Time, Speed and Distance : Question Types

Illustration 4: Suraj drives first 120 km in 2 hrs and next 180 km in next 4 hrs. What is his average speed for the entire trip in km per hour?

Sol: Total Distance travelled = 120 + 180 = 300 km. Total Time taken = 2 + 4 = 6 hrs.
Average Speed =Total Distance Travelled/Total Time Taken = 300/6 = 50 km/hr.

Illustration 5: A train covered first 120 km at a speed of 20 km an hour and then covered the remaining 180 km at a speed of 45 km an hour. Find its average speed.

Sol: Total distance = 120 + 180 = 300 km. Time taken for the first 120 km = 120/20 → 6 hrs.
Time taken for the next 180 km = 180/45 → 4 hrs. Total time taken = 6 + 4 = 10 hrs.
Average Speed =Total Distance Travelled/Total Time Taken = 300/10 = 30 km/hr.

Illustration 6: A cyclist travels at 10 km/hr for 2 hours and then at 13 km/hr for 1 hour. Find his average speed.

Sol: Distance travelled in first 2 hours =10 × 2 = 20 km. Distance travelled in next 1 hour =13 × 1 = 13 km. Total Distance travelled = 20 + 13 = 33 km. Total time taken = 2 + 1 = 3 hrs.
Average Speed =Total Distance Travelled/Total Time Taken =33/3 = 11 km/hr.

Illustration 7: A motorist travels one hour at an average speed of 45 kmph and the next hour at an average speed of 65 kmph. Then what is his average speed?

Sol: (45 + 65) ÷ 2 = 55 kmph. The total distance traveled by the motorist in these two hours = 65 + 45 = 110 km and he has taken two hours. Therefore, his average speed = 55 kmph.

Illustration 8: On my way from the office to the Pimpri class, I drive at 30 kmph and on the return journey I drive at 45 kmph. What is my average speed of travel?

Sol: 37.5 kmph is incorrect as the time traveled is different in both the cases and only the distances are same. Let the distance between the office and Pimpri class be x km.
∴Time taken on my onward journey = x/30 hours and time taken on my return journey = x/45 .
∴The total time taken for my onward and return journey = x/30 + x/45 = 5x/90 hours.
The total distance traveled both ways = 2x km ∴ Average speed = 2x/(5x/90) = 36 kmph.

Let the two speeds be a kmph and b kmph. Let the distance traveled in each of the speeds be x km.

Time taken to cover x km at ‘a’ kmph = x/a & at ‘b’ kmph = x/b.
Total time taken =
and the total distance covered = 2x.
Therefore, average speed =

Illustration 9: What is the distance covered by a car traveling at a speed of 40 kmph in 15 minutes?

Sol: 40 x 15/60 = 10 km. The important point to note is that time given was in minutes, whereas the speed was in kmph. Therefore, either speed will have to be expressed as km/min or time will have to be expressed in hours to apply the relationship.
In this case we converted time into hours to get the answer. Conversely, converting speed into km/min, we get 40 kmph = 40/60 km/min = 2/3 km/min. Therefore, distance traveled = 15 x 2/3 = 10 km.

Illustration 10:Traveling at a speed of 50 kmph, how long is it going to take to travel 60 km?

Sol: Time = Distance ÷ Speed → 60/50 = 1.2 hours = 1 hour and 12 minutes.
Note: While converting decimal hours into minutes, these are to be multiplied with 60 and not by hundred.

Illustration 11: Walking 5/6^th of his usual speed, Mike reached his destination 10 minutes late. Find his usual time, and the time taken on this occasion?

Sol: Let his usual speed be x km/hr and his usual time be t hours. His time on this occasion is 5/6x , The time taken is (t + 10/60) hours.
Since the distance travelled on both occasions is the same, xt = 5x/6 x (t + 10/60) .
Solving for t, we get t = 5/6 hours = 50 minutes, and the time taken on this occasion = 50 + 10 = 60 minutes.

Illustration 12: If the distance traveled by Mike be 60 km, then what was his usual speed and what was the speed on this occasion?

Sol: Usual time taken = 50 minutes = 5/6 hours. The distance = 60 km. Usual Speed = Distance ÷ Usual Time → 60/(5/6) = 72 kmph.
Speed on this occasion = Distance ÷ Time on this occasion = 60/1 = 60 kmph.
The ratio between the usual speed to the speed on this occasion = 72/60 = 6/5
The ratio of the usual time taken to the time taken on this occasion = 50min/60min = 5/6 .

Note: In general, speed and time have an inverse relationship. Therefore, if the speed becomes, say 0.5 times the original speed, then the time taken becomes twice as much as taken originally for the same distance. Or if the ratio of the speed of two moving objects is in the ratio of 3:4, the time taken by them to cover identical distance will be in the ratio of 4:3.

Relative Speed and Trains: Relative speed is basically defined as the speed of one object with respect to the other.

Illustration 13: A train traveling at 60 kmph crosses a man in 6 seconds. What is the length of the train?

Sol: Speed in m/sec = 60 x 5/18 = 50/3 m/sec. Time taken to cross the man = 6 seconds. Therefore, distance traveled = 50/3 x 6 = 100 m = length of the train.

Illustration 14: A train traveling at 60 kmph crosses another train traveling in the same direction at 24 kmph in 30 seconds. What is the combined length of both the trains?

Sol: As both the trains are moving in the same direction, the relative speed of the faster train is 60 – 24 = 36 kmph. The relative speed in m/sec = 36 x 5/18 = 10 m/sec. Time taken = 30 sec.
Therefore, distance traveled = 10 × 30 = 300 m = Combined length of two trains.