{2.5}=8.0\ \text{s}$; $x=\tfrac12(5.0+25)(8.0)=120\ \text{m}$.
2 $v=3.0\times6.0=18\ \text{m/s}$; $x=\tfrac12(3.0)(36)=54\ \text{m}$.
3 $u=20\ \text{m/s}$: $a=-\dfrac{u^2}{2x}=-\dfrac{400}{800}=-0.50\ \text{m/s}^2$; $t=\dfrac{20}{0.50}=40\ \text{s}$.
4 $a=-\dfrac{(200)^2}{2(0.050)}=-4.0\times10^5\ \text{m/s}^2$; $t=\dfrac{0.050}{\tfrac12(200)}=5.0\times10^{-4}\ \text{s}$.
5 $s_5=2.0+\tfrac{4.0}{2}(9)=20\ \text{m}$.
6 $20=u+1.5a$ and $40=u+3.5a$: subtracting, $2a=20$, $a=10\ \text{m/s}^2$, $u=5.0\ \text{m/s}$.
7 From (1), $t=\dfrac{v-u}{a}$. Into (2): $x-x_0=\dfrac{u(v-u)}{a}+\dfrac{(v-u)^2}{2a}=\dfrac{2uv-2u^2+v^2-2uv+u^2}{2a}=\dfrac{v^2-u^2}{2a}$.
8 First stage: $\left(\tfrac u2\right)^2=u^2-2|a|(60)$, so $u^2=160|a|$. Second: $0=\left(\tfrac u2\right)^2-2|a|s$, so $s=\dfrac{u^2}{8|a|}=20\ \text{m}$.
II
Stopping distances and chases.
  • 1 Find the stopping distance of a car at $72\ \text{km/h}$ if the reaction time is $0.80\ \text{s}$ and the deceleration is $6.0\ \text{m/s}^2$.
  • 2 Under the same conditions, what is the highest speed (in km/h) from which the car can stop within $40\ \text{m}$?
  • 3 How does the braking distance (not the reaction distance) at $80\ \text{km/h}$ compare with that at $40\ \text{km/h}$, for the same deceleration?
  • 4 A car waiting at a signal starts with $a=2.0\ \text{m/s}^2$ just as a truck passes it at a steady $10\ \text{m/s}$. When and where does the car overtake the truck, and how fast is it moving then?
  • 5 In part 4, what is the largest distance by which the truck leads the car?
1 $20(0.80)+\dfrac{20^2}{2(6.0)}=16+33.3=49\ \text{m}$.
2 $0.80u+\dfrac{u^2}{12}=40\Rightarrow u^2+9.6u-480=0\Rightarrow u=\dfrac{-9.6+\sqrt{92.16+1920}}{2}=17.6\ \text{m/s}=63\ \text{km/h}$.
3 Braking distance $\propto u^2$: four times as long.
4 $t^2=10t$ (car $\tfrac12(2.0)t^2$, truck $10t$): $t=10\ \text{s}$, at $100\ \text{m}$; car speed $20\ \text{m/s}$.
5 Equal speeds at $t=5.0\ \text{s}$: gap $=50-25=25\ \text{m}$.
III
Free fall. Use $g=9.8\ \text{m/s}^2$ and ignore air resistance.
  • 1 A stone is dropped from a height of $44.1\ \text{m}$. Find the time to reach the ground and its speed then.
  • 2 A ball is thrown straight up at $19.6\ \text{m/s}$. Find its maximum height, the time to reach it, and the time of flight back to the hand.
  • 3 A ball is thrown up at $12\ \text{m/s}$. Find its velocity and height $2.0\ \text{s}$ later. Is it rising or falling?
  • 4 A ball rises to a maximum height of $20\ \text{m}$. With what speed was it thrown?
  • 5 A stone dropped into a well is heard to splash $2.0\ \text{s}$ later. (a) Estimate the depth ignoring the travel time of sound. (b) Taking the speed of sound as $343\ \text{m/s}$, is the true depth larger or smaller? Estimate it.
  • 6 Find the distances fallen from rest in the 1st, 2nd and 3rd seconds, and their ratio.
  • 7 A ball is thrown vertically downwards at $5.0\ \text{m/s}$ from a bridge $30\ \text{m}$ above a river. Find the time to reach the water and the speed then.
  • 8 How high would the ball of part 2 rise on the Moon ($g=1.62\ \text{m/s}^2$)?
  • 9 Sketch the v–t graph of a ball thrown up and caught at the same height. Why are the areas above and below the time axis equal?
1 $t=\sqrt{\dfrac{2(44.1)}{9.8}}=3.0\ \text{s}$; $v=gt=29.4\ \text{m/s}$.
2 $H=\dfrac{(19.6)^2}{19.6}=19.6\ \text{m}$; $t_{\text{up}}=2.0\ \text{s}$; time of flight $4.0\ \text{s}$.
3 $v=12-19.6=-7.6\ \text{m/s}$ (falling); $y=24-4.9(4.0)=4.4\ \text{m}$ above the launch point.
4 $u=\sqrt{2gH}=\sqrt{392}=19.8\ \text{m/s}$.
5 (a) $h=\tfrac12(9.8)(2.0)^2=19.6\ \text{m}$. (b) Some of the $2.0\ \text{s}$ is spent by sound coming up, so the fall time is shorter and the well is shallower: solving $\sqrt{2h/g}+h/343=2.0$ gives $h\approx18.6\ \text{m}$.
6 $4.9\ \text{m}$, $14.7\ \text{m}$, $24.5\ \text{m}$: ratio $1:3:5$.
7 Down positive: $30=5.0t+4.9t^2\Rightarrow t=\dfrac{-5.0+\sqrt{25+588}}{9.8}=2.0\ \text{s}$; $v=\sqrt{25+2(9.8)(30)}=24.8\ \text{m/s}\approx25\ \text{m/s}$.
8 $H=\dfrac{(19.6)^2}{2(1.62)}=119\ \text{m}$ — about six times higher.
9 A straight line of slope $-9.8\ \text{m/s}^2$ from $+u$ to $-u$, crossing the axis at $t=u/g$. The ball ends where it started, so the net displacement — the signed area — is zero: the triangle above balances the triangle below.
IV
Calculus, graphs and proofs.
  • 1 Show by differentiating that $x=x_0+ut+\tfrac12at^2$ gives $\dfrac{dx}{dt}=u+at$ and $\dfrac{d^2x}{dt^2}=a$.
  • 2 $x=3.0+12t-2.0t^2$ (SI units). Identify $x_0$, $u$ and $a$. When is the body at rest, and what is its largest $x$?
  • 3 Show that for constant acceleration the average velocity over $[0,t]$ equals the instantaneous velocity at $t/2$. Check with $u=4.0\ \text{m/s}$, $a=2.0\ \text{m/s}^2$, $t=6.0\ \text{s}$.
  • 4 Prove that for a body starting from rest with constant acceleration, the distances covered in successive equal intervals $\tau$ are in the ratio $1:3:5:\ldots$
  • 5 A ball is dropped from a height of $40\ \text{m}$. At the same instant a second ball is thrown straight up at $20\ \text{m/s}$ from the ground directly below. When and at what height do they meet?
1 Result 2.1: $\dfrac{d}{dt}(x_0)=0$, $\dfrac{d}{dt}(ut)=u$, $\dfrac{d}{dt}(\tfrac12at^2)=at$; so $v=u+at$ and then $\dfrac{dv}{dt}=a$.
2 $x_0=3.0\ \text{m}$, $u=12\ \text{m/s}$, $\tfrac12a=-2.0$ so $a=-4.0\ \text{m/s}^2$. $v=12-4.0t=0$ at $t=3.0\ \text{s}$; $x_{\max}=3.0+36-18=21\ \text{m}$.
3 $\bar v=\dfrac{ut+\tfrac12at^2}{t}=u+\tfrac12at=v\!\left(\tfrac t2\right)$. Check: $\bar v=\dfrac{24+36}{6.0}=10\ \text{m/s}$; $v(3.0)=4.0+6.0=10\ \text{m/s}$ ✓.
4 $x(n\tau)=\tfrac12a n^2\tau^2$, so the $n$th interval covers $\tfrac12a\tau^2\left[n^2-(n-1)^2\right]=\tfrac12a\tau^2(2n-1)$, proportional to $1,3,5,\ldots$
5 Up positive from the ground: $y_1=40-4.9t^2$, $y_2=20t-4.9t^2$. Equal when $40=20t$: $t=2.0\ \text{s}$, at $y=40-19.6=20.4\ \text{m}$. (The second ball is then moving up at $0.4\ \text{m/s}$.)
🎯 Exam corner BoardJEE MainJEE AdvancedNEET

E1. Board · 3 marks A body moves with initial velocity $u$ and constant acceleration $a$. Using its velocity–time graph, derive $x=ut+\tfrac12at^2$ for the displacement in time $t$.

Graph The v–t graph is a straight line from $(0,u)$ to $(t,v)$ with $v=u+at$ (slope $a$). [1]
Area Displacement = area under the graph = rectangle $u\,t$ + triangle $\tfrac12\,t\,(v-u)$. [1]
Result $v-u=at$, so $x=ut+\tfrac12t(at)=ut+\tfrac12at^2$. [1]
Exam tip Draw and label the graph (axes, $u$, $v$, $t$, the rectangle and triangle); the diagram carries a mark in most schemes.

E2. NEET A ball thrown vertically upwards returns to the thrower's hand after $6.0\ \text{s}$. Taking $g=9.8\ \text{m/s}^2$, the maximum height reached is (a) $44.1\ \text{m}$ (b) $29.4\ \text{m}$ (c) $88.2\ \text{m}$ (d) $176.4\ \text{m}$

Time up Half the flight: $3.0\ \text{s}$. So $u=g\times3.0=29.4\ \text{m/s}$.
Height $H=\tfrac12g(3.0)^2=44.1\ \text{m}$.
Answer (a).
Exam tip (b) is the launch speed, not a height; (d) uses the full $6.0\ \text{s}$ in $\tfrac12gt^2$. Think of the descent alone: a drop from rest for $3.0\ \text{s}$.

E3. JEE Main NEET A body starts from rest with uniform acceleration. The ratio of the distances it covers in the 3rd second and in the 5th second is (a) $3:5$ (b) $5:9$ (c) $9:25$ (d) $1:2$

Formula From rest, $s_n=\tfrac a2(2n-1)$.
Ratio $\dfrac{s_3}{s_5}=\dfrac{5}{9}$.
Answer (b).
Exam tip (c) is the ratio of the total distances in $3\ \text{s}$ and $5\ \text{s}$ ($\propto t^2$). Read "in the 3rd second" versus "in 3 seconds" carefully.

E4. JEE Main (Numerical value; take $g=10\ \text{m/s}^2$) A body dropped from rest from the top of a tower covers $45\ \text{m}$ in the last second of its fall. Find the height of the tower in metres.

Last second If the fall lasts $n$ seconds, $s_n=\tfrac{g}{2}(2n-1)=5(2n-1)=45$, so $n=5\ \text{s}$.
Height $h=\tfrac12(10)(5)^2=125\ \text{m}$.
Answer $125$.
Check In $4\ \text{s}$ it falls $80\ \text{m}$; $125-80=45\ \text{m}$ ✓. If the fall time is not a whole number of seconds, use $h(t)-h(t-1)=45$ instead — same algebra, same answer here.

E5. JEE Advanced (One or more correct; take $g=10\ \text{m/s}^2$) A ball is thrown vertically upwards at $10\ \text{m/s}$ from the top of a tower $40\ \text{m}$ high and lands at the foot of the tower. Which statements are correct? (A) Its greatest height above the ground is $45\ \text{m}$. (B) It reaches the ground $4\ \text{s}$ after being thrown. (C) It hits the ground at $30\ \text{m/s}$. (D) Its average speed over the whole flight is $10\ \text{m/s}$.

(A) Rise above the tower $\dfrac{u^2}{2g}=5\ \text{m}$: greatest height $45\ \text{m}$ ✓.
(B) Up positive from the top: $-40=10t-5t^2\Rightarrow t^2-2t-8=0\Rightarrow t=4\ \text{s}$ ✓.
(C) $v=10-10(4)=-30\ \text{m/s}$: speed $30\ \text{m/s}$ ✓.
(D) Path length $5+45=50\ \text{m}$ in $4\ \text{s}$: average speed $12.5\ \text{m/s}$ ✗. ($10\ \text{m/s}$ is the magnitude of the average velocity, $40\ \text{m}$ down in $4\ \text{s}$.)
Answer (A), (B), (C).

E6. JEE Main NEET Two trains, $200\ \text{m}$ and $150\ \text{m}$ long, run on parallel tracks in the same direction at $90\ \text{km/h}$ and $54\ \text{km/h}$. The time the faster train takes to pass the slower one completely is (a) $14\ \text{s}$ (b) $35\ \text{s}$ (c) $20\ \text{s}$ (d) $8.75\ \text{s}$

Note Relative velocity is beyond the current NCERT/board syllabus but is examined in JEE Main and NEET (see the 🚀 box).
Relative motion In the slower train's frame the faster train moves at $25-15=10\ \text{m/s}$ and must cover $200+150=350\ \text{m}$.
Answer (b) $35\ \text{s}$.
Exam tip (d) uses the sum of the speeds (trains in opposite directions); (a) and (c) use only one train's length. Without relative velocity: in time $t$ the fast train goes $25t$ and the slow one $15t$, and the $350\ \text{m}$ that must be gained gives $25t-15t=350$ — the same equation.
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes

Research question. How much does using a phone slow a person's reactions — and what does that cost in stopping distance?

Why it matters. Road-safety researchers in many countries have measured drivers' reaction times in simulators and found that conversations and texting lengthen them noticeably. Because reaction distance is speed × reaction time, every extra tenth of a second is travelled at full speed before the brakes even start. A falling ruler turns free fall into a simple, cheap reaction timer.

What to do.

  1. A partner holds a $30\ \text{cm}$ ruler vertically with its zero mark between your open thumb and finger and drops it without warning. Record the distance $d$ it falls before you catch it. Show that your reaction time is $t=\sqrt{2d/g}$.
  2. Take 10 trials, find the mean $d$ and the mean $t$. Repeat (sitting safely) while holding a conversation, and again while typing a message.
  3. For each condition, find the extra reaction distance at $50\ \text{km/h}$ and at $100\ \text{km/h}$ compared with your undistracted value.
  4. Use Example 2's method to find how much each condition adds to the total stopping distance at $100\ \text{km/h}$ with a deceleration of $7.0\ \text{m/s}^2$.

Tools: A $30\ \text{cm}$ ruler, a partner, a notebook or spreadsheet.

Hand in: A results table (mean $d$, mean $t$ and their spread for each condition), your stopping-distance calculations, and a paragraph on why a ruler test underestimates reaction time on a real road.

Going further. Look up how road-safety agencies choose the reaction time used in official stopping-distance tables, and why it is much longer than the ruler test gives.

Instructor note: Typical catches are $d\approx15$–$20\ \text{cm}$, so $t\approx0.18$–$0.20\ \text{s}$; distraction often adds $0.05$–$0.15\ \text{s}$. Each extra $0.10\ \text{s}$ adds $1.4\ \text{m}$ at $50\ \text{km/h}$ and $2.8\ \text{m}$ at $100\ \text{km/h}$; the braking part ($55\ \text{m}$ at $100\ \text{km/h}$, $7.0\ \text{m/s}^2$) is unchanged. Real driving reaction times ($\approx1\ \text{s}$) are longer because the driver must first notice and interpret the hazard.

🧪 Self-check quiz

0/5 correct

Q1. Which equation of motion would you use if the time is neither known nor wanted?

Q2. At the highest point of a ball thrown straight up,

Q3. If a car's speed is doubled, its braking distance (same deceleration) becomes

Q4. A stone dropped from rest falls distances $s_1$, $s_2$, $s_3$ in the 1st, 2nd and 3rd seconds. Then $s_1:s_2:s_3$ is

Q5. A ball is thrown straight up at $9.8\ \text{m/s}$. It returns to the thrower's hand after

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