Motion problems — SSC CGL Quant Notes

Concepts and rules

Speed = distance ÷ time. To change km/h into m/s multiply by 5/18; to change m/s into km/h multiply by 18/5, because 1 km/h = 1000 m ÷ 3600 s.

Relative speed: two bodies moving in opposite directions approach (or separate) at the sum of their speeds; in the same direction, at the difference of their speeds.

Distance a train covers while passing a pole, a standing man or a signal post = its own length. While passing a platform, bridge or tunnel = its length + the length of the platform.

Two trains crossing each other cover the sum of their lengths at the relative speed: time = (L1 + L2)/(S1 + S2) in opposite directions and (L1 + L2)/(S1 − S2) in the same direction.

A train passing a walking man covers only its own length, but at the relative speed of the train and the man.

Meeting and chasing: two bodies d apart moving towards each other meet after d/(S1 + S2); a chaser closes a gap d in d/(S1 − S2).

Average speed = total distance ÷ total time. For two equal distances at speeds u and v it is 2uv/(u + v), never (u + v)/2 unless u = v.

If two trains start at the same time from A and B towards each other and, after meeting, take t1 and t2 hours to reach B and A, then speed of the first : speed of the second = √t2 : √t1.

Method and shortcuts

Step 1: Convert all speeds to one unit, using m/s when lengths are in metres and times are in seconds.

Step 2: Decide the distance to be covered (which lengths are involved) and the relative speed (sum or difference).

Step 3: Time = distance ÷ relative speed. With unknowns, write two such equations and subtract one from the other.

Shortcut: if a train passes a pole in t1 seconds and a platform of length P in t2 seconds, its speed = P/(t2 − t1) and its length = speed × t1.

Learn these pairs: 36 km/h = 10 m/s, 54 km/h = 15 m/s, 72 km/h = 20 m/s, 90 km/h = 25 m/s and 108 km/h = 30 m/s.

When the options are in km/h, work in m/s and convert only at the end by multiplying by 18/5.

Example 1: Train passing a pole

Question: A train 180 m long runs at 54 km/h. How long does it take to pass a pole?

Solution: 54 km/h = 54 × 5/18 = 15 m/s. The distance is only the train's length, 180 m, so time = 180 ÷ 15 = 12 seconds.

Answer: 12 seconds.

Example 2: Train crossing a platform

Question: A train 240 m long running at 72 km/h crosses a platform 360 m long. How long does it take?

Solution: 72 km/h = 20 m/s. Distance = 240 + 360 = 600 m, so time = 600 ÷ 20 = 30 seconds.

Answer: 30 seconds.

Example 3: Two trains, both directions

Question: Two trains 150 m and 100 m long run at 60 km/h and 30 km/h on parallel tracks. How long do they take to cross each other when moving in opposite directions, and when moving in the same direction?

Solution: Distance = 150 + 100 = 250 m. Opposite directions: relative speed = 60 + 30 = 90 km/h = 25 m/s, so time = 250 ÷ 25 = 10 seconds.

Same direction: relative speed = 60 − 30 = 30 km/h = 25/3 m/s, so time = 250 × 3/25 = 30 seconds.

Answer: 10 seconds in opposite directions; 30 seconds in the same direction.

Example 4: Length and speed from two crossings

Question: A train passes a man standing on a platform in 8 seconds and passes the whole platform, 264 m long, in 20 seconds. Find its length and speed. (a) 176 m, 79.2 km/h (b) 160 m, 72 km/h (c) 192 m, 79.2 km/h (d) 176 m, 72 km/h

Solution: The extra 20 − 8 = 12 seconds are spent covering only the platform, so speed = 264 ÷ 12 = 22 m/s = 22 × 18/5 = 79.2 km/h.

Length = 22 × 8 = 176 m. Check: (176 + 264) ÷ 22 = 440 ÷ 22 = 20 seconds.

Answer: (a) 176 m, 79.2 km/h

Example 5: Times after meeting

Question: Two trains start at the same time from stations A and B towards each other. After meeting, the first train takes 9 hours to reach B and the second takes 16 hours to reach A. If the first train runs at 80 km/h, what is the speed of the second? (a) 50 km/h (b) 45 km/h (c) 70 km/h (d) 60 km/h

Solution: Let them meet after t hours and let S be the second train's speed. The first train covers in 9 hours what the second covered in t hours: 80 × 9 = S × t, so S × t = 720.

The second covers in 16 hours what the first covered in t hours: S × 16 = 80 × t, so t = S/5. Then S × S/5 = 720, S × S = 3600 and S = 60 km/h (they meet after 12 hours).

Shortcut: first : second = √16 : √9 = 4 : 3, so the second = 80 × 3/4 = 60 km/h.

Answer: (d) 60 km/h

Example 6: Train overtaking two walkers

Question: A train overtakes two men walking in its direction at 2 km/h and 4 km/h in 9 seconds and 10 seconds respectively. What is the length of the train? (a) 45 m (b) 60 m (c) 50 m (d) 55 m

Solution: Let the train's speed be v km/h. The relative speeds are (v − 2) and (v − 4) km/h, and in both cases the distance is the train's own length.

So (v − 2) × 9 = (v − 4) × 10, giving 9v − 18 = 10v − 40 and v = 22 km/h.

Length = (22 − 2) × 5/18 × 9 = 50 m. Check with the second man: 18 × 5/18 × 10 = 50 m.

Answer: (c) 50 m

Common mistakes

Leaving speeds in km/h while lengths are in metres and times in seconds. Never mix units inside one equation.

Adding the platform length when the train passes a pole or a standing man, or forgetting it when the train passes a platform or bridge.

Using the sum of speeds for trains moving in the same direction. Same direction means the difference of speeds.

Taking average speed as (u + v)/2 over equal distances. It is 2uv/(u + v), which is smaller unless u = v.

When a train passes a walking man, using the train's speed alone instead of the relative speed.

Practice questions

Q1. A policeman sees a thief 200 m ahead. The thief runs at 10 km/h and the policeman chases at 12 km/h. How far will the thief have run when he is caught? (a) 800 m (b) 1000 m (c) 1200 m (d) 900 m

Q2. Ravi drives from Delhi to Agra at 40 km/h and returns by the same road at 60 km/h. What is his average speed for the whole journey? (a) 50 km/h (b) 52 km/h (c) 45 km/h (d) 48 km/h

Q3. Two trains 120 m and 180 m long cross each other in 12 seconds when moving in opposite directions, and the faster overtakes the slower in 60 seconds when they move in the same direction. What is the speed of the faster train? (a) 54 km/h (b) 60 km/h (c) 45 km/h (d) 72 km/h

Answers

Q1. (b) The gap of 200 m = 0.2 km closes at 12 − 10 = 2 km/h, taking 0.1 hour. In 0.1 hour the thief runs 10 × 0.1 = 1 km = 1000 m.

Q2. (d) Average speed = 2 × 40 × 60/(40 + 60) = 4800/100 = 48 km/h.

Q3. (a) Sum of speeds = 300/12 = 25 m/s and difference = 300/60 = 5 m/s, so the faster = (25 + 5)/2 = 15 m/s = 54 km/h.

Quick revision

km/h × 5/18 = m/s and m/s × 18/5 = km/h.

Opposite directions: add speeds; same direction: subtract them.

Distance = train's length, plus the length of whatever it passes if that object has length.

Equal distances: average speed = 2uv/(u + v).

Times t1 and t2 after meeting: S1 : S2 = √t2 : √t1.

Concepts and rules

Let u = speed of the boat (or swimmer) in still water and v = speed of the stream. Downstream speed D = u + v and upstream speed U = u − v; the boat can go upstream only if u is greater than v.

From D and U: u = (D + U)/2 and v = (D − U)/2.

Time to go d km and come back: d/(u + v) + d/(u − v) = 2du/(u² − v²).

Average speed for a round trip = (u + v)(u − v)/u = (u² − v²)/u, which is less than u whenever the stream flows.

If going upstream takes k times as long as going downstream over the same distance, then u + v = k(u − v), so u : v = (k + 1) : (k − 1).

Anything floating (a log, a hat) moves at the stream speed v. Measured relative to the water, the boat moves at u in both directions, so the stream cancels out in such chase problems.

If a boat covers 4 km downstream in the same time as 3 km upstream, then D : U = 4 : 3, because in equal times distances are in the ratio of speeds.

Method and shortcuts

Step 1: Turn every statement into an equation in D and U (or u and v); distance ÷ time gives D or U directly.

Step 2: When two mixed trips are given, put x = 1/U and y = 1/D; each trip becomes a linear equation: time = (upstream distance) × x + (downstream distance) × y.

Step 3: Find u and v as the half-sum and half-difference of D and U.

Shortcut: over the same distance, time upstream : time downstream = D : U, since time is inversely proportional to speed.

For floating-object problems, work relative to the water: the time taken to return to the object equals the time spent moving away from it.

Example 1: Speeds from two trips

Question: A boat goes 48 km downstream in 3 hours and 30 km upstream in 3 hours. Find the speed of the boat in still water and the speed of the stream.

Solution: D = 48 ÷ 3 = 16 km/h and U = 30 ÷ 3 = 10 km/h. So u = (16 + 10)/2 = 13 km/h and v = (16 − 10)/2 = 3 km/h.

Answer: Boat 13 km/h, stream 3 km/h.

Example 2: Round trip time

Question: A boat whose speed in still water is 9 km/h goes 36 km downstream and comes back. The stream flows at 3 km/h. How long does the whole trip take?

Solution: D = 9 + 3 = 12 km/h and U = 9 − 3 = 6 km/h. Time = 36/12 + 36/6 = 3 + 6 = 9 hours.

Answer: 9 hours.

Example 3: Upstream takes twice as long

Question: A man takes twice as long to row a distance upstream as to row the same distance downstream. If the stream flows at 2 km/h, find his speed in still water.

Solution: Twice the time means u + v = 2(u − v), so u = 3v. With v = 2 km/h, u = 6 km/h. Check: D = 8, U = 4, and 8 = 2 × 4.

Answer: 6 km/h.

Example 4: Two mixed trips

Question: A boat goes 30 km upstream and 44 km downstream in 10 hours. It also goes 40 km upstream and 55 km downstream in 13 hours. What is the speed of the boat in still water? (a) 7 km/h (b) 8 km/h (c) 9 km/h (d) 6 km/h

Solution: Let x = 1/U and y = 1/D. Then 30x + 44y = 10 and 40x + 55y = 13.

Multiply the first by 4 and the second by 3: 120x + 176y = 40 and 120x + 165y = 39. Subtracting, 11y = 1, so y = 1/11 and D = 11 km/h.

Then 30x = 10 − 44/11 = 6, so x = 1/5 and U = 5 km/h. Hence u = (11 + 5)/2 = 8 km/h and v = 3 km/h.

Answer: (b) 8 km/h

Example 5: Ratio of distances and total time

Question: A man rows to a place 48 km away and back in 14 hours. He finds that he can row 4 km downstream in the same time as 3 km upstream. What is the speed of the stream? (a) 2 km/h (b) 1.5 km/h (c) 2.5 km/h (d) 1 km/h

Solution: In equal times distances are in the ratio of speeds, so D : U = 4 : 3. Let D = 4k and U = 3k.

48/(4k) + 48/(3k) = 14 ⇒ 12/k + 16/k = 14 ⇒ 28/k = 14 ⇒ k = 2. So D = 8 km/h and U = 6 km/h.

Stream speed v = (8 − 6)/2 = 1 km/h, and the man's speed in still water is u = 7 km/h.

Answer: (d) 1 km/h

Example 6: The floating hat

Question: A man rowing upstream drops his hat under a bridge. After 20 minutes he notices, turns back at once and picks up the hat 2 km downstream of the bridge. If his rowing speed in still water does not change, what is the speed of the stream? (a) 2 km/h (b) 4 km/h (c) 3 km/h (d) 6 km/h

Solution: Relative to the water the hat is at rest, and the man moves away from it at speed u for 20 minutes; at the same relative speed u he needs another 20 minutes to get back to it.

So the hat floats for 20 + 20 = 40 minutes = 2/3 hour and travels 2 km with the stream.

Stream speed = 2 ÷ 2/3 = 3 km/h. The rowing speed u is not needed; the answer is the same for every u greater than the stream speed.

Answer: (c) 3 km/h

Common mistakes

Taking the round-trip average speed as u or as (D + U)/2. The average speed is total distance ÷ total time = (u² − v²)/u.

Mixing up the two formulas: (D + U)/2 is the boat's speed and (D − U)/2 is the stream's speed.

Accepting an answer with v greater than or equal to u; the boat could not then move upstream at all, so recheck the working.

In two-trip problems, writing equations in U and D directly, which gives fractions with unknowns in the denominator. Put x = 1/U and y = 1/D to make the equations linear.

In floating-object problems, giving the hat the boat's speed. The hat moves only with the stream.

Practice questions

Q1. A boat whose speed in still water is 15 km/h goes 54 km downstream and returns. The stream flows at 3 km/h. How many hours does the round trip take? (a) 7 (b) 7.5 (c) 7.2 (d) 8

Q2. A boat's speed in still water is 10 km/h and the stream flows at 4 km/h. What is its average speed for a journey to a point and back? (a) 10 km/h (b) 8 km/h (c) 8.4 km/h (d) 9.6 km/h

Q3. A boat's speed in still water is 12 km/h. It takes 3 times as long to go a distance upstream as to go the same distance downstream. What is the speed of the stream? (a) 4 km/h (b) 5 km/h (c) 3 km/h (d) 6 km/h

Answers

Q1. (b) D = 18 and U = 12 km/h, so time = 54/18 + 54/12 = 3 + 4.5 = 7.5 hours.

Q2. (c) (100 − 16)/10 = 84/10 = 8.4 km/h. Check with 21 km each way: 21/14 + 21/6 = 1.5 + 3.5 = 5 hours for 42 km, which is 8.4 km/h.

Q3. (d) u : v = (3 + 1) : (3 − 1) = 2 : 1, so v = 12/2 = 6 km/h. Check: D = 18, U = 6 and 18 = 3 × 6.

Quick revision

D = u + v, U = u − v; u = (D + U)/2, v = (D − U)/2.

Round trip of d each way: time = 2du/(u² − v²); average speed = (u² − v²)/u.

Upstream takes k times as long: u : v = (k + 1) : (k − 1).

Two mixed trips: substitute x = 1/U and y = 1/D, then eliminate.

Floating objects move with the stream; relative to the water the stream cancels.