04-04-2011, 01:30 PM

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TIME AND DISTANCE

IMPORTANT FACTS AND FORMULAE

Distance Distance

1. Speed = Time , Time= Speed , Distance = (Speed * Time)

2. x km / hr = x * 5

18

3. x m/sec = (x * 18/5) km /hr

4. 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: 1 a b

or b:a.

5. Suppose a man covers a certain distance at x km/ hr and an equal distance at y km / hr . Then , the average speed during the whole journey is 2xy km/ hr.

x+y

SOLVED EXAMPLES

Ex. 1. How many minutes does Aditya take to cover a distance of 400 m, if he runs at a speed of 20 km/hr?

Sol. Aditya’s speed = 20 km/hr = {20 * 5} m/sec = 50 m/sec

18 9

Time taken to cover 400 m= { 400 * 9 } sec =72 sec = 1 12 min 1 1 min.

50 60 5

Ex. 2. A cyclist covers a distnce of 750 m in 2 min 30 sec. What is the speed in km/hr of the cyclist?

Sol. Speed = { 750 } m/sec =5 m/sec = { 5 * 18 } km/hr =18km/hr

150 5

Ex. 3. A dog takes 4 leaps for every 5 leaps of a hare but 3 leaps of a dog are equal to 4 leaps of the hare. Compare their speeds.

Sol. Let the distance covered in 1 leap of the dog be x and that covered in 1 leap of the hare by y.

Then , 3x = 4y => x = 4 y => 4x = 16 y.

3 3

Ratio of speeds of dog and hare = Ratio of distances covered by them in the same time

= 4x : 5y = 16 y : 5y =16 : 5 = 16:15

3 3

Ex. 4.While covering a distance of 24 km, a man noticed that after walking for 1 hour and 40 minutes, the distance covered by him was 5 of the remaining distance. What was his speed in metres per second?

7

Sol. Let the speed be x km/hr.

Then, distance covered in 1 hr. 40 min. i.e., 1 2 hrs = 5x km

3 3

Remaining distance = { 24 – 5x } km.

3

5x = 5 { 24 - 5x } 5x = 5 { 72-5x } 7x =72 –5x

3 7 3 3 7 3

12x = 72 x=6

Hence speed = 6 km/hr ={ 6 * 5 } m/sec = 5 m/sec = 1 2

18 3 3

Ex. 5.Peter can cover a certain distance in 1 hr. 24 min. by covering two-third of the distance at 4 kmph and the rest at 5 kmph. Find the total distance.

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

2 x 1 x

3 + 3 = 7 x + x = 7 7x = 42 x = 6

4 5 5 6 15 5

Ex. 6.A man traveled from the village to the post-office at the rate of 25 kmph and walked back at the rate of 4 kmph. If the whole journey took 5 hours 48 minutes, find the distance of the post-office from the village.

Sol. Average speed = { 2xy } km/hr ={ 2*25*4 } km/hr = 200 km/hr

x+y 25+4 29

Distance traveled in 5 hours 48 minutes i.e., 5 4 hrs. = { 200 * 29 } km = 40 km

5 29 5

Distance of the post-office from the village ={ 40 } = 20 km

2

Ex. 7.An aeroplane files along the four sides of a square at the speeds of 200,400,600 and 800km/hr.Find the average speed of the plane around the field.

Sol. :

Let each side of the square be x km and let the average speed of the plane around the field by y km per hour then ,

x/200+x/400+x/600+x/800=4x/y25x/25004x/yy=(2400*4/25)=384

hence average speed =384 km/hr

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

7

Sol. :New speed =5/6 of the usual speed

New time taken=6/5 of the usual time

So,( 6/5 of the usual time )-( usual time)=10 minutes.

=>1/5 of the usual time=10 minutes.

usual time=10 minutes