Time and Distance Formula, Shortcuts, Methods & Short Tricks to Solve Problems

Time and Distance Formula
Aspirants
who are going to appear in competitive exams should know the Time and
Distance Formula. In quantitative aptitude section, candidates spend more
time as they don’t know the formulas, shortcuts and short tricks to solve
problems. So they find it very difficult but the fact is quantitative aptitude
section is very easy. If you know the shortcuts, methods and short tricks then
you can solve any problem in very less time. Competitive exams need accuracy
and speed; if you have both of them then you can crack any tough exam.
Formula
of distance, speed and time is correlated with each other, so don’t get
confused among them. Aspirants can prepare from the sample papers and previous
year question papers for the upcoming exams, you can buy it from the nearest
books shop or download it from their official website. We the team of www.privatejobshub.in are providing some
time and distance formulas and some useful short ticks below for your
convenience. Take a look and read them carefully.
Methods &
Short Tricks to Solve Problems
Generally students don’t know the basic tricks and shortcuts
to solve the questions and that’s why they were unable to score good marks. Time
and Distance formula and some shortcuts are available below so you can attempt
all the questions and score better marks.
Formulas
and how to convert them
Basics:

Convert
kilometers per hour (km/hr) to meters per second (m/s):
x km/hr = x X 5/18 m/s
Convert meters per second (m/s) to kilometers per hour
(km/hr):
x m/s= x X 185 km/hr
Speed and relation between speed, distance
and time
Speed is a measure of how quickly an object moves from one
place to another. It is equal to the distance traveled divided by the time.
Types of speed are mentioned below:
Average Speed:
If an object covers a certain distance at x kmph
and an equal distance at y kmph, the average speed of the whole
journey
=2xy/x+y kmph
Relation between
Distance, Speed and Time:
Speed and time are inversely proportional (when distance is
constant)
⟹speed∝1/time (when distance is constant)
If the ratio of the speeds of A and B is a:b , a:b, then,
the ratio of the time taken by them to cover the same distance is
1/a:1/b=b:a
Assume two objects A and B start at the same time in
opposite directions from P and Q respectively. After passing each other, A
reaches Q in a seconds and B reaches P in b seconds. Then,
Speed of A : Speed of B =√b:√a
An object covered a certain distance at a speed of v kmph.
If it had moved v1 kmph faster, it would have
taken t1 hours less. If it had moved v2 kmph slower, it
would have taken t2 hours more. Then,

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If two objects are moving in the same direction
at v1 m/s and v2 m/s respectively
where v1>v2, then their relative speed = (v1−v2) m/s
Consider two objects A and B separated by a distance of d meter. Suppose A and B start moving in the same direction at the same time such that A moves towards B at a speed of a meter/second and B moves away from A at a speed of b meter/second where a>b. Then,
relative speed =(a−b) meter/second
time needed for A to meet B =d/a−b seconds
If two objects are moving in opposite directions at v1 m/s and v2 m/s respectively, then their relative speed = (v1+v2) m/s
Consider two objects A and B separated by a distance of d meter. Suppose A and B start moving towards each other at the same time at a meter/second and b meter/second respectively. Then,
relative speed =(a+b) meter/second time needed for A and B to meet each other =d/a+b seconds
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Tips: Practice
Which Helps In Preparation
Illustrations
1) A dog runs from one side of a park to the other. The park
is 80.0 meters across. The dog takes 16.0 seconds to cross the park. What is
the speed of the dog?
Answer: The distance the dog travels and the time it
takes are given. The dog’s speed can be found with the formula:

s = 5.0 m/s
The speed of the dog is 5.0 meters per second.
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2) A golf cart is driven at its top speed of 27.0 km/h for
10.0 minutes. In meters, how far did the golf cart travel?
Answer: The first step to solve this problem is to
change the units of the speed and time so that the answer found will be in
meters, since this is what the question asks for. The speed is:
s = 27.0 km/h

s = 7.50 m/s
Converting the units, the speed is 7.50 m/s. The time the
cart traveled for was:
t = 10.0 min

t = 600s
The speed of the cart and the time of travel are given, so
the distance traveled can be found using the formula:
d = st
d = (7.50 m/s) (600 s)
d = 4500 m
The golf cart traveled 4500 m, which is equal to 4.50 km.
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3) A person goes from A to B at the speed of 40 kmph and
comes back at the speed of 60 kmph. What is his average speed for the whole
journey?
Answer: Since the distance travelled on both sides is
the same, we can use the formula of harmonic mean of speeds
Average Speed: 2xy/(x+y) where, x is the speed while going
from A to B and y is the speed while coming back.
So using this formula, we get the answer as 48 kmph.
4) Moving at 50 kmph, a person reaches his office 10 min
late. Next day, he increases his speed and moves at 60 kmph and reaches his
office 5 min early. What is the distance from his home to his office?
Answer: We can observe that difference in timings on
both days is 15 min (and not 5 min, as one day he is late and on the other day
he is early)
Let the required distance = D km. As time taken at the speed
of 50 kmph is more than time taken at 60 kmph, so equation can be formed as
D/50-D/60=15/60 .
Solving this equation, we get the answer as 75 km.
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Final words
Dear students, read all the time
and distance formula, remember the shortcuts and methods at the time of
examination. We hope that the above information that we provided it will be
helpful for you. At last we wish to say all the best to the aspirants who are
going to attend the exams.
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