Unit 4: Time, Speed and Moving Objects

I. Orientation — Governing Relationship

Motion problems connect three quantities: distance travelled, time taken, and speed of travel. Speed measures the rate at which distance changes with time. Most questions in this unit are solved by expressing all quantities in compatible units, identifying whether speeds are constant or variable, and applying proportional or relative-motion relationships.

• Defining relationship: Distance equals speed multiplied by time.

D = S × T
S = D / T
T = D / S

• D = distance travelled.
• S = speed of the moving object.
• T = time taken.

• Standard units: Distance may be measured in metres (m) or kilometres (km); time in seconds (s), minutes, or hours (h); speed in m/s or km/h.
• Uniform motion assumption: Unless variation is stated, speed is treated as constant throughout a specified part of the journey.
• Unit consistency: The units of distance, speed, and time must belong to the same system before a formula is applied.
• Reference frame: Relative-motion questions measure one object's velocity from the viewpoint of another object.
• Direction convention: Speeds in the chosen positive direction may be treated as positive, while speeds in the opposite direction may be treated as negative.

II. Foundations of Motion — Quantities, Units, and Central Measures

A. Concept of time, speed and distance

Time, speed, and distance form the basic quantitative model for describing motion.

• Time: Time measures the duration of motion; for example, 2.5 hours = 2 hours 30 minutes.
• Distance: Distance is the total path covered and is a scalar quantity, so it has magnitude but no direction.
• Speed: Speed is distance travelled per unit time; a car covering 150 km in 3 h has speed 50 km/h.
• Direct relationship: At fixed time, distance is directly proportional to speed.

D1 / D2 = S1 / S2        when T is constant

• D1, D2 = distances covered.
• S1, S2 = corresponding speeds.

• Inverse relationship: For a fixed distance, time is inversely proportional to speed.

T1 / T2 = S2 / S1        when D is constant

• T1, T2 = times taken.
• S1, S2 = corresponding speeds.

• Journey decomposition: For a journey completed in stages, calculate each stage using Dᵢ = SᵢTᵢ, then add the relevant distances or times.

B. Conversion of units and proportionality

Unit conversion places all measurements on a common scale, while proportionality provides shortcuts for comparing journeys.

• Kilometres per hour to metres per second: Multiply by 5/18.

S(m/s) = S(km/h) × 5/18

• Metres per second to kilometres per hour: Multiply by 18/5.

S(km/h) = S(m/s) × 18/5

• Reason for the conversion: Since 1 km = 1000 m and 1 h = 3600 s, then 1 km/h = 1000/3600 m/s = 5/18 m/s.
• Time conversions: 1 h = 60 min = 3600 s; therefore, 15 min = 1/4 h.
• Direct proportion: If speed is fixed, doubling time doubles distance: D ∝ T.
• Inverse proportion: If distance is fixed, increasing speed in the ratio a:b changes time in the ratio b:a.
• Worked example: Convert 72 km/h to m/s.

72 × 5/18 = 20 m/s

C. Average speed concept

Average speed is total distance divided by total time, not generally the arithmetic mean of the given speeds.

• General formula:

Average speed = Total distance / Total time
Savg = (D1 + D2 + ... + Dn) / (T1 + T2 + ... + Tn)

• Savg = average speed.
• D1 ... Dn = distances of individual stages.
• T1 ... Tn = corresponding times.

• Equal time intervals: If speeds S1 and S2 are maintained for equal times, their average is the arithmetic mean.

Savg = (S1 + S2) / 2

• Equal distances: If the same distance is covered at speeds S1 and S2, average speed is the harmonic mean.

Savg = 2S1S2 / (S1 + S2)

• Stops and delays: Waiting time is included in total time when the average speed is measured over the complete trip.
• Worked example: A vehicle covers 60 km at 30 km/h and another 60 km at 60 km/h. Total time is 2 + 1 = 3 h, so average speed is 120/3 = 40 km/h, not 45 km/h.

III. Structured Motion Problems — Changes, Ratios, and Competition

A. Advanced time and speed based questions

Advanced problems usually describe a change in speed that produces an early or late arrival over a fixed distance.

• Fixed-distance principle: When the route remains unchanged, equate the distance expressions for the different conditions.

S1T1 = S2T2

• Delayed arrival: If travelling at speed S1 causes a delay of a hours, actual travel time is scheduled time plus a.
• Early arrival: If travelling at speed S2 causes arrival b hours early, actual travel time is scheduled time minus b.
• Combined model: If t is the scheduled journey time and D is fixed:

D = S1(t + a) = S2(t - b)

• a = delay.
• b = time saved.
• t = scheduled time.

• Percentage speed change: For fixed distance, a speed increase of p% changes time by p/(100 + p) × 100%.
• Worked example: Increasing speed from 40 to 50 km/h reduces the time for 200 km from 5 h to 4 h, saving 1 h.

B. Ratio based problems

Ratio methods compare motion quantities without requiring their exact numerical values.

• Same time: Distances are in the ratio of speeds.

D1 : D2 = S1 : S2

• Same distance: Times are in the inverse ratio of speeds.

T1 : T2 = S2 : S1

• General relationship: Since D = ST, the distance ratio is:

D1 / D2 = (S1 / S2) × (T1 / T2)

• Percentage interpretation: If speeds are in the ratio 5:4, the first speed is 25% greater than the second because (5 - 4)/4 × 100 = 25%.
• Time-saving interpretation: At a fixed distance, increasing speed from ratio 4 to 5 reduces time from ratio 5 to 4, a reduction of 1/5 = 20%.
• Worked example: Two cyclists ride for equal times at speeds in the ratio 3:5; their distances are also in the ratio 3:5.

C. Races

Race problems compare competitors over a fixed course by measuring differences in distance or finishing time.

• Distance lead: In a D-metre race, if A beats B by x metres, then when A covers D, B covers D - x.

Speed of A / Speed of B = D / (D - x)

• Time lead: If A finishes t seconds before B, B's finishing time exceeds A's by t.
• Head start: If B receives a head start of h metres in a D-metre race, B needs to cover only D - h.
• Dead heat condition: A head start is fair when both competitors reach the finish simultaneously.
• Successive comparisons: If A:B = a:b and B:C = c:d, equalise B's terms before finding A:C.
• Worked example: In a 100 m race, A beats B by 20 m; their speed ratio is 100:80 = 5:4.

IV. Relative Motion — Movement Between Observers

A. Relative speed concept

Relative speed is the rate at which the distance between two moving objects changes.

1. Opposite directions: Speeds are added because the separation closes or grows through both motions.

Relative speed = S1 + S2

2. Same direction: The smaller speed is subtracted from the larger because one object gains only by the difference.

Relative speed = |S1 - S2|

• Meeting time: If initial separation is D, then:

T = D / Relative speed

• Velocity form: Relative velocity of A with respect to B is vA/B = vA - vB; signs represent direction.
• Stationary reference: If B is stationary, A's relative speed with respect to B equals A's ordinary speed.
• Worked example: Two vehicles 210 km apart approach each other at 60 km/h and 45 km/h; meeting time is 210/(60 + 45) = 2 h.

B. Application based questions on relative speed

Relative speed simplifies train, platform, overtaking, meeting, and separation situations.

• Crossing a point object: A train crossing a pole or person covers its own length.

T = L / Srel

• L = train length.
• Srel = relative speed.

• Crossing an extended object: A train crossing a platform covers the sum of train and platform lengths.

T = (Ltrain + Lplatform) / S

• Two trains crossing: The distance required for complete crossing is L1 + L2; use summed speeds in opposite directions and speed difference in the same direction.
• Overtaking: A faster object catches a slower one using T = initial gap/(Sf - Ss).
• Circular tracks: For circumference C, first meeting time is C/(S1 + S2) in opposite directions and C/|S1 - S2| in the same direction.
• Worked example: A 150 m train moving at 54 km/h = 15 m/s crosses a 90 m platform in (150 + 90)/15 = 16 s.

V. Motion in Currents and Algebraic Models

A. Downstream and upstream

Boat-and-stream problems combine the boat's speed in still water with the speed of the current.

1. Downstream motion: The current assists the boat, so the speeds are added.

Sd = B + C

2. Upstream motion: The current opposes the boat, so its speed is subtracted.

Su = B - C

• Sd = downstream speed.

• Su = upstream speed.

• B = boat speed in still water.

• C = current speed.

• Recovering component speeds:

B = (Sd + Su) / 2
C = (Sd - Su) / 2

• Time calculation: For distance D, downstream time is D/Sd, while upstream time is D/Su.
• Feasibility condition: The boat must satisfy B > C to make upstream progress.
• Worked example: If downstream and upstream speeds are 15 km/h and 9 km/h, then boat speed is 12 km/h and current speed is 3 km/h.

B. Two variable problems

Two-variable problems translate two independent motion conditions into simultaneous equations for two unknown quantities.

• Choice of variables: Let the required unknowns be quantities such as distance D and normal speed S, or boat speed B and current speed C.
• Equation formation: Each distinct condition must produce one independent equation, usually based on D = ST.
• Fixed-distance model: If two speeds produce two travel times:

D = S1T1
D = S2T2

• Changed-speed model: If normal time is t, a delay is a, and an early arrival is b:

D = S1(t + a)
D = S2(t - b)

• Solution methods: Use substitution when one variable is easily isolated; use elimination when corresponding coefficients can be matched.
• Validation: Substitute the values into both original conditions and reject negative speed, distance, or time values.
• Worked example: A journey takes 6 h at speed S and 5 h at speed S + 10. Since the distance is fixed, 6S = 5(S + 10), giving S = 50 km/h and D = 300 km.