Time, Speed & Distance

Time, Speed & Distance

Concept

Speed, time, and distance are related by the fundamental equation: Distance = Speed × Time. From this single equation, you can derive everything. The trick in SSC is knowing which form to use and when. Average speed deserves special attention — many students wrongly take arithmetic mean of two speeds, but this is only correct when equal time is spent at each speed, not equal distance.

Train Problems: The most distinctive TSD question type in SSC. When a train crosses a stationary object (pole, person, tree), the distance covered equals the train’s own length. When it crosses a platform or another train, distance = sum of both lengths. Always convert speed to m/s when mixing with lengths in metres.

Conversions: 1 km/h = (1000m)/(3600s) = 5/18 m/s. Conversely, 1 m/s = 18/5 = 3.6 km/h.

Boats and Streams: Upstream speed = Boat speed in still water − Stream speed. Downstream speed = Boat speed + Stream speed. The current (stream) subtracts upstream and adds downstream.

Key Points

• When same distance d is covered at speeds a and b: Average speed = 2ab/(a+b).
• When same time t is spent at speeds a and b: Average speed = (a+b)/2.
• For boats: Downstream speed − Upstream speed = 2 × Stream speed. Downstream speed + Upstream speed = 2 × Boat speed.
• If speed changes in arithmetic progression over equal time intervals, average speed = (first + last)/2.
• For two trains moving in opposite directions: relative speed = S₁ + S₂. Same direction: |S₁ − S₂|.

Worked Example

Q: A boat goes downstream in 4 hours and upstream in 6 hours, covering a total distance of 72 km. Find the speed of the stream. Approach: Downstream speed = D, Upstream speed = U. D + U = total/??? Wait, we need individual distances. Let downstream distance = d, upstream distance = (72−d). But we only know times. D = d/4, U = (72−d)/6. Also D − U = 2 × stream speed (call s). And D + U relationship from distances: We have 2 unknowns, 2 equations. d/4 + (72−d)/6 = combined? Hmm let me use: D − U = 2s. Also from D/U = (d/4) / ((72−d)/6) = 6d / 4(72−d) = 3d / 2(72−d). Actually: D = d/4, U = (72−d)/6. And we know D + U = d/4 + (72−d)/6. But we also know D+U = (total distance)/(???). No. Let’s solve: D+U = d/4 + (72−d)/6 = (3d + 2(72−d))/12 = (3d + 144 − 2d)/12 = (d + 144)/12. And D−U = d/4 − (72−d)/6 = (3d − 2(72−d))/12 = (3d − 144 + 2d)/12 = (5d−144)/12 = 2s. And d/4 − (72−d)/6 also = (5d−144)/12. Let me just use the formula: (d+144)/12 + (5d−144)/12 = 2D? Not helpful. Let me try a different approach: Since D = d/4, U = (72-d)/6. We can find d: Actually let me set d = downstream distance. Then d/4 = downstream speed, (72-d)/6 = upstream speed. The ratio of these is: (d/4) / ((72-d)/6) = 3d/2(72-d). Without another relationship this is underdetermined. But wait — I can find stream speed directly using formula: Stream speed = (Downstream distance/time_down − Upstream distance/time_up)/2? No that’s not right. Let me solve with d: D = d/4, U = (72-d)/6. Also we know boat speed in still water b: D = b+s, U = b-s. So b = (D+U)/2, s = (D−U)/2. And d + (72-d) = 72. Now D and U expressed in terms of d. b = (d/4 + (72-d)/6)/2 = (d+144)/24. And s = (d/4 − (72-d)/6)/2 = (5d−144)/24. We need another condition… Actually, there might be a missing piece: the question gives total distance and times, we can solve for d. (d+144)/24 × 2 = b? No. Actually let me use the fact that D/U = (d/4) / ((72-d)/6) = 3d/(72-d)/2? Let me try d = 36 as a guess… Actually I’ll use the formula: D = d/4, U = (72-d)/6. If I multiply D×U = (d/4)((72-d)/6) = d(72-d)/24. This doesn’t help. Let me try solving directly: set the equation based on the ratio of D to U? We don’t know that. Let me reconsider: Actually maybe the total 72 km includes both downstream and upstream journeys separately, not round trip? If downstream 4 hours and upstream 6 hours covering 72 km total in the two directions: Let downstream speed = d/4, upstream speed = (72-d)/6. And we can also use: d/4 + (72-d)/6 = total distance / ??? Actually no. Let me just set up with b and s: Let b = boat speed in still water, s = stream speed. Then downstream speed = b+s, upstream = b-s. Distance downstream = 4(b+s), distance upstream = 6(b-s). Total = 4(b+s) + 6(b-s) = 4b+4s+6b-6s = 10b - 2s = 72. That’s one equation with two unknowns. Another equation comes from the fact that the ratio of downstream to upstream distance = 4(b+s) / 6(b-s). But we don’t know that ratio either. Hmm, the question as stated seems underdetermined. Let me modify it to make it solvable: “A boat goes downstream and covers a distance of 48 km in 4 hours. Going upstream, it covers the same distance in 6 hours. Find the speed of the stream.” Then: D = 48/4 = 12, U = 48/6 = 8. s = (12-8)/2 = 2 km/h. This is a standard problem.

SSC Pattern / Tips

• Always draw a diagram for train problems — label train length, platform length, and the gap between trains.
• For upstream/downstream, the key is that the current’s effect is additive in one direction and subtractive in the other.
• When a train overtakes another, relative speed = difference of speeds (same direction).
• Average speed formula 2ab/(a+b) only applies for equal distances at speeds a and b.