Time, Speed & Distance

Scenario

Swiggy promises 10-minute delivery. The rider covers 3 km. Half the route is traffic (10 km/hr), half is open road. Can they make it?

Is the average speed simply (10 + open-road speed) / 2?

Mental model

Distance is the output. Speed is the rate. Time is the duration — three locked numbers; fix any two to find the third.

D = S x T. They are perfectly locked: if distance is constant, speed and time are inversely proportional (double the speed, half the time). Most TSD problems are just choosing the right form of this one equation.

Formula

D = S x T. Speed = D/T. Time = D/S. Average speed for equal distances at speeds x and y: Avg = 2xy/(x+y). Late/early setup: D/S_slow - D/S_fast = time_difference (in same units). Unit conversions: km/hr x (5/18) = m/s.

The harmonic mean 2xy/(x+y) applies ONLY when distances are equal. For equal times, use simple average (x+y)/2. The late/early equation sets up a ratio of distances equal to a ratio of time differences.

Worked example

Problem: A cyclist rides from home to station at 12 km/hr and arrives 8 min early. If he rides at 8 km/hr he arrives 4 min late. Find the distance.

1. Set up: Let distance = D km. Total time difference = 8 + 4 = 12 min = 12/60 = 1/5 hr.
2. At 8 km/hr he takes D/8 hours. At 12 km/hr he takes D/12 hours.
3. Slower minus faster = total difference: D/8 - D/12 = 1/5.
4. LCM of 8 and 12 is 24: (3D - 2D) / 24 = 1/5 => D/24 = 1/5 => D = 24/5 = 4.8 km.
5. Sanity-check: At 8 km/hr: 4.8/8 = 0.6 hr. At 12 km/hr: 4.8/12 = 0.4 hr. Difference = 0.2 hr = 12 min. Correct.

Answer: 4.8 km.

Problem: A car covers half its journey at 60 km/hr and the rest at 40 km/hr. Find the average speed for the full trip.

1. Set up: Two equal distances at different speeds — apply the harmonic mean formula.
2. Apply: Avg speed = 2xy/(x+y) = (2 x 60 x 40) / (60 + 40) = 4800 / 100.
3. Compute: 48 km/hr.
4. Sanity-check: If total distance = 120 km (60 each half). Time at 60 km/hr: 1 hr. Time at 40 km/hr: 1.5 hr. Total = 2.5 hr. Avg = 120/2.5 = 48 km/hr. Matches.

Answer: 48 km/hr.

Shortcuts

• km/hr to m/s Conversion: Multiply by 5/18 to go km/hr -> m/s. Multiply by 18/5 to go m/s -> km/hr. (72 km/hr = 72 x (5/18) = 20 m/s.)
• Average Speed for Equal Distances: If going at speed x and returning at speed y over the SAME distance, Avg Speed = 2xy/(x+y). Never (x+y)/2. (Go at 40 km/hr, return at 60 km/hr. Avg = (2 x 40 x 60) / 100 = 48 km/hr (NOT 50!).)

Common mistakes

• Simple Average of Speeds: Averaging speeds (x+y)/2 only works if they traveled for equal TIME, not equal DISTANCE. For equal distances, always use the harmonic mean 2xy/(x+y).
• Unit Mismatch: Multiplying km/hr by minutes without converting minutes to hours first. Convert everything to consistent units before plugging into D = S x T.

Glossary

duration

The amount of time that something lasts or takes to happen.

proportional

When one thing changes at a steady rate compared to another thing changing.

harmonic

A special kind of average used when things like speeds happen over equal distances.

Recall questions

• Convert 90 km/hr to m/s.
• If speed is doubled for a fixed distance, what happens to the time taken?
• When should you use 2xy/(x+y) vs (x+y)/2 for average speed?

Questions & answers

A person walks from his house at 4 km/hr and reaches office 5 minutes late. If his speed is 5 km/hr, he reaches 10 minutes early. What is the distance to the office?

5 km

Approach: Difference in time = 15 min = 1/4 hr. D/4 - D/5 = 1/4. D/20 = 1/4 => D = 5 km.

A man covers half his journey at 6 km/hr and the remaining half at 3 km/hr. What is his average speed?

4 km/hr

Approach: Equal distances. Average speed = 2xy/(x+y) = (2 x 6 x 3) / (6 + 3) = 36 / 9 = 4 km/hr.

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