## Time and Distance Part 1

Important Formula:

$Speed=\frac{Distance}{time}$

m/s——->km/h=>$\frac{18}{5}$            km/h—–>m/s=>$\frac{5}{18}$

Average speed=$\frac{Total&space;distance}{Total&space;time}$

Average speed=$\frac{2ab}{a+b}$

Problem 1: A car driver travels from the plains to the hill station which is 200km apart at an average speed of 40 km/h in return trip he covers the same distance at an average speed 20 km/h the average speed of the car over the entire distance of 400km?

Solution: $\frac{200}{40}=5hrs,&space;\frac{200}{20}=10hrs$

Average speed=$\frac{Total&space;distance}{Total&space;time}$=$\frac{400}{10+5}=\frac{80}{3}=26\frac{2}{3}km/hr$

Problem 2: During a journey, a man covered 50 km at a speed of 25km/hr. He covered the remaining 70km at a speed of 35km/hr. Find the average speed of the man?

Solution:

50km———>25km/hr———>$\frac{50}{25}=2hrs$

70km———>35km/hr——–>$\frac{70}{35}=2hrs$

Average speed=$\frac{Total&space;distance}{Total&space;time}$=$\frac{50+70}{2+2}=\frac{120}{4}=30$

Problem 3: A man covered $\frac{1}{3}$rd of his at a speed of 20km/hr, $\frac{1}{4}$th of remaining at the speed of 25km/hr, and the remaining journey at a speed of 30km/hr. Find the average speed of the man?

Solution:

$\frac{1}{3}$——–>20km/hr

$\frac{1}{4}$——–>25km/hr

Remaining———>30km/hr

Assume distance is 300 which is divisible by 3 and 4

5+2+5=12

Average speed = $\frac{300}{12}=25$

## Time and Distance Part 2

Problem 1: The average speed of a bus excluding the stoppage is 54km/hr and the average speed including stoppage is 45km/hr. Then how many minutes did the bus stop per hour?

Solution:

1hr———->54km

1hr———–>45km

54-45=9

$\frac{9}{54}\times&space;60=10min$

Problem 2: A man traveled from the village to the post office at the rate of 25km/hr and walked back at the rate of 4km/hr. If the whole journey took 5hours 48 minutes. Find the distance of the post office from the village?

Solution:

Given time =5 hours 48 minutes

=$5\frac{48}{60}=5\frac{4}{5}=\frac{29}{5}$

distance=Speed * time

= $\frac{2\times&space;25\times&space;4}{29}\times&space;\frac{29}{5}=\frac{40}{2}=20$

Problem 3: Walking at 3km/hr, a boy reaches school 5 min late. If he walks at 4km/hr he reaches his destination 5 min early find the distance between his home and school?

Solution:

Assume distance=x km

5+5=10 min

$\frac{x}{3}-\frac{x}{4}=\frac{10}{60}$

$\frac{x}{12}=\frac{1}{6}$

x=2km

Formula :$d=\frac{s_{1}\times&space;s_{2}}{s_{1}\sim&space;s_{2}}\times&space;\Delta&space;t$

$\frac{3\times&space;4}{1}\times&space;\frac{10}{60}=2km$

## Time and Distance Part 3

Problem 1: A starts running at the speed of 50km/hr. After each half hour, its speed increased by 10km/hr. How much distance will be covered by it in 3 hours?

Solution:

50+60+70+80+90+100=450

450*0.5=225

Problem 2: An express train traveled at an average speed of 100km/hr stopping for 3 min after every 75km. How long it takes to reach its destination 600 km from starting point?

Solution: $\frac{600}{75}=8-1=7$

7*3=21min

Time=$\frac{distance}{speed}$=$\frac{600}{100}=6$hours

Total time=6 hours 21 minutes

Problem 3: Two trains X and y starts from station A and B towards B to A respectively. After passing each other they take 4 hours 48 min and 3 hours 28 min to reach B to A respectively. If train X is moving 60km/hr then find the speed of train ‘y’?

Solution:

Formula= $\sqrt{\frac{T_{2}}{T_{1}}}=\frac{S_{1}}{S_{2}}$

4*60+48=288

3*60+20=200

$\sqrt{\frac{200}{288}}=\frac{60}{s_{2}}$

$\sqrt{\frac{100}{144}}=\frac{60}{s_{2}}$

$\frac{10}{12}=\frac{60}{s_{2}}$

$s_{2}=72km/hr$

## Time and Distance Part 4

Problem 1: A former traveled a distance of 61 km in 9 hours. He traveled part on foot at the rate of 4km/hr and partly in bicycle at the rate of 9 km/hr. The distance traveled on foot is?

Solution:

Distance=Speed*Time

61=4*x +9(9-x)

61= 4x + 81-9x

x=4

d=4*4=16

Problem 2: Gopal can walk a certain distance in 40 days when he rests 9 hours a day. How long will he take to walk twice the distance twice as fast and rest twice as long each day?

Solution:

Time = $\frac{Distance}{Speed}$

In 24 hours he takes 9 hours rest

24-9=15

15-9=6

$\frac{1}{1}\times&space;15\times&space;40=\frac{2}{2}\times&space;6\times&space;x$

x=100 days

Problem 3: A policeman starts to chase a thief when the thief goes to 10 steps the police moves 8 steps. 5 steps of policeman are equal to 7 steps of the thief the ratio of the speed of the policeman and thief is?

Solution:

let thief=t ,  policeman=p

Time=10*t=p*8

$\frac{t}{p}=\frac{8}{10}$

t : p = 8 : 10

Distance=5*p=7*t

$\frac{p}{t}=\frac{7}{5}$

t : p =5 : 7

Speed=$\frac{Distance}{Time}$

$\frac{5}{8}=\frac{7}{10}$

28: 25

## Time and Distance Part 5

Problem 1: In a flight of 600 km, an aircraft slowed down due to bad weather. Its average speed for the trip was reduced by 200km/hr and the time of flight increased by 30 minutes. The duration of the flight is?

Solution:

Let speed of flight is x

Time=$\frac{Distance}{speed}$

$\left&space;[&space;\frac{600}{x-200}&space;\right&space;]-\left&space;[&space;\frac{600}{x}&space;\right&space;]=\frac{1}{2}$

x=600

Problem 2: 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 90kmph. This train catches up the goods train in 4 hours. Find the speed of the goods train?

Solution:

Distance=Speed*Time

6*x +x*4=10x

10x=90*4

x=36

Problem 3: A monkey tries to ascend a greased pole 14m high. He ascends 2meters in the first minute and slips down one meter in the second minute. If he continues to ascend in this fashion how long does he takes to reach the tom?

Solution:

2 minutes———->1 meter

12*2 minutes——->1 meter *12

24 minutes———->12 meter

1 minute————>2 meter

25 minutes ——->14 meter

In 25 minutes the monkey reached 14 meters tomb.