**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.

**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.

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