Solved Examples Time & Distance - General Aptitude for GATE - Mechanical

The Basics of Time, Speed & Distance

Time

Time is a measure of duration or the interval between two events. It is quantified in units such as seconds, minutes, hours, days, etc. In problems involving motion and travel, understanding time and its units is essential for converting speeds and distances correctly.

• Units of time: seconds (s), minutes (min), hours (hr), days, weeks, months, years.
• Notation: 24-hour notation (e.g., 14:30) or 12-hour AM/PM notation (e.g., 2:30 PM).

Speed

Speed is the rate at which an object covers distance with respect to time. It describes how fast an object is moving relative to a reference.

• Formula: Speed = Distance ÷ Time.
• Units: metres per second (m/s), kilometres per hour (km/h), miles per hour (mph), etc.
• Types of speed: average speed is total distance divided by total time; instantaneous speed is the speed at a specific instant.

Distance

Distance is the measure of how far an object has moved - the length of the path travelled between two points.

• Units: metres (m), kilometres (km), miles (mi), feet (ft), etc.
• Calculation: use geometric formulas for straight-line distance (e.g., Pythagoras) or route distances for real-world journeys.

Relationship between Time, Speed and Distance

The three quantities are related by the fundamental formula:

• Distance = Speed × Time

From this, any one quantity can be found when the other two are known: Speed = Distance ÷ Time, Time = Distance ÷ Speed.

Applications

• Transportation: route planning, travel-time estimation, scheduling.
• Sports: timing races, analysing athletes' performance.
• Physics: basic kinematics, motion analysis.
• Engineering: vehicle and traffic design, logistics and planning.

Rishi Kapoor can swim a certain course against the river flow in 84 minutes; he can swim the same course with the river flow in 9 minutes less than he can swim in still water. How long would he take to swim the course with the river flow?

Solution:

Let the time to swim the course in still water be T minutes.

Then time downstream = T − 9 minutes.

Time upstream = 84 minutes.

For the same course length L, speed in still water v = L/T.

Upstream speed v − r = L/84.

Downstream speed v + r = L/(T − 9).

Eliminate v and r by using ratios. From v = L/T, we have

1/(v − r) = 84/L and 1/(v + r) = (T − 9)/L, and 1/v = T/L.

Equating the expression for r obtained from upstream and downstream relations yields

(84 − T)/(84T) = 9/(T(T − 9)).

Solving this equation gives

T^2 − 93T + 1512 = 0.

The roots are T = 72 minutes and T = 21 minutes.

We require the physically reasonable root where still-water time lies between downstream and upstream times. For T = 72 minutes, downstream time = 72 − 9 = 63 minutes, and upstream = 84 minutes, so 63 < 72 < 84 holds.

Therefore, the time to swim the course with the river flow is 63 minutes.

Time and Distance - Examples

The fundamental relation between distance (s), speed (v) and time (t) is given by:

• s = v × t

If x1 and x2 are distances covered at velocities v1 and v2 respectively, then the average velocity over the entire distance (x1 + x2) is given by the harmonic-type relation for equal distances and the weighted mean for unequal distances when required.

For the same distance, time is inversely proportional to speed. Problems of relative percentage change in speed are frequently converted to changes in time using this inverse relation.

Ex.1 If I decrease my speed by 20% of original speed, I reach office 7 minute late. What is my usual time and new time of reaching office?

Sol. 

Let usual time be T minutes and distance be D, usual speed be V.

Then D = V × T.

New speed = 80% of V = 0.8V.

New time = D ÷ (0.8V) = (V × T) ÷ (0.8V) = T ÷ 0.8 = 1.25 T.

Increase in time = 1.25T − T = 0.25T.

Given 0.25T = 7 minutes.

So T = 7 × 4 = 28 minutes.

New time = 1.25 × 28 = 35 minutes.

Ex.2 A train leaves Calcutta for Mumbai, a distance of 1600 kms at the same time a train leaves Mumbai to Calcutta. The trains meet at Nagpur which is at a distance of 700 kms from Mumbai. What is the ratio of the speeds of the trains?

Sol.

Let S1 be speed of train from Calcutta to Mumbai and S2 be speed of train from Mumbai to Calcutta.

Distance from Mumbai to Nagpur = 700 km.

Distance from Calcutta to Nagpur = 1600 − 700 = 900 km.

Both trains start at the same time and meet at the same instant, so time to meeting is equal for both.

Therefore, S1 : S2 = distance from Calcutta to meeting point : distance from Mumbai to meeting point = 900 : 700 = 9 : 7.

Ex.3 A train 110 m long travels at 60 kmph. How long does it take?

(a) To pass a telegraph post by the side of the track?

(b) To pass a man running at 6 kmph in the same direction as the train?

(c) To pass a man running at 6 kmph in the opposite direction?

(d) To pass a station platform 240 m long?

(e) To pass another train 170 m long, running at 40 kmph in the same direction?

(f) To pass another train 170 m long, running at 60 kmph in the opposite direction?

Sol.

Convert train speed to m/s. 60 km/h = 60 × (5/18) = 50/3 m/s.

(a) Time to pass a telegraph post = Length of train ÷ speed of train.

Time = 110 ÷ (50/3) = 110 × 3/50 = 6.6 seconds.

(b) Speed of man = 6 km/h = 6 × (5/18) = 5/3 m/s.

Relative speed (same direction) = (50/3 − 5/3) = 45/3 = 15 m/s.

Time = 110 ÷ 15 = 22/3 ≈ 7.333... seconds.

(c) Relative speed (opposite direction) = (50/3 + 5/3) = 55/3 m/s.

Time = 110 ÷ (55/3) = 110 × 3/55 = 6 seconds.

(d) To pass a platform of length 240 m, total distance = 110 + 240 = 350 m.

Time = 350 ÷ (50/3) = 350 × 3/50 = 21 seconds.

(e) Speed of second train = 40 km/h = 40 × (5/18) = 100/9 m/s.

Relative speed (same direction) = (50/3 − 100/9) = (150 − 100)/9 = 50/9 m/s.

Total length = 110 + 170 = 280 m.

Time = 280 ÷ (50/9) = 280 × 9/50 = 252/5 = 50.4 seconds.

(f) Second train speed (opposite) = 60 km/h = 50/3 m/s.

Relative speed = 50/3 + 50/3 = 100/3 m/s.

Time = 280 ÷ (100/3) = 280 × 3/100 = 8.4 seconds.

Boats & Streams

Let the speed of the boat in still water be b km/h and the speed of the stream be w km/h.

• Downstream speed (with the stream), D = b + w
• Upstream speed (against the stream), U = b − w

Ex.4 A man can row 4.5 km/hr in still water. It takes him twice as long to row upstream as to row downstream. What is the rate of the current?

Sol. 

Let b = 4.5 km/h and let w be the rate of the current in km/h.

Time upstream is twice the time downstream, so

1/(b − w) = 2 × 1/(b + w).

Therefore, b + w = 2(b − w).

Thus, b + w = 2b − 2w.

Rearrange: 3w = b.

So w = b/3 = 4.5/3 = 1.5 km/h.

Circular Motion

Circular Motion with two people

When two people run on a circular track, meeting times depend on whether they run in the same direction or opposite directions, and on the difference or sum of their speeds. Convert all speeds and the circumference to consistent units before calculation.

Ex.5 Sachin and Saurav, as a warm-up exercise, are jogging on a circular track. Saurav is a better athlete and jogs at 18km/hr while Sachin jogs at 9 km/hr. The circumference of the track is 500m (i.e. 1⁄2 km). They start from the same point at the same time and in the same direction. When will they be together again for the first time?

Sol.

Speeds: Saurav = 18 km/h, Sachin = 9 km/h. Relative speed (same direction) = 18 − 9 = 9 km/h.

Distance to gain for first meeting = one full lap = 0.5 km.

Time = distance ÷ relative speed = 0.5 ÷ 9 hours = 1/18 hour = (1/18) × 3600 = 200 seconds.

Alternative reasoning: speed ratio 18 : 9 = 2 : 1. For every full round Sachin completes, Saurav completes two rounds; hence Saurav gains one full lap over Sachin in the time Sachin takes for one lap. Sachin's lap time = 0.5/9 h = 1/18 h = 200 s.

Ex.6 Suppose in the earlier problem when would the two meet for the first time if they are moving in the opposite directions?

Sol.

Convert speeds to m/s if working with metres: Saurav = 18 km/h = 5 m/s, Sachin = 9 km/h = 2.5 m/s.

Relative speed (opposite directions) = 5 + 2.5 = 7.5 m/s.

Track circumference = 500 m.

Time for first meeting = 500 ÷ 7.5 = 200/3 ≈ 66.67 seconds.

Ex.7 If the speeds of Saurav and Sachin were 8 km/hr and 5 km/hr, then after what time will the two meet for the first time at the starting point if they start simultaneously?

Sol.

Take circumference = 0.5 km.

Lap time for Saurav = 0.5 ÷ 8 = 1/16 hour = 225 seconds.

Lap time for Sachin = 0.5 ÷ 5 = 1/10 hour = 360 seconds.

They will be together at the starting point when both have completed an integer number of laps simultaneously; time required = LCM of individual lap times = LCM(225 s, 360 s) = 1800 seconds = 0.5 hour = 30 minutes.

Circular Motion with three people

When three runners are on the same track, determine pairwise meeting intervals and then take the least common multiple (LCM) to find when all three meet together.

Ex.8 Let us now discuss the cases of circular motion with three people: Laxman joins Saurav and Sachin, and all of them run in the same direction from the same point simultaneously in a track of length 500 m. Laxman moves at 3 km/hr, Sachin at 5 km/hr and Saurav at 8 km/hr. When will all of them be together again?

a. for the first time?

b. for the first time at the starting point?

Sol.

Work with lap length 0.5 km.

Relative speed between Saurav and Sachin = 8 − 5 = 3 km/h → time to meet = 0.5/3 h = 1/6 h = 10 minutes.

Relative speed between Saurav and Laxman = 8 − 3 = 5 km/h → time to meet = 0.5/5 h = 1/10 h = 6 minutes.

To have all three meet together at some point, take LCM of pairwise meeting times: LCM(10, 6) minutes = 30 minutes.

Hence they meet for the first time (anywhere on the track) after 30 minutes.

For them to be together at the starting point simultaneously, find individual lap times:

Saurav's lap time = 0.5/8 h = 225 seconds.

Sachin's lap time = 0.5/5 h = 360 seconds.

Laxman's lap time = 0.5/3 h = 600 seconds.

LCM(225, 360, 600) = 1800 seconds = 30 minutes.

Therefore, they will be at the starting point together for the first time after 30 minutes.

Ex.9 A thief is spotted by a policeman from a distance of 200 m. When the policeman starts a chase, the thief starts running. Speed of thief is 10 Kmph and that of policeman is 12 kmph. After how many hours will the policeman catch the thief?

Sol. 

Relative speed = 12 − 10 = 2 km/h.

Convert relative speed to m/s: 2 km/h = 2 × (5/18) = 5/9 m/s.

Distance = 200 m.

Time = 200 ÷ (5/9) = 200 × 9/5 = 360 seconds = 6 minutes = 0.1 hour.

Ex.10 A man steals a car at 1 : 30 pm & drives at 40 kmph. At 2 pm the owner starts chasing his car at 50 kmph. At what time does he catch the man?

Sol. 

Time difference when owner starts chasing = 2:00 pm − 1:30 pm = 0.5 hours.

Distance covered by thief in 0.5 hours = 40 km/h × 0.5 h = 20 km.

Relative speed = 50 − 40 = 10 km/h.

Time to catch = Distance ÷ Relative speed = 20 ÷ 10 = 2 hours.

Owner catches thief at 2:00 pm + 2 hours = 4:00 pm.