Time speed distance

1. Basic relationship and mathematical assumptions

Time–speed–distance problems describe how quickly an object covers a length. Speed is a scalar quantity: it measures distance travelled per unit time without specifying direction. Distance is the actual length of the path, whereas displacement describes the change in position. CSAT arithmetic normally uses distance and speed. A person returning to the starting point has zero displacement but a positive distance travelled, so average speed is not zero.

The fundamental relationship is Distance = Speed × Time. It applies directly when speed is constant. For a journey with changing speeds, divide the journey into constant-speed segments, calculate each segment separately, and then combine the distances and times. Do not assume that a speed stated for one segment applies to the entire journey.

Before calculating, identify the starting positions, starting times, direction of movement and relevant finishing event. A meeting, an overtaking and the complete crossing of a platform are different events. Unless the question specifies acceleration or changing conditions, standard arithmetic problems generally assume uniform speeds within each stated segment.

• Write the three quantities with units before substituting into a formula.
• Use a distance–speed–time table when there are several journey segments.
• Draw a short route diagram when direction or starting position could be misunderstood.

2. Units, ratios and percentage changes

Units must be consistent throughout an equation. Since one kilometre equals 1,000 metres and one hour equals 3,600 seconds, 1 km/h equals 5/18 m/s. Thus, 54 km/h equals 15 m/s, and a vehicle travelling at this speed covers 450 metres in 30 seconds. Also distinguish decimal hours from clock notation: 1.5 hours means 1 hour 30 minutes, not 1 hour 50 minutes.

For a fixed distance, time varies inversely with speed. If two speeds are in the ratio 3:4, the corresponding times are in the ratio 4:3. For a fixed time, distances are directly proportional to speeds. These relationships often eliminate the need to calculate the actual distance. For example, increasing speed from 40 to 50 km/h changes journey time in the ratio 5:4.

If speed increases by p per cent, the percentage reduction in time for the same distance is 100p/(100 + p). If speed decreases by p per cent, the percentage increase in time is 100p/(100 − p), with p below 100. A 25 per cent speed increase therefore reduces time by 20 per cent; it does not reduce time by 25 per cent.

Arrival-time questions use the difference between journey durations. If covering distance D at speed u takes longer than at speed v, the difference is D/u − D/v. When one speed results in arriving 10 minutes late and another results in arriving 5 minutes early for the same appointment, the journey-time difference is 15 minutes. Convert this into hours before using speeds in km/h.

• For the same distance: speed ratio and time ratio are reciprocals.
• Convert minutes to hours by dividing by 60, not by 100.
• Apply percentage shortcuts only when the distance is unchanged.

3. Average speed and journeys with stoppages

Average speed is total distance divided by total time. It is not generally the arithmetic mean of the stated speeds. If equal amounts of time are spent travelling at speeds u and v, the average speed is (u + v)/2. If equal distances are covered at those speeds, the average speed is 2uv/(u + v), the harmonic mean of the two speeds.

Consider a 120 km outward journey at 40 km/h and a 120 km return journey at 60 km/h. The outward leg takes 3 hours and the return leg 2 hours. Average speed is therefore 240/5 = 48 km/h, not 50 km/h. The slower leg receives greater weight because the traveller spends more time on it. For unequal segments, calculate each duration using distance divided by speed.

Stoppages count when a question asks for average speed over the entire elapsed journey. Suppose a bus travels 180 km at a running speed of 60 km/h and stops for a total of 30 minutes. Its moving time is 3 hours, but its elapsed time is 3.5 hours. Its overall average speed is approximately 51.43 km/h.

If running speed is v and average speed including stops is a, the fraction of total elapsed time spent moving is a/v. Consequently, stoppage time per hour of elapsed journey is 60(1 − a/v) minutes. This shortcut assumes the stated running speed is uniform or represents the average speed during movement.

• Equal times: use the arithmetic mean of speeds.
• Equal distances: use the harmonic-mean formula.
• Unequal distances or durations: return to total distance divided by total time.

4. Relative speed, trains and circular tracks

Relative speed describes how quickly the separation between two moving objects changes. Two travellers moving towards each other at speeds u and v close their initial gap at u + v. If a faster traveller follows a slower one in the same direction, the gap closes at u − v. Meeting or catch-up time equals the initial gap divided by the appropriate closing speed.

A delayed start creates a head start. If a cyclist travels at 12 km/h for 30 minutes before another cyclist begins pursuit, the initial lead is 6 km. If the pursuer travels at 18 km/h, the relative speed is 6 km/h, so catching up takes 1 hour after the pursuer starts. The first cyclist has then travelled for 1.5 hours.

For train questions, identify the length that must pass the reference point. A train crossing a pole covers its own length. A train completely crossing a platform covers the train length plus the platform length. Two trains completely crossing each other cover the sum of their lengths at their relative speed. Add their speeds for opposite directions and subtract them for the same direction.

On a circular track of circumference C, two runners starting together meet again after C/|u − v| when running in the same direction at unequal speeds. In opposite directions, their first subsequent meeting occurs after C/(u + v). Meeting anywhere on the track is different from returning together to the starting point; the latter requires a common multiple of their individual lap times, expressed in compatible units.

• Measure pursuit time from the later start unless the question asks otherwise.
• For complete overtaking of trains, include both train lengths.
• For runners starting at different track positions, account for the initial gap rather than automatically using one full circumference.

5. Boats, races and an efficient CSAT approach

In boat-and-stream problems, let b be the boat's speed in still water and c the current speed. Downstream speed is b + c, while upstream speed is b − c. Therefore, b is half the sum of downstream and upstream speeds, and c is half their difference. These formulas assume motion along the stream and uniform current. Upstream progress requires b to exceed c. A freely drifting raft moves with the current.

Race questions compare distances covered during the same time interval. If A beats B by 20 metres in a 100-metre race, then when A finishes, B has covered 80 metres. Their speed ratio is 100:80, or 5:4, assuming simultaneous starts and uniform speeds. A distance head start and a time head start are not interchangeable: translate a time advantage into distance using the relevant runner's speed.

In CSAT, begin by classifying the question: fixed-distance comparison, segmented journey, pursuit, complete crossing or stream movement. Select the smallest necessary equation and simplify ratios before multiplying. Options can help through estimation or substitution, but only after the physical situation is correctly understood. Finally, check whether the result is plausible: an overall average cannot exceed the fastest segment speed, and including stoppages cannot increase average speed.

• Avoid introducing several variables when a ratio or one equation is sufficient.
• Keep fractions exact until the final step if options are close.
• Check whether the requested answer is distance, elapsed time, moving time or speed.

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