APTITUDE PREPARATION: Time and Distance


INTRODUCTION:

Speed = Distance Travelled/ Time Taken
Total distance travelled = Average speed x Time taken
1 km/hr = 1000 / (60 x 60) = (5/18) m/s
1 m/s = (18/5) km/hr
If two objects are moving in the same direction then, the relative speed is the difference of the individual speeds
If two objects are moving in the opposite directions then, the relative speed is the sum of the individual speeds
Suppose a person covers a certain distance at a speed of X m/s and covers the SAME DISTANCE at a speed of Y m/s (both X and Y should be in the same units) while returning then,
Average Speed = (2.X.Y)/(X+Y) m/s
If the speed is changed at a ratio of x : y then, to cover the same distance, the ratio of time taken will change with a ratio of y:x
Let, the speed of the stream (river) is v km/hr and speed of the boat in still water is x km/hr.
Upstream: the speeds will oppose each other and resultant speed = (x-v)
Downstream: the speeds will get added and resultant speed = (x+v)

TRAIN PROBLEMS:

1. If a train has to cross a pole (or) a stationary man standing nearby (ie., anything that does not have a length), it has to cover its own length only
2. If a train has to cross a bridge (or) railway platform (ie., anything having a length), it has to cover the length of the bridge (or) platform also in addition to its own length
3. When two trains are moving either in the same direction (or) opposite directions then, they have to cover distance equal to the sum of the length of two trains.

Example 1:

A train which is moving at a speed of 72 km/hr passes a telegraph post in 10 seconds. What is the length of the train?

Solution:

Train passes a stationary post.
Speed of train = 72 km/hr = (72 x 5/18) m/s
Train length = Distance covered in 10 seconds = 10 x (72 x 5/18) = 200 m.

Example 2:

A train which is moving at a speed of 72 km/hr passes completely a bridge 400 m long in 30 seconds. What is the length of the train?

Solution:

Train passes a platform of length 400 m.
Speed of train = 72 km/hr = (72 x 5/18) m/s
(Train length + Platform length) = Distance covered in 30 seconds = 30 x (72 x 5/18)
(Train length + 400) = 600 m
Train length = (600-400) = 200 m.



Example 3:

In a train track, electric poles are at a distance of 90 m apart. A passenger travelling in the train counted 2000 poles during his 4 hours travel. Find the speed of the train.

Solution:

Total distance covered in m = Speed of train in m/s x Time taken in seconds
= Number of poles x Distance between two poles in m
Speed of train x (4x60x60) = 2000 x 90
Speed of train = (2000 x 90)/ (4x60x60)
= 12.5 m/s = (12.5 x 18/5) km/hr
= 45 km/hr

Example 4:

Two trains of length 200 m each are travelling in the same direction at speeds 36 km/hr and 72 km/hr. In how many seconds will they cross each other?

Solution:

Relative speed = (72-36) = 36 km/hr = 36x(5/18) m/s = 10 m/s
Total distance to be covered = Sum of train lengths = (200 + 200) = 400 m
Time taken = Distance / Speed = (400/10) = 40 seconds

Example 5:

If I walk at (5/7)th of my usual speed, I will reach my office 16 minutes late. What is my usual time taken?

Solution:

Distance covered is same in both the cases.
Usual speed = v m/minute
Usual time taken = T minutes
Distance covered = speed x time taken
(v x T) = [(5/7)v x (T+16)] … ( v gets cancelled)
T = (5/7)T + (5×16/7)
(2/7)T = (5×16/7)
T = (5x16x7)/(7×2) = 40 minutes

Example 6:

If you walk at a speed of 20 km/hr from your house, you will reach your office at 1005 hrs. If you run at a speed of 30 km/hr, you can reach your office at 0955 hrs. Find the distance between your house and office.

Solution:

If the speed is 30 km/hr, you can reach 10 minutes earlier = (10/60) = (1/6) hr
Distance covered is same in both the cases.
Time taken at 20 km/hr = T hrs
Distance covered = speed x time taken
(20 x T) = 30 x [T-(1/6)] = 30T –(30/6) = 30T – 5
10 T = 5 and T = (5/10) = 0.5 hrs
Distance = 20T = 20 x 0.5 = 10 km

Example 7:

In a 200 m race, A can beat B by 50 m and B can beat C by 8 m. In the same race, by how many metres can A beat C?

Solution:

Speeds are directly proportional to distances covered
If A covers 200 m, B covers 150 m
If B covers 200 m, C covers 192 m
If B covers 150 m, C covers (150/200)x192 = 144 m
The speeds ratios: A:B:C = 200:150:144
That is, A can beat C by (200-144) = 56 m.



Example 8:

The diameter of the wheel of a bicycle is 3.5 m. To cover a distance of 11 km, find the number of revolutions the bicycle has to make.

Solution:

While cycling, one revolution of the wheel will cover a distance equal to the circumference (perimeter).
Circumference of circle = 2πR = πD = (22/7) x 3.5 m ….. (π=3.14=22/7 approx.)
Distance covered in m= (11×1000) = Circumference in m x Number of revolutions
Number of revolutions = (11×1000)/(22/7)x3.5 = 1000.

Example 9:

A train which is travelling at constant speed crosses an electric post in 18 seconds and it takes 27 seconds to cross a bridge 99 m long. Find the train length.

Solution:

Train speed = v m/s
Crossing a post:
Distance covered = Train length = 18v … (i)
Crossing a bridge 99 m long:
Distance covered = (Train length + Bridge length) = 27v
Train length = 27v – Bridge length = 27v – 99 … (ii)
Equating: 27v – 99 = 18v
9v = 99
v = 11 m/s
Train length = 18 v = 18 x11 = 198 m.

Example 10:

A siren blows every 34 seconds. Anitha traveling in a bus moving towards the siren hears the second blow 33 seconds after she hears the first blow. Find the speed of the bus in km/hr. (Assume the speed of sound as 330 m/s)

Solution:

This can be well tackled by assuming the bus is stationary and the siren moves towards the bus.

Case (i):

Bus is stationary
Siren moves with a velocity of 330 m/s (velocity of sound).
Time taken = 34 s
Distance travelled = 34 x 330 = 11220 m

Case (ii):

Speed of bus = v m/s is added to the siren
Siren moves with a velocity of (330+v) m/s (velocity of sound + velocity of bus) towards the bus.
Time taken = 33 s
Distance travelled = 33 x (330+v)
Distance covered is same in both the cases.
Equating:
33 x (330+v) = 11220
v = (11220/33) – 330 = 10 m/s = 10 x(18/5) = 36 km/hr

About Author

Dr. BASKAR .A,

a former Research Scientist Government of India, presently working as a Professor in the Dept of Mechanical Engineering, Authored books on Kinematics and Dynamics of Machinery, Also Published Numerous Research papers moreover a Geek in Mathematics

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