# Time And Distance

Posted on 4:13 AM in

•                                                              Time and Distance

(i) Speed = [distance / time] ,
• (ii) Time = [ distance/ speed]
(iii) Distance = (Speed x Time).
(iv) 1 km/hour = 5/18 m/sec.
(v) 1 m/sec. = 18/5 km/hr.
(vi) If the ratio of the speeds of A and B is a : b, then the ratio of the times taken by them to cover the same distance is 1/a : 1/b or b: a .
(vii) Suppose a man covers a certain distance at x Kmph and an equal distance at y Kmph.
Then, the average speed during the whole journey is [ 2 xy/ (x + y)] Kmph
Solved Problems

Ex. 1. A scooterist covers II certain distance at 36 kmph. How many metres does he cover in 2 minutes?
Sol. Speed = 36 kmph = (36 x 5/18 ) m/sec. = 10m/sec.
:. Distance covered in 2 min. = (10 x 2 x 60) m = 1200 m.

Ex. 2. If a man runs at 3 metres per second, how many kilometres does he run in 1 hour 40 minutes.

Sol. Speed of the man = [ 3 x 18/5] km/hr = 54/5 km/hr.
Distance covered in 5/3 hours = (54/5 x 5/3) km = 18 km

Ex. 3. There are two townsA and B. Anil goes from A toB at40kmp! and comes back to the starting point at 60 kmph. What is his average speed during the whole journey ?

Sol: Average speed = [2xy / ((x+y) ] Kmph = ( 2X40X60)/(40+60)

Ex. 4. If a man walks at the rate of 5 kmph, he misses a train by only 7 minutes. However, ifhe walksattherateof6 kmph,.hereac/;les the station 5 minutes before the arrival of the train. Find the distance covered by him to reach the station.

Sol. Let the required distance be x km.
Difference in the times taken at two speeds = 12 min. =1/5 hr.
X/5 – x/6 = 1/5 or 6x – 5x=6 or x=6km
Hence, the required distance is 6 km.

Ex. 5. A man travels for 5 hrs 15 min. If he covers the first half of the journey at 60 kmph and the rest at 45 kmph. Find the total distance traveled by him.

Sol. Let the total distance be x km. Then,

(x/2)/60 + (x/2)/45 = 21/4 or x/120 + x/90 = 21/4

or 3x+4x=21 x90 or x=270.
:. Required distance = 270 km.

Ex. 6. A and B are two stations 330 km apart. A train starts from A at 8 a.m. and travels towards B at 60 kmph. Another train starts from B at 9. a.m. and travels towards A at 75 kmph. At what time do they meet?

Sol. Suppose they meet x hrs after 8 a.m. Then,
(Distance moved by first in x hrs)
+ [Distance moved by second in (x - 1) hrs] = 330
:. 60x+75 (x-1)=330 or x=3.
So, they meet at 11 a.m..

Ex. 7. A goods train leaves a station at a certain time and at a fixed speed. After 6 hours, an express train leaves the same station and moves in the same direction at a uniform speed of 90 kmph. This train catches up the goods train in 4 hours. Find the speed of the goods train.

Sol. Let the speed of the goods train be x kmph.
Distance covered by goods train in 10 hours
= Distance covered by the express train in 4 hours
. :. 10x = 4 x 90 or x = 36.
:. Speed of goods train = 36 kmph.

Ex. 8. Walking 5/6 of its usual speed, a train is 10 minutes too late. Find its usual time to cover the journey.

Sol. New speed =5/6 of the usual speed.
:. New. time taken = 6/5 of the usual time.
... (6/5 of the usual time)- (usual time) = 10 min.
:. 1/5 of the usual time = 10 min. or usual time = 50 min.

Ex. 9. Two bicyclists cover the same distance at 15 kmph and 16 kmph respectively. Find the distance traveUed by each, if one takes 16 minutes longer than the other.

Sol. Let the required distance be x km. Then,

X/15 – x/16 = 16/60 or 16x - 15x = 64 or x = 64 .
Hence, the required distance = 64 km.

Ex. 10. A thief is spotted by a policeman from a distance of 200 m. When the policeman starts the chase, the thief also starts running. If the speed of the thief be 10 km/hr and that of policeman 12 km/hr, how far the thief will have run before he is overtaken?

Sol. Relative speed of the policeman = 2 kmph.
Time taken by policeman to cover 200 m = [200/1000 X1/2]km = 1/10 hrs
In 110 hrs, the thief covers a distance of (10 x 1/10 ) km = 1 km.