TIME AND DISTANCE

[attachment=11664]
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/y25x/25004x/yy=(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
 « Next Oldest | Next Newest »

## Important Note..!

 Popular Searches: little leaps, pdf file mathematics of time and distance, aditya**n telecommunications piracy, rv sol, sec schools by, training for speed and distance, time of flight distance measurements using smart phones using matlab,  (case insensitive)