Relative Velocity and Relative Speed

Speed · velocity · relative motion — maths and physics together

7 Part11 worked examples30 practice questions

A passenger sitting next to you in a moving train looks at rest to you, but to a man standing on the platform the same passenger is rushing at 60 km/h. Both are right. Motion is always measured from the point of view of an observer. This idea is called relative velocity (or, along a straight line, relative speed).

In “Basic Science and Mathematics” papers speed, velocity and relative velocity come up in both the maths and the physics section. So this chapter joins the two: first the physics of speed, velocity and acceleration, then straight-line relative-speed problems (the maths side, time-speed-distance), then two-dimensional problems (boat and river, aircraft and wind, rain and umbrella), and finally the idea of a frame of reference.

Part	What you learn	Key idea
1. Foundations	distance-displacement, speed-velocity, units, graphs, equations, free fall	$v = \frac{s}{t}$ , $v = u + a t$
2. Straight line	meeting, chasing, walkways, the shuttling dog	same way: subtract; towards: add
3. Circles	laps, $n$ -th meeting, clock hands, back-and-forth path	lap length ÷ relative speed
4. Two dimensions	vector subtraction, boat-river, aircraft-wind, rain-umbrella, closest approach	$v_{A B} = v_{A} - v_{B}$
5. With acceleration	overtaking, braking, objects dropped from a moving vehicle	$s = u t + \frac{1}{2} a t^{2}$
6. Physics link	frames of reference, inertial frames, Galilean addition	$v_{A C} = v_{A B} + v_{BC}$

Here: $u$ = initial velocity, $v$ = final velocity, $a$ = acceleration, $s$ = displacement, $t$ = time; $v_{A B}$ = velocity of $A$ relative to $B$ ; $v_{A}, v_{B}$ = velocities relative to the ground.

How to startBefore any formula write three things: (1) who is moving and in which direction? (2) relative to whom is the speed asked? (3) are all the units the same?

Part 1 — Foundations of motion

distance · velocity · units · graphs · equations

A firm grip on speed, velocity and acceleration before we compare them.

Relative velocity compares two motions. The comparison is right only when each motion has a clear meaning, unit and direction. This part lays that foundation.

Distance, displacement, speed and velocity

scalar versus vector

Path length versus straight-line gap, speed versus velocity, average speed versus average velocity.

The idea

Understand two words carefully. Distance is what your feet (or the wheels) actually covered — the full length of the path. Displacement is the straight line from the starting point to the end point, stated with its direction. Suppose you go 3 km east from home and then 4 km north: the distance is 7 km but the straight gap from home (the displacement) is only 5 km.

◭ Diagram

Distance is the length of the path; displacement is the arrow joining start and end

A quantity that has only a size (distance, speed, time) is a scalar. A quantity that has size and direction (displacement, velocity, acceleration) is a vector. This is exactly what separates speed from velocity:

Quantity	Meaning	Type	Zero or negative?
Distance	total length of the path	scalar	never negative, never decreases
Displacement	straight arrow from start to end	vector	can be zero or negative
Speed	distance ÷ time	scalar	never negative
Velocity	displacement ÷ time	vector	can be zero or negative

∑ Formula

average speed = \frac{total distance}{total time}, v_{avg} = \frac{total displacement}{total time}

Average speed and average velocity

$v_{avg}$: average velocity (km/h or m/s, with direction)
$e x t t o t a l d i s t an ce$: length of the whole path (km or m)
$e x t t o t a l d i s pl a ce m e n t$: straight start-to-end gap (km or m)
$e x t t o t a lt im e$: time of the whole journey including halts

Spot itWords such as “from the starting point”, “displacement”, “velocity”, “returns” point to displacement. “How much in all” and “speed” point to distance.

Steps

Add the lengths of all legs to get the total distance.
Join start and end by a straight line to get the displacement: in one line use addition and subtraction with direction signs, at a right angle use Pythagoras.
Put the time in the right unit (hours or seconds).
Average speed $=$ distance ÷ time; average velocity $=$ displacement ÷ time.

Worked examples

Example 1 · Basic

Amit walks 6 km east from a crossing and then 8 km north. The whole trip takes 30 min. Find his average speed and the magnitude of his average velocity.

Solution, step by step

1Distance is the total length of the path, whatever the direction: 6 + 8 = 14 km.
2Displacement is the straight line from start to end. The east and north legs make a right angle, so by Pythagoras $6^{2} + 8^{2} = 100 = 10$ km.
3Speeds are wanted in km/h, so write the time in hours: 30 min $= 0.5$ h.
4Average speed $= \frac{total distance}{total time} = \frac{14}{0.5} = 28$ km/h.
5Average velocity $= \frac{displacement}{total time} = \frac{10}{0.5} = 20$ km/h (towards the north-east side).

Answer 28 km/h and 20 km/h

Average speed is never smaller than the magnitude of average velocity; here the path turned, so it is larger.

Example 2 · Standard

Joseph starts running from a point on the edge of a round park of radius 14 m and completes one full lap in 40 s. (Take

π = \frac{22}{7}

.) Find the average speed and the magnitude of the average velocity.

Solution, step by step

1One lap is $2 π r = 2 \times \frac{22}{7} \times 14 = 88$ m. For one full lap the distance is $1 \times 88 = 88$ m.
2After whole laps he is back at the starting point, so start and end are the same place: displacement 0.
3The time is 40 s, so the speeds come out in m/s.
4Average speed $= \frac{88}{40} = 2.2$ m/s.
5Average velocity $= \frac{0}{40} = 0$ m/s.

Answer 2.2 m/s and 0 m/s

Trap: the average velocity is 0 but the average speed is not — speed is a scalar, velocity a vector.

Practice (try it yourself)

BasicTry it yourself first

Imran walks 9 km east from a crossing and then 12 km north. The whole trip takes 45 min. What is his displacement?

BasicTry it yourself first

Kishan walks 120 m east along a straight road and then comes back 70 m west. It takes 25 s in all. What total distance did he cover?

StandardTry it yourself first

Ravi walks 15 km east from a crossing and then 20 km north. The whole trip takes 1 h 15 min. What is the magnitude of his average velocity?

StandardTry it yourself first

Suresh starts running from a point on the edge of a round park of radius 21 m and completes one and a half laps in 66 s. (Take

π = \frac{22}{7}

.) What is his average speed?

AdvancedTry it yourself first

Deepak walks 150 m east along a straight road and then comes back 110 m west. It takes 16 s in all. What is the magnitude of his average velocity?

AdvancedTry it yourself first

Anil starts running from a point on the edge of a round park of radius 7 m and completes one and a half laps in 7 s. (Take

π = \frac{22}{7}

.) What is the magnitude of his average velocity?

Units, conversions and uniform / non-uniform motion

km/h ⇄ m/s

Units of speed and how to convert them; uniform versus non-uniform motion.

The idea

Speed means distance covered in one unit of time. So its unit is always “unit of distance ÷ unit of time”: km/h, m/s, m/min. If the question gives distance in km and time in seconds, bring them to one system first — this is the most common slip.

◭ Diagram

The bridge between km/h and m/s

From	To	Multiply by	Example
km/h	m/s	$\frac{5}{18}$	$72 \to 20$
m/s	km/h	$\frac{18}{5}$	$15 \to 54$
m/min	km/h	$\frac{3}{50}$	$250 \to 15$
km/h	m/min	$\frac{50}{3}$	$12 \to 200$
m/s	m/min	60	$4 \to 240$
knot (nautical mile/h)	km/h	1.852	$10 \to 18.52$

Here: the factor is what you multiply the old number by to get the number in the new unit. 1 nautical mile $= 1852$ m; 1 (land) mile $\approx 1.609$ km, so 1 mph $\approx 0.447$ m/s.

Uniform motion means equal distances in equal times (steady speed). Non-uniform motion means unequal distances in equal times (the speed changes). Everyday journeys are almost always non-uniform, so we use the average speed = total distance ÷ total time.

∑ Formula

S = \frac{D}{T} D = S T T = \frac{D}{S}

Speed, distance, time

$S$: speed (km/h or m/s)
$D$: distance covered (km or m)
$T$: time taken (h or s)

Spot itMatch the unit trio: km ↔ h ↔ km/h, or m ↔ s ↔ m/s. If they are mixed, convert first.

Steps

Write the units of distance, time and speed and check whether they match.
If they do not match, apply the factor from the table (decide first whether the answer should be bigger or smaller).
Now apply $D = ST$ .
Give the answer in the unit asked.

Worked examples

Example 3 · Basic

Convert 15 m/s into km/h.

Solution, step by step

1Going the other way we use the reverse of $\frac{5}{18}$ , i.e. $\frac{18}{5}$ (1 m/s $= 3.6$ km/h).
215 $\times \frac{18}{5} = 54$ km/h.

Answer 54 km/h

Handy: 36 km/h = 10 m/s, 54 = 15, 72 = 20, 90 = 25, 108 = 30.

Example 4 · Advanced

A trolley covers 12, 20, 28, 40 m in 4 successive intervals of 5 s each. Find the average speed over the whole trip.

Solution, step by step

1Speed in each interval $= \frac{distance}{5}$ : $2.4$ , $4$ , $5.6$ , $8$ m/s. The speed keeps changing, so the motion is non-uniform.
2Total distance $= 100$ m, total time $= 4 \times 5 = 20$ s.
3Average speed $= \frac{100}{20} = 5$ m/s.

Answer 5 m/s

Practice (try it yourself)

BasicTry it yourself first

Convert 126 km/h into m/s.

BasicTry it yourself first

Convert 450 m/min into km/h.

StandardTry it yourself first

Convert 15 knots into km/h. (1 knot

= 1.852

km/h)

StandardTry it yourself first

A rocket climbs 2 km in 40 seconds at a uniform rate. Find its speed in m/s and in km/h.

AdvancedTry it yourself first

Kishan first moves at 36 km/h for 10 minutes and then at 8 m/s for 15 minutes. How many metres did he cover in all and what was his average speed in km/h?

AdvancedTry it yourself first

A trolley covers 18, 36, 63 m in 3 successive intervals of 6 s each. What was his speed in the last interval, in km/h?

Graphs of motion

slope = speed, area = distance

Reading distance-time and speed-time graphs.

The idea

A graph lets you read motion like a picture. In a distance-time graph time is along the horizontal axis and distance along the vertical axis. The steeper the line, the greater the speed; a horizontal line means the body is at rest.

Shape of the graph	In a distance-time graph	In a speed-time graph
horizontal line	at rest (speed 0)	steady speed
straight sloping line	steady speed; slope = speed	steady acceleration; slope = acceleration
curve bending up	speed rising (acceleration)	acceleration rising
falling line	position falling (returning)	speed falling (retardation)
area under the line	no meaning here	distance covered

∑ Formula

speed = \frac{Δ d}{Δ t} (d-t slope), a = \frac{Δ v}{Δ t} (v-t slope), d = area

Three rules for reading graphs

$Δ d$: change in distance (km or m)
$Δ t$: change in time (h or s)
$a$: acceleration (m/s²)
$Δ v$: change in speed (m/s)

◭ Diagram

A journey: moving, resting, then moving faster. The slope of each coloured part is its speed

◭ Diagram

Speed-time graph: rising slope is acceleration, flat part steady speed, falling slope retardation; the area below is the distance

TrapWhen reading a slope from a distance-time graph, turn the time into hours (if the axis is in minutes use

Δ t /60

). In a speed-time graph the unit of area is (unit of speed) × (unit of time).

Steps

Read the (time, distance) points at the two ends of the part.
For speed take $Δ d$ ÷ $Δ t$ (in hours); for distance (speed-time graph) take the area, splitting the shape into rectangles and triangles.
For average speed take total distance ÷ total time (the halt included).

Worked examples

Example 5 · Basic

The distance-time graph of Amit’s journey is a broken line through (0 min, 0 km), (30 min, 20 km), (50 min, 20 km), (110 min, 80 km). What is his speed in part 3?

Solution, step by step

1The slope $= \frac{change in distance}{change in time}$ is the speed (with the time in hours).
2Part 3: 60 km in 60 min $\Rightarrow$ $60$ km/h

Answer 60 km/h

Example 6 · Advanced

A train leaves a station, reaches 72 km/h in 10 s, runs at that speed for 50 s, then stops in 20 s (its speed-time graph is a trapezium). Find the average speed of the whole run in km/h.

Solution, step by step

1Speed in m/s: 72 $\times \frac{5}{18} = 20$ m/s, because the time is in seconds.
2Areas of the three parts: $\frac{1}{2} \times 10 \times 20 = 100$ , $50 \times 20 = 1, 000$ , $\frac{1}{2} \times 20 \times 20 = 200$ m.
3Total distance $= 1, 300$ m, total time $= 80$ s, average speed $= \frac{1 , 300}{80} = 16.25$ m/s.
4In km/h: $16.25 \times \frac{18}{5} = 58.5$ km/h.

Answer 58.5 km/h

The average speed is below the top speed because starting and stopping are slow parts.

Practice (try it yourself)

BasicTry it yourself first

The distance-time graph of Imran’s journey is a broken line through (0 min, 0 km), (40 min, 30 km), (100 min, 30 km), (160 min, 90 km). What is his speed in part 1?

BasicTry it yourself first

The distance-time graph of Kishan’s journey is a broken line through (0 min, 0 km), (40 min, 30 km), (100 min, 30 km), (160 min, 90 km). For how long did he rest?

StandardTry it yourself first

The distance-time graph of Ravi’s journey is a broken line through (0 min, 0 km), (20 min, 12 km), (50 min, 12 km), (80 min, 42 km). Find the average speed of the whole journey, including the halt.

StandardTry it yourself first

A train leaves a station, reaches 12 m/s in 6 s, runs at that speed for 14 s, then stops in 10 s (its speed-time graph is a trapezium). What is the acceleration while starting and the retardation while stopping? (in m/s²)

AdvancedTry it yourself first

The distance-time graph of Deepak’s journey is a broken line through (0 min, 0 km), (45 min, 36 km), (75 min, 12 km). (The graph shows position measured from the start of the road.) Find the average speed from the total path and the magnitude of the average velocity from the final position.

AdvancedTry it yourself first

A train leaves a station, reaches 72 km/h in 12 s, runs at that speed for 30 s, then stops in 18 s (its speed-time graph is a trapezium). What total distance did it cover?

Acceleration and the three equations of motion

v = u + a t

s = u t + \frac{1}{2} a t^{2}

v^{2} = u^{2} + 2 a s

Acceleration, retardation, the sign rule and the three equations.

The idea

When the speed changes the motion is accelerated. Acceleration tells by how much the velocity changes every second. 2 m/s² means the velocity grows by 2 m/s each second. When the velocity falls, the same quantity is negative and is called retardation (deceleration). Acceleration is a vector too; along a straight line its direction is shown by its sign.

◭ Diagram

Uniform acceleration: the speed goes from u to v and the displacement in time t is s

∑ Formula

a = \frac{v - u}{t}

Definition of acceleration

$a$: acceleration (m/s²)
$u$: initial velocity (m/s)
$v$: final velocity (m/s)
$t$: time (s)

When the acceleration is constant, three equations hold. Remember them with the variable each one leaves out. Sign rule: take the direction of the starting motion as positive. For a retardation $a$ is negative; on stopping $v = 0$ .

∑ Formula

v = u + a t s = u t + \frac{1}{2} a t^{2} v^{2} = u^{2} + 2 a s s = \frac{u + v}{2} t

Equations for constant acceleration

$s$: displacement (m)
$u$: initial velocity (m/s)
$v$: final velocity (m/s)
$a$: acceleration (m/s²); negative for retardation
$t$: time (s)

Equation	Does not contain	Choose it when
$v = u + a t$	$s$	distance is not involved
$s = u t + \frac{1}{2} a t^{2}$	$v$	final speed is not involved
$v^{2} = u^{2} + 2 a s$	$t$	time is not involved
$s = \frac{u + v}{2} t$	$a$	acceleration is not involved
$s_{n} = u + \frac{a}{2} (2 n - 1)$		distance in the $n$ th second only

How to chooseChoosing the equation: of the five quantities

u, v, a, s, t

three are known and two are wanted. The one neither given nor asked is the missing one; pick the equation that leaves it out.

Steps

Write every quantity in SI units (turn km/h into m/s).
Choose the direction: motion positive; the $a$ of a retardation negative.
Spot the quantity that is neither given nor asked and pick the matching equation.
Solve, write the answer with its unit and check that it is sensible.

Worked examples

Example 7 · Basic

A scooter moving at 4 m/s speeds up uniformly at 2 m/s² for 8 s. Find the final speed and the distance covered.

Solution, step by step

1Take the direction of motion as positive: $u$ , $a$ and $t$ are all positive.
2Using $v = u + a t$ : $v = 4 + 2 \times 8 = 20$ m/s. (The speed grows by 2 m/s each second, so by 16 in 8 s.)
3Using $s = u t + \frac{1}{2} a t^{2}$ : $s = 4 \times 8 + \frac{1}{2} \times 2 \times 8^{2} = 32 + 64 = 96$ m. The first term is the distance at the steady speed $u$ , the second the extra gained from the rising speed.

Answer 20 m/s and 96 m

Example 8 · Standard

Gurpreet’s car runs at 72 km/h. On braking it stops with a uniform retardation of 4 m/s². What distance and time does it need to stop?

Solution, step by step

1The time is in seconds, so convert: 72 $\times \frac{5}{18} = 20$ m/s.
2At rest the final speed is $v = 0$ and retardation is negative: $a = - 4$ . From $v^{2} = u^{2} + 2 a s$ : $0 = 400 - 2 \times 4 \times s$ , so $s = \frac{400}{8} = 50$ m.
3From $v = u + a t$ : $0 = 20 - 4 t$ , so $t = 5$ s.

Answer 50 m, 5 s

Trap: doubling the speed makes the stopping distance four times (

s \propto u^{2}

Example 9 · Advanced

A ball starts rolling down a slope at 3 m/s and speeds up at 2 m/s². How far does it go in the 7th second alone?

Solution, step by step

1Subtract the distance in $(n - 1)$ seconds from that in $n$ seconds. In 7 s: $3 \times 7 + \frac{1}{2} \times 2 \times 7^{2} = 70$ m.
2In 6 s: $3 \times 6 + \frac{1}{2} \times 2 \times 6^{2} = 54$ m. Difference $= 70 - 54 = 16$ m.
3Direct formula: $s_{n} = u + \frac{a}{2} (2 n - 1) = 3 + 1 \times 13 = 16$ m.

Answer 16 m

Trap: the distance in the

n

th second and in

n

seconds are different things.

Practice (try it yourself)

BasicTry it yourself first

A scooter moving at 6 m/s speeds up uniformly at 1.5 m/s² for 10 s. What is its final speed?

BasicTry it yourself first

A cyclist moving at 2 m/s speeds up uniformly at 3 m/s² for 6 s. How far does it travel in this time?

StandardTry it yourself first

Suresh’s car runs at 54 km/h. On braking it stops with a uniform retardation of 3 m/s². What distance and time does it need to stop?

StandardTry it yourself first

A bus starts from rest, accelerates at 2 m/s² for 8 s and then cruises at that speed for 12 s. What total distance does it cover?

AdvancedTry it yourself first

A train speeds up from 12 m/s to 28 m/s over 80 m. What is its acceleration?

AdvancedTry it yourself first

A ball starts rolling down a slope at 5 m/s and speeds up at 4 m/s². How far does it go in the 9th second alone?

Free fall and throwing upward

g = 10

9.8

m/s²

Motion under gravity alone: falling, throwing up, the $n$ th second.

The idea

When a body moves under gravity alone (air resistance ignored) it is in free fall. Near the Earth every body — light or heavy — has the same acceleration: the acceleration due to gravity $g \approx 9.8$ m/s² downward (exams often take 10). So these are the same three equations with $a = \pm g$ .

◭ Diagram

A ball thrown up: equal time up and down; equal speed at the same height

Situation	$u$	Formula
dropped from height $h$	0	$t = \frac{2 h}{g}, v = 2 g h$
thrown straight up	$u$	$H = \frac{u ^{2}}{2 g}, t_{u p} = \frac{u}{g}, T = \frac{2 u}{g}$
after $t$ s (up positive)	$u$	$s = u t - \frac{1}{2} g t^{2}, v = u - g t$
$n$ th second from rest	0	$s_{n} = \frac{g}{2} (2 n - 1)$

Here: $h$ = height of release, $H$ = maximum height, $T$ = total time of flight, $t_{u p}$ = time to rise, $u$ = speed of throw, $v$ = speed at that moment.

TrapOn the way down the object has the same speed at each height as on the way up, just reversed. At the top the velocity is zero, not the acceleration — that is still

g

Steps

Choose the positive direction (up or down) and fix $a = \pm g$ .
Spot $u$ : 0 if dropped, the given speed if thrown; at the top $v = 0$ .
Pick the right equation of motion and solve.

Worked examples

Example 10 · Basic

A stone from a bridge is dropped, without a push, from a height of 80 m. (Take g = 10 m/s².) How long does it take to hit the ground and what is its speed on impact?

Solution, step by step

1Dropping means $u = 0$ ; take downward as positive, $a = g$ . From $h = \frac{1}{2} g t^{2}$ : $t = \frac{2 h}{g} = \frac{2 \times 80}{10} = 4$ s.
2Impact speed $v = g t = 10 \times 4 = 40$ m/s. (Check: $v^{2} = 2 g h = 1, 600$ .)

Answer 4 s, 40 m/s

The time depends only on the height, not on the weight of the object (ignoring air).

Example 11 · Advanced

A stone starts falling from rest. (Take g = 9.8 m/s².) How far does it fall in the 4th second?

Solution, step by step

1With $u = 0$ : $s_{n} = \frac{g}{2} (2 n - 1) = \frac{9.8}{2} \times 7 = 34.3$ m.
2Check: in 4 s $\frac{1}{2} g n^{2} = 78.4$ , in 3 s $44.1$ ; difference $= 34.3$ m.

Answer 34.3 m

Galileo’s law: from rest the distances in successive seconds are

1 : 3 : 5 : 7 \dots

Practice (try it yourself)

BasicTry it yourself first

An iron ball from a tower is dropped, without a push, from a height of 45 m. (Take g = 10 m/s².) How long does it take to hit the ground and what is its speed on impact?

BasicTry it yourself first

Kishan throws a ball straight up at 20 m/s. (Take g = 10 m/s².) How high does it rise and what is the total time from throw to return to his hand?

StandardTry it yourself first

An apple falls from a tree and hits the ground at 30 m/s. (Take g = 10 m/s².) From what height did it fall, and in what time?

StandardTry it yourself first

A stone is thrown straight up at 25 m/s. (Take g = 10 m/s².) What are its height and velocity after 2 s? (State the direction of the velocity too.)

AdvancedTry it yourself first

An iron ball from a tower is dropped, without a push, from a height of 78.4 m. (Take g = 9.8 m/s².) How long does it take to hit the ground and what is its speed on impact?

AdvancedTry it yourself first

A stone is thrown straight up at 40 m/s. (Take g = 10 m/s².) What are its height and velocity after 6 s? (State the direction of the velocity too.)

Part 2 — Relative speed along a straight line

add and subtract · meeting · chasing · walkway · the bird

When all motions lie on one straight line, relative speed simply adds or subtracts.

🔒

This part is in All Access.

With the pass, every example, practice question and solution in this chapter opens.

See the pass · ₹100.00 Already have a pass? Sign in

Relative speed: add and subtract

same way: subtract · towards each other: add

Relative speed of two moving things, the gap growing and shrinking, and passing long objects.

🔒

This part is in All Access.

With the pass, every example, practice question and solution in this chapter opens.

See the pass · ₹100.00 Already have a pass? Sign in

Meeting from two places

time and place of meeting

Meeting after starting from two places, together or at different times.

🔒

This part is in All Access.

With the pass, every example, practice question and solution in this chapter opens.

See the pass · ₹100.00 Already have a pass? Sign in

Head start and chasing

same road, one starts later

When one body starts first and a faster one tries to catch it.

🔒

This part is in All Access.

With the pass, every example, practice question and solution in this chapter opens.

See the pass · ₹100.00 Already have a pass? Sign in

Walkways, staircases and walking inside a train

ground = belt + person

When someone walks on something that itself moves: the two velocities add.

🔒

This part is in All Access.

With the pass, every example, practice question and solution in this chapter opens.

See the pass · ₹100.00 Already have a pass? Sign in

The bird, the dog and the messenger

total time = time to meet

The distance of something running back and forth: find the time just once.

🔒

This part is in All Access.

With the pass, every example, practice question and solution in this chapter opens.

See the pass · ₹100.00 Already have a pass? Sign in

Part 3 — Circular tracks and clock hands

laps ·

n

th meeting · the clock

When the path is a circle, the gap is one lap and the laps keep repeating.

🔒

This part is in All Access.

With the pass, every example, practice question and solution in this chapter opens.

See the pass · ₹100.00 Already have a pass? Sign in

The first meeting on a circular track

\frac{L}{v _{1} \pm v _{2}}

The first meeting of two runners going the same or opposite ways.

🔒

This part is in All Access.

With the pass, every example, practice question and solution in this chapter opens.

See the pass · ₹100.00 Already have a pass? Sign in

$n$ th meeting, meeting at the start and meeting points

interval · LCM · ratio

Meeting again and again on a circle: when, where and how often.

🔒

This part is in All Access.

With the pass, every example, practice question and solution in this chapter opens.

See the pass · ₹100.00 Already have a pass? Sign in

Clock hands: a race on a circle

minute hand 6°/min, hour hand 0.5°/min

The two hands are runners going the same way round a circular track.

🔒

This part is in All Access.

With the pass, every example, practice question and solution in this chapter opens.

See the pass · ₹100.00 Already have a pass? Sign in

Part 4 — Relative velocity in two dimensions

vector subtraction · boat · aircraft · rain

When the motions are in different directions, direction becomes part of the relative velocity.

🔒

This part is in All Access.

With the pass, every example, practice question and solution in this chapter opens.

See the pass · ₹100.00 Already have a pass? Sign in

Velocity is a vector: vector subtraction

v_{A B} = v_{A} - v_{B}

Components, the velocity of $A$ relative to $B$ , magnitude and direction.

🔒

This part is in All Access.

With the pass, every example, practice question and solution in this chapter opens.

See the pass · ₹100.00 Already have a pass? Sign in

Bodies moving at a right angle: how the gap grows

relative speed

a^{2} + b^{2}

Two bodies leaving one point along perpendicular directions such as east and north.

🔒

This part is in All Access.

With the pass, every example, practice question and solution in this chapter opens.

See the pass · ₹100.00 Already have a pass? Sign in

Velocities at an angle: the cosine rule

a^{2} + b^{2} - 2 ab cos θ

When the angle between two velocities is something other than 90°.

🔒

This part is in All Access.

With the pass, every example, practice question and solution in this chapter opens.

See the pass · ₹100.00 Already have a pass? Sign in

Boat and river: quickest crossing and shortest path

v_{bank} = u + w

Velocity relative to the water and relative to the bank when crossing a river.

🔒

This part is in All Access.

With the pass, every example, practice question and solution in this chapter opens.

See the pass · ₹100.00 Already have a pass? Sign in

Aircraft and wind

ground velocity = air velocity + wind velocity

An aircraft flies through air that itself moves: the two velocities add.

🔒

This part is in All Access.

With the pass, every example, practice question and solution in this chapter opens.

See the pass · ₹100.00 Already have a pass? Sign in

Rain and umbrella

apparent rain

=

rain

-

man

Why rain looks slanted to a walking man or a moving train.

🔒

This part is in All Access.

With the pass, every example, practice question and solution in this chapter opens.

See the pass · ₹100.00 Already have a pass? Sign in

Closest approach and collision

d_{m i n} = \frac{∣ b x _{0} - a y _{0} ∣}{a ^{2} + b ^{2}}

How near two moving bodies come: the perpendicular on the relative path.

🔒

This part is in All Access.

With the pass, every example, practice question and solution in this chapter opens.

See the pass · ₹100.00 Already have a pass? Sign in

Part 5 — Relative motion with acceleration

overtaking · braking · falling objects

When speeds are not steady, relative velocity and relative acceleration change with time.

🔒

This part is in All Access.

With the pass, every example, practice question and solution in this chapter opens.

See the pass · ₹100.00 Already have a pass? Sign in

A car with acceleration overtaking

s_{rel} = u_{rel} t + \frac{1}{2} a_{rel} t^{2}

A car starting from rest or speeding up catching a steadily moving one.

🔒

This part is in All Access.

With the pass, every example, practice question and solution in this chapter opens.

See the pass · ₹100.00 Already have a pass? Sign in

Braking and avoiding a collision

safe gap

\geq \frac{( u _{1} - u _{2} ) ^{2}}{2 a}

The rear vehicle’s brake: the distance needed to bring the relative speed to zero.

🔒

This part is in All Access.

With the pass, every example, practice question and solution in this chapter opens.

See the pass · ₹100.00 Already have a pass? Sign in

Objects falling from a moving vehicle

horizontal and vertical motions are independent

A packet from an aircraft and a ball tossed in a train: two different pictures from two frames.

🔒

This part is in All Access.

With the pass, every example, practice question and solution in this chapter opens.

See the pass · ₹100.00 Already have a pass? Sign in

Part 6 — The physics link: frames of reference

frame of reference · inertial frame · Galilean addition

The physics behind relative velocity.

🔒

This part is in All Access.

With the pass, every example, practice question and solution in this chapter opens.

See the pass · ₹100.00 Already have a pass? Sign in

Frames of reference and Galilean addition of velocities

v_{A C} = v_{A B} + v_{BC}

Relative to what do we measure motion? Inertial frames and why velocities add.

🔒

This part is in All Access.

With the pass, every example, practice question and solution in this chapter opens.

See the pass · ₹100.00 Already have a pass? Sign in

Summary, mixed practice, revision

All formulas in one place

One table of all formulas, exam-style mixed questions and revision.

Summary of all formulas

Which formula when

One page to pick the formula as soon as you read a question.

🔒

This part is in All Access.

With the pass, every example, practice question and solution in this chapter opens.

See the pass · ₹100.00 Already have a pass? Sign in

Mixed practice — set 1

Basic to standard

All topics mixed; spot the method first, then solve.

🔒

This part is in All Access.

With the pass, every example, practice question and solution in this chapter opens.

See the pass · ₹100.00 Already have a pass? Sign in

Mixed practice — set 2

Standard to advanced

Mixed questions from the hard end of the exam.

🔒

This part is in All Access.

With the pass, every example, practice question and solution in this chapter opens.

See the pass · ₹100.00 Already have a pass? Sign in

Revision — cards and traps

Flip and recall

Read the front, think, then flip.

🔒

This part is in All Access.

With the pass, every example, practice question and solution in this chapter opens.

See the pass · ₹100.00 Already have a pass? Sign in