Time, Speed & Distance
Speed or Velocity
If the location of an object
changes with time then it is said to be in
motion. A bus running on a road, an ant crawling on a floor, etc. — all are examples of objects in motion
because the locations of these objects keep on changing w.r.t. their
surroundings.
Let an
object move from a point A to the point B through any path, then the actual length of the path followed by the object
is called the distance travelled by the object.
The rate at which any moving body covers a
particular distance is called its speed.
Speed = Distance travelled
Time taken
if the distance is constant, Speed inversly proportionate to 1/Time
Time = Distance
Speed
if the time is constant,
Distance directly proportionate to Speed.
Distance = Time x
Speed,
if the speed is
constant, Distance directly proportionate to Time.
We can say that for constant distance travelled, speed
is inversely proportional to the time taken.
This can be explained by a simple example. To cover a distance
of 100 km, if a person goes at the speed of
25 kmph, he will require 4 hours to complete the journey and travelling
at a speed of 50 kmph, 2 hours will be required.
Uniform Speed
If the object covers equal distances in equal time intervals, howsoever small the interval may be, then its
speed is called the Uniform speed.
Variable Speed
If the object travels different distances
in equal intervals of time, then its speed is called
a Variable speed. In this case the speed changes from instance to instance.
Units of measurement
Time is measured in seconds (s), minutes (min) or hours (hr).
Distance is usually measured in metres (m),
kilometres (km), miles, yards or feet.
Speed is usually measured in metres per
second (mps), kilometres per hour (kmph) or miles per
hour (mph).
Conversion of units
1 hour = 60 minutes = 60 x 60 seconds.
1 kilometre = 1000 metres.
1 kilometre = 0.6214 mile.
1 mile = 1.609 kilometre. i.e.
8 kilometres = 5 miles.
1 yard = 3 feet.
5
1 km/hr = 18 m/sec.
(To convert kmph to m/s we multiply by 5/18)
1 m /sec = 18 /5 km/hr.
(To convert m/s in to kmph
we multiply by 18/5)
1km = 5 miles =
1 miles = 22 ft
hr 8 hr hr 15 sec
If a man changes his speed in the ratio m : n, then the
ratio of time taken becomes n : m.
El. Walking at 5/6 of his usual rate, a
boy reaches his school 12 min late. Then
find his usual time taken to reach the school.
Sol. If the boy is
walking at 5/6th of his usual speed, then the time taken should become 6/5 times the original time
required. Thus, we can say that the boy is going to
require 1/5th more/extra time as compared to the usual time required (as
6 T-1T=
1 T).
5 5
Thus,1/ 5 T 12 min
1T X
Solve to get X = 60 min.
13
E2.
Walking at 11 th of his
usual rate, a boy takes 3 min less to reach
his school. Then find his usual time taken to reach the school.
Sol. If the boy is walking at 13/11 th of
his usual speed, then the time taken should become 11/13 times the
original time required.
Thus, we can say that the boy is going to require 2/13 th less
time as compared to the standard time required
Thus , 2/13 T 3 min.
1T X
Solve to get X
= 39/2 = 19.5 minutes.
If three men cover the same distance with speeds in
the ratio a : b : c, the times taken by these three will be respectively in the ratio 1/a :
1/b : 1/c. respectively in the ratio 1/a : 1/b : 1/c.
Average Speed
If A goes from X to Y at U kmph and comes back
from Y to X at V kmph, then Average speed during the whole journey
2U V / U+V kmph.
The
average speed is not (U + V)/2, but it is given by
Average Speed = Total Distance
Covered
Total Time Taken
E3.
If a person goes around a equilateral triangle shaped field at speeds of X, Y and Z kmph on the first, second
and the last side respectively and
reaches back to his starting point, then find his average speed during
the journey.
Sol. Let the measure of each side of A triangle
is D km. The person travelled the distance from A to B with X kmph, B to
C with Y kmph and C to A with Z kmph..
If
TAB
= Time taken by the person to travel from A to B,
TBC = Time taken by the person to travel from
B to C,
TCA = Time taken by the person to travel from
A to B,
Then, total time = TAB + TBC +
TCA
= D +D +D = D
(YZ+XZ+XY)
X Y
Z X Y Z
Total
distance travelled = D + D + D = 3D. Hence average speed
3D = 3XYZ kmph
D(YZ+XZ+XY) XY+YZ+ZX
XYZ
Similarly, other
questions involving average speed calculation in case of 4, 5 or more speeds
given can be found out.
Important
If two persons (or vehicles or trains) start at
the same time from two points A & B
towards each other and after crossing, they take X and Y hours in
reaching B & A respectively, then
Speed of first =
Y
Speed of second √ X
E4. A man starts from B to K, another from K to B at the same
time. After passing each other they complete their journeys in
10/3 and 24/5 hours, respectively. If the speed of the first is 12
kmph, then find the speed of the second man
…
1st man's speed =
√y/x = √ 24/5
= √ 24/5 x √3/10 = √36/25 =
6/5
2nd man's speed √10/3
12 =
6/5
2nd man's speed
2nd man's speed = 60 / 6 = 10 kmph.
Important
If a person/moving body
moves at a average speed of V1 kmph to cover a distance of D km without stopping and moves at a average speed
of V2 kmph to cover the same distance with stoppages,then
the stoppage
time per hour is given by V1
- V2
V1
E5. A train travels at a speed of 60 kmph between two
stations A and B, 240 km apart, when it does
not stop between any other station between them. But it goes at an average
speed of 40 kmph when it stops. What is the
average stoppage time per hour of the train?
Sol.
Time taken, when it does not stop = 240/60 = 4 hrs. Time taken, when it stops = 240/40 = 6 hrs. Total
stoppage time = 2 hrs.
Average
stoppage time per hour = 2/6 = 1/3 hrs
Short-cut: Using the formula directly,
we get 60-40 = 20 = 1
hrs.
60 60 3
Relative Speed
Suppose two trains A and B are moving with
same speed in the same direction on two
parallel tracks. To an observer sitting in train A, the train B appears to be stationary. This observation is expressed by saying that the relative speed of train A is
zero w.r.t. train B and the relative speed of the train B is zero w.r.t.
train A.
It is only due to the relative motion that to an observer sitting
in a moving train, the trees and the
telephone poles along the railway track appear to the moving in the
backward direction. If another train moving in the opposite direction, crosses
the train under consideration, it does so
with enormous speed, much larger than its actual speed. It is also due
to the relative speed.
The relative speed of a body A w.r.t. a body
B is the rate of change of position of
body A w.r.t. body B.
The relative speed of a body A w.r.t.
another body B when both are in motion, can be
determined by adding to the speed of A as well as to the speed of B.
Important
The time taken by a train L metres long,
travelling at a speed of X
m/sec in passing a signal post or a telegraph pole (or a standing man) is the same as the time taken by the
train to cover a distance of L metres i.e. its
own length because the length of the pole is nearly
zero with respect to the length of the train though its height may be significant.
Thus, T = L /X seconds.
The time taken by a train "L1"
metres long, travelling at a speed of X m/sec in passing a
stationary object (a bridge or a tunnel or a train at rest or a platform) of
length "L2" metres is the same as the time taken by the
train to cover a distance of (L1 + L2) metres.
Thus, T = (L1 + L2) seconds.
X
If two trains of lengths L1 and L2 metres
respectively, are moving in the same
direction with a speed of X and Y m/sec (where X > Y), then (X — Y)
m/sec is called their Relative Speed and time taken by faster train to pass the
slower train
= (L1 + L2) seconds.
(X — Y )
If two trains of lengths L1 and L2 metres
respectively, are moving in opposite directions (towards each other or away from each other) with a speed of X and Y m/sec,
then (X + Y) m/sec is called their Relative
Speed and
time taken by the trains to pass each
other
L, + L2 seconds.
X + Y
If a man is running at a speed of X m/sec
in the same direction in
which a train of length L metres is running at a speed Y m/ sec, then (Y — X) m/sec is called the speed
of the train relative
to man. Then the time taken by the train to cross the man
L seconds.
Y - X
If a man is running at a speed of X m/sec in a direction opposite to that in which a train of length L
metres is running with a speed Y m/sec, then X + Y is called the speed
of the train relative to man. Then the time taken by the train to cross the man
L seconds.
Y + X
If two trains or moving bodies are crossing
each other such that
Length of the first train or object = L1 metres,
Length of the second train or object = L2
metres,
Time taken by the two when crossing each other in opposite
direction = X sec,
Time taken by the two when crossing each
other in same direction = Y sec,
Then speed of the faster
train = L1 /2 + L2 /2 (1/X
+ 1/Y)
Speed of the slower train = L1
/2 + L2 /2 (1/X
- 1/Y)
E6. A train, 110 m long, travels at 60 kmph.
How long does it take to cross
(a)
a
telegraph post?
(b)
a
man running at 6 kmph in the same direction?
(c)
a
man running at 6 kmph in the opposite direction?
(d)
a
platform 240 m long?
(e) another train 170 m long, standing on another parallel track?
(f)
another train 170 m long, running at 54
kmph in same direction?
(g) another
train 170 metre long, running at 80 kmph in opposite direction?
Sol. Since 1 kmph = 5/18
m/s
Speed
of train = 60 kmph = 60 x 5/18 m/s.
(a) The telegraph post is a stationary object, so, the time taken by the train is the same as the time taken
by the train to cover a distance equal to its own length.
Required time = 110+ 0 = 6.6
seconds.
60 x 5 /18
60 x 5 /18
The man is moving in same direction, so length to be covered =
Length of the train and relative speed = speed of train — speed of man. So,
required time
110 = 110
= 7.33 second
(60 – 6) x 5/18
15
(b) The man is moving, in opposite direction, so length to be
covered = Length of the train and relative speed = speed of train + speed of
man So, required time
110 = 110 x 18 = 6 seconds.
(60+6) x 5/18 330
(c)
The platform is stationary of length 240
m. Length to be covered
= Length of the train + Length of the platform
= 110 +
240 = 350 m and relative speed = speed of train. So, required time
= 350/300
x 18 = 21 seconds.
(e
Another train is stationary
Length to be covered
= Length of the train + Length
of the other train
= 110 + 170 = 280 m and relative
speed = 60 kmph.
So, required time
280 x 3 /5 = 16.8 seconds.
Another train is moving in same direction.
Length to be covered
= Length of the train + Length of the other train
= 110 + 170 = 280 m and relative speed =
60 — 54 = 6 kmph. So, required time
=
280 x 3 / 5 = 168 seconds.
Here, another train is moving in opposite direction. Length to be
covered
= Length
of the train + Length of the other train = 110 + 170 = 280 m and
relative speed
= 60 + 80 = 140 kmph. So, required
time =280/140 x 18/5
= 7.2 seconds
E7. Two trains 110 m and 88 m long respectively are running in same
direction. The first runs at the rate of 35.2 kmph and the second at the rate of 44 kmph. How long will they take to cross each
other?
Sol. It is clear that the trains will cross each
other when they have travelled a
distance equal to the sum of their lengths =
110 + 88 = 198 m. Since they are moving in the same direction, we can find the
Relative Speed = 44 — 35.2 = 8.8 km/hr
Time required = distance/speed = 198 x 18
= 81 second.
8.8 5
For students practice
E8. A train in motion, 66 m long overtakes a train 88 m long
travelling @ 30 kmph in the opposite direction in 0.168 min. The speed of the
first train is
(1) 60 kmph (2)
25 kmph
(3) 45 kmph (4)
36 kmph
E9.
A
train travelling with constant speed crossed a 96 m long platform in 12 sec and
another 141 m long platform
in 15 sec. Find the length and speed of the
train.
Boats and Streams
A few important terminologies
The following terms will be used often while
discussing boats and streams.
Stream: It implies that the water in the river is moving
or flowing.
Upstream: Going
against the flow of the river.
Downstream: Going with the flow of the river.
Still water: It
implies that the speed of water is zero (generally in a lake).
Downstream (With the stream) Rowing
It
indicates that the stream favours the man's rowing (or boating). i.e. direction of rowing and direction of flow (stream)
is same.
Upstream (Against the stream) Rowing
It indicates that the stream
flows against the man's rowing (or boating)
i.e. direction of rowing and direction of stream (current) are opposite.
Important
Let the speed of a boat (or man) in still water be X
m/sec and the speed of the stream (or current) be Y m/sec.
Speed of boat with the stream (or
Downstream or D/S) =
(X + Y) m/sec.
Speed of boat against the stream (or Upstream or U/S) = (X - Y) m/sec.
Speed of man / boat in still water ,
= downstream
– upstream
2
E1O.A boat is rowed
down a river 28 km in 4 hours and up a river 12 km in
6 hours. Find the speed of the boat and the river.
Sol.
Downstream
speed is 28/4 = 7 kmph,
Upstream speed is12/ 6
= 2 kmph.
Speed
of Boat = 1/2 [downstream
+ upstream speed]
=
1/2 (7 + 2) = 4.5 kmph.
Speed
of current =1/ 2 (downstream - upstream speed)
= ½
( 7- 2 ) = 5/2 = 2.5 kmph
E11.A man rows 18 km down a
river in 4 hours with the stream and returns
in 12 hours. Find his speed and also the speed of the stream.
Sol. Speed
with the stream = 18/4 = 4.5 kmph.
Speed
against the stream = 18/12 = 1.5 kmph.
=
Speed of the stream = 1/2 {(4.5 - 1.5)} = 1.5 kmph and speed of the man = 4.5 -
1.5 = 3 kmph.
If a man capable of rowing at the
speed of X m/sec in still water, rows the
same distance up and down a stream flowing at a rate of Y m/sec, then his average speed throughout the
journey is
= Speed Upstream x Speed
Downstream = (X - Y)(X + Y)
Man's rate in still water X
Man's rate in still water X
Important
·
When downstream distance
= upstream distance, then
Man's
rate in still water = tup + tdown
speed of stream tup
- tdown
·
Average speed for total journey
(UP + DOWN) =
Upstream
rate x Downstream rate
Man's
rate in still water
·
Total Journey time (tup +
tdown)
= Man's
rate in still water x Total distance
Upstream rate x Downstream rate
Upstream rate x Downstream rate
E12.A man rows 10 km upstream and back again
to the starting point in 55 min. If the speed of stream is 2 kmph, then find
the speed of rowing in still water.
Sol.
Let x be the speed of rowing in still water.
y =
speed of stream = 2 kmph. Total time T =55/60 h. Hence,
Total time = Speed
in still water x Total distance
Upstream
rate x Downstream rate
= 55 =
X x 2 x 10 =
55 (X2 - 22 ) = 2 x X x 10
( X + 2) (X – 2) 60
= 11X2 – 240X – 44 =
0 =
(X-22)(11X+2) =0
X = 22 , since (-) ve
value is not admissible
Total Distance = Downstream distance + Upstream distance = 2 x any one side distance.
E13.A man who can swim 48 m/min in still water swims 200
m against the current and 200 m with the current. If the difference between these two
times is 10 min, then find the speed of the current
in m per min.
Sol. Let
the speed of stream be x kmph. The equation becomes
200 - 200 = 10
48- x 48+ x
= 200(48
+ x) - 200(48 - x) = 10[482 - x2]
xz +
40x - 2304 = 0.
On solving it we
get x = 32 and x = -72 (not acceptable) i.e.
speed of stream is 32 m/min.
DIRECTIONS: Read
the information given below and answer the question that follow.
An aeroplane
takes half an hour longer to fly a journey of 300 miles against the wind
than it takes to fly back with the wind (assumed constant throughout). The
speed of the plane in still air is 110 mph and that of the wind is U mph.
E14.
What
is the speed of the plane against the wind (in mph)?
(1) 100 (2) 130
(3) 120 (4) 140
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