Quant Notes | Speed , Distance and and time
This topic is easy if basic concept clear. Here use
only one formula to solve all question related to speed, distance and time.
1. When any one quantity among D, S or T ;
become constant; then we could correlate the entire situation to ratio and
which would lessen the burden of heavy calculation
2. Suppose two men are moving in same direction at u m/s and v m/s respectively , where u >v then relative speed=(u-v)m/s.
3. Suppose two men are moving in opposite direction at u m/s and v m/s respectively , where u >v then relative speed=(u-v)m/s.
We take some example-
Type1
E.g. A cyclist cover a distance of 450 m in 2 min 30 sec. what is the speed in km/hr. of the cyclist?
Solution-
2 min 30sec=150sec
Speed=450/150 m/sec
=3m/sec
=(5*18/5)km/hr.
=18km/hr.
E.g. How many minutes does Anish take to cover a distance of 400m, if he runs at a speed of 20 km/hr.
Solution-
Anish speed=20km/hr.
(20*5/18)m/sec=50/9 m/sec.
Time taken to cover 400 m =(400*9/50)sec
= 72 sec=1 and 12/60min = 1 1/5min
Type2
E.g. Moving 4/9 of his usual speed of a person complete his race in 36 min. how much time he could save by moving his usual speed.
Solution-
4/9--------------36
9-----------------36
1-----------------36/9=4
4 multiply by numerator
4*4=16
36-16=20
E.g. Moving 4/9 of his usual speed of a person complete his race in 36 min. how much time he could waste by moving his usual speed.
Solution:-
13/7---------------------35
7--------------------------35
1--------------------------5
13*5=65
65-35=30
E.g. Moving 5/6 th of his usual speed a person reaches his office 20min. late then find the usual time?
Solution-
Type3
Type4
E.g. A man is walking at a speed of 12km after every km he takes a rest for 12 min. then the total time taken by him to occur a distance of 36 km.
Solution-
Suppose like this
36-1=35
Actual time=36/12=3
Waste time=35*12=420min.
420/60=7hr.
Total time=3+7=10hr.
Type5
Train
1st object train
Train length
Train speed
2nd object
1. Pole, tree , standing man---------------speed=0 length=0
2. Tunnel, bridge, standing train---------length speed=0
E.g. Train 150km in length running at the rate of 30 km/hr. how much time will it take to cross a telephone pole.
Solution-Train------l1=150km s1=30
Pole------l2=0 s2=0
t=d/s
t=150/30
=5hr.
E.g. Train 70 km in length running at rate of 20km/hr. how much time will it take to cross a man who is running 18km/hr. in same and opposite direction.
solution -tain------------------------l1=70km s1=20km/hr.
moving man------------l2=0 s2=18km/hr.
s1-s2 = d/t
20-18=150/t
70/2=35 hr.
E.g. Two train are moving in opposite direction at 50m/s. and 40m/s. The faster train cross a man in slower train in 3 sec .then find the length of faster train.
Solution-L1=t*(s1+s2)
=3*(50+40)
=270
=l1+0/270=270
1. Change 1km/hr to m/s
1. Change 1m/s to km/hrs
2. Average speed
1. Suppose a man cover a certain distance at x km/hr and equal distance at y km/hr. then the average speed during the whole journey is ( 2xy/x+y )km/hr.2. Suppose two men are moving in same direction at u m/s and v m/s respectively , where u >v then relative speed=(u-v)m/s.
3. Suppose two men are moving in opposite direction at u m/s and v m/s respectively , where u >v then relative speed=(u-v)m/s.
We take some example-
Type1
E.g. A cyclist cover a distance of 450 m in 2 min 30 sec. what is the speed in km/hr. of the cyclist?
Solution-
2 min 30sec=150sec
Speed=450/150 m/sec
=3m/sec
=(5*18/5)km/hr.
=18km/hr.
E.g. How many minutes does Anish take to cover a distance of 400m, if he runs at a speed of 20 km/hr.
Solution-
Anish speed=20km/hr.
(20*5/18)m/sec=50/9 m/sec.
Time taken to cover 400 m =(400*9/50)sec
= 72 sec=1 and 12/60min = 1 1/5min
Type2
E.g. Moving 4/9 of his usual speed of a person complete his race in 36 min. how much time he could save by moving his usual speed.
Solution-
4/9--------------36
9-----------------36
1-----------------36/9=4
4 multiply by numerator
4*4=16
36-16=20
E.g. Moving 4/9 of his usual speed of a person complete his race in 36 min. how much time he could waste by moving his usual speed.
Solution:-
13/7---------------------35
7--------------------------35
1--------------------------5
13*5=65
65-35=30
E.g. Moving 5/6 th of his usual speed a person reaches his office 20min. late then find the usual time?
Solution-
Type3
E.g. A man travelled from the village to the post-office at the rate of 25km/hr. and walked back at the rate of 4 km/hr. If whole journey took 5hr. 48 min. ,find the distance of the post-office from the village.
Solution -
average speed=( 2xy/x+y )km/hr.
=(2*25*4/25+4)km/hr.
=200/29km/hr.
Distance travelled in 5 hr. 48 minutes
i.e. 5 and 4/5hrs = ( 200/29*29/5 )km
=40km.
Distance of the post-office from the village = (40/2)
=20km.
Solution -
average speed=( 2xy/x+y )km/hr.
=(2*25*4/25+4)km/hr.
=200/29km/hr.
Distance travelled in 5 hr. 48 minutes
i.e. 5 and 4/5hrs = ( 200/29*29/5 )km
=40km.
Distance of the post-office from the village = (40/2)
=20km.
Type4
E.g. A man is walking at a speed of 12km after every km he takes a rest for 12 min. then the total time taken by him to occur a distance of 36 km.
Solution-
36-1=35
Actual time=36/12=3
Waste time=35*12=420min.
420/60=7hr.
Total time=3+7=10hr.
Type5
Train
1st object train
Train length
Train speed
2nd object
1. Pole, tree , standing man---------------speed=0 length=0
2. Tunnel, bridge, standing train---------length speed=0
E.g. Train 150km in length running at the rate of 30 km/hr. how much time will it take to cross a telephone pole.
Solution-Train------l1=150km s1=30
Pole------l2=0 s2=0
t=d/s
t=150/30
=5hr.
E.g. Train 70 km in length running at rate of 20km/hr. how much time will it take to cross a man who is running 18km/hr. in same and opposite direction.
solution -tain------------------------l1=70km s1=20km/hr.
moving man------------l2=0 s2=18km/hr.
s1-s2 = d/t
20-18=150/t
70/2=35 hr.
E.g. Two train are moving in opposite direction at 50m/s. and 40m/s. The faster train cross a man in slower train in 3 sec .then find the length of faster train.
Solution-L1=t*(s1+s2)
=3*(50+40)
=270
=l1+0/270=270
