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Time, Speed, and Distance Simplified

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Fundamentals of Time, Speed, and Distance

Distance = Speed X Time


Proportionalities implicit to the above equation.



1) Direct Proportionality between Time and Distance (When Speed is constant)


When speed is constant distance will vary as time

Car A moves for 2 hours @ 25kmph ---------> d = s X t -------> 50 = 25 X 2
Car B moves for 3 hours @ 25kmph ---------> d = s X t --------> 75 = 25 X 3
Here (tA / tB) = (dA / dB) -----> time ratio = distance ratio = 2 : 3


2) Direct Proportionality between Speed and Distance (When Time is constant)

If time of two bodies or motion is constant, then distance will vary as speed
Here (S1 / S2) = (d1 / d2)

If two cars start simultaneously from opposite ends towards each other. They meet at point ‘C’
In this case distance covered by each car will vary as their speeds since they travelled for equal period.

Suppose the distance AB = 900 km Speed A = 25kmph Speed B = 20kmph
In this case their meeting point (C) will be (S1 / S2) = (d1 / d2) ------> 25/20 ----> 5/4 = 500 / 400 i.e. 500 kms from A


3) Inverse Proportionality between Speed and Time (when Distance is constant)

If the distance to be covered is constant, then time will vary inversely as speed i.e. as speed increases, time decreases and vice versa.
(S1 / S2) = (t2 / t1)

Example 1 :- A Train meets with an accident and moves at ¾ of its original speed. Because of this it delayed by 20 minutes. What is the original time for the journey beyond the point of accident.
-----> Speed becomes ¾ hence time becomes 4/3 -------------------- remember (S1 / S2) = (t2 / t1))
-----> increased time is 1/3 = 20 minutes. So original time = 60 minutes. (1 belongs to 20mins, then 3 belongs to 60mins)

In other words

------> speed dropped to 75% and time increased to 133.33% ----> increased time = 33.33% = 20 minutes
Now 33.33 % belongs to 20 minutes, So 100% belongs to 60 minutes.

Example 2 :- A Man walked from his house to office at 5kmph and got 20 minutes late. if he had travelled at 7.5kmph, he would have reached 12 minutes early. The distance from his house to office is?

Here s1 = 5kmph t1 = t2 + 20 + 12 s2 = 7.5kmph t2 = t2
(S1 / S2) = (t2 / t1) ------> (5/7.5) = (t2 / t2 + 32) ---------> 5t2 + 160 = 7.5t2 -------> t2 = 64 minutes
So distance = 7.5 X 1.7 (64 mins) = 8km


Relative Speed


Relative Speed can be viewed as a movement of one body relative to another moving body.


1. Relative Speed of bodies moving in same direction.

In the case of the bodies moving to and fro between two points A and B, The faster body will reach the end first and will meet the second body on its way back. The relative speed S1 – S2 will apply till the point of reversal of the faster body and after that the two bodies will start to move in the opposite directions at a relative speed of S1 + S2. The relative speed governing the movement of the two bodies will alternate between S1 – S2 and S1 + S2 everytime anyone of the bodies reverses the direction. However, if both the bodies reverse their direction at the same instant, there will be no change in the relative speed equation.
In this case, the description of the motion of the two bodies between two consecutive meetings will also be governed by the proportionality between speed and distance – since the time of movement between any two meetings will be constant.

In this case, for every meeting, the total distance covered by the two bodies will be 2d (d = distance between the extreme points). The respective coverage of the distance will in the ratio of the individual speeds. Thus for the 9th meeting the total distance covered will be 9 X 2d = 18d


2. Relative Speed of bodies moving in opposite direction.

In the case of the bodies moving to and fro between two points A and B starting from opposite ends of the path, The two bodies will meet at a point in between A and B, then move apart away from each other. The faster body will reach its extreme point first followed by the slower body reaching its extreme point next. Relative speed will change every time ; one of the bodies reverses direction.
In this case, the position of the meeting point will be determined by the ratio of speeds of the bodies – since the time travelled of both the bodies is same. (Remember (S1 / S2) = (d1 / d2))
In this case, for the first meeting, the total distance covered by the two bodies will be d (d = distance between the extreme points). The respective coverage of the distance will in the ratio of the individual speeds. Thereafter, as the bodies separate and start coming together, the combined distance to be covered will be 2d. Thus for the 9th meeting the total distance covered will be d + 8 X 2d = 17d

NOTE :- TRY THIS CLASSIC EXAMPLE. IT CAN BE SOLVED USING ABOVE CONCEPT IN A SHORT TIME.
m03-09-ps-trains-71044.html


Example 1:- Two bodies A and B start from opposite ends P and Q of a straight road. They meet at a point 0.6d from P. Find the point of their fourth meeting.
-----> Since time is constant we have speed ratio as 3:2
-----> Total distance to be covered by the two together for the fourth meeting is d + 3d = 7d. This distance is divided in a ration of 3:2 thus we have that A will cover 4.2d and B will cover 2.8d
-----> We can find out the fourth meeting point by either tracking A’s movement or that of B’s
-----> A, having moved a distance of 4.2d, will be at a point 0.2d from P.


Example 2:- A start walking from a place at a uniform speed of 2 kmph in a particular direction. After half an hour, B starts from the same place and walks in the same direction as A at a uniform speed and overtakes A after 1 hour 48 minutes. Find the speed of B.
-----> A is walking at 2kmph and B started chasing him after half an hour so A must have covered 1 km distance till B starts chasing.
This distance of 1 km is covered by B in 1(48/60) i.e. in 1.8 hours.
So the equation is (SB – SA) X t = Distance -----> (SB – 2) X 1.8 = 1 ---------> SB – 2 = 1/1.8 -----> SB = 47 / 18.



Trains

Points to remember while solving such problems.

Train crosses a stationary object without length. St X t = Lt

Here the train has to cover its own length to completely cross any stationary object that have a negligible length - a standing person, Electric Pole, Tree etc – so here the distance covered( d) by the train while crossing will be equivalent to its own length
So the equation will be---------> speed of train X time to cross object = length of train


Train crosses a stationary object with length. St X t = Lt + Lo


Here the train has to cover its own length and the length of stationary object to completely cross that stationary object that have a certain length - Platform, another stationary train, bridge etc – so here the distance covered ( d) by the train while crossing will be equivalent to its own length + the length of stationary object
So the equation will be----------> speed of train X time to cross object = length of train + length of object

Example 1 :- A Train crosses a pole in 8 seconds. If the length of the train is 200 meters, find the speed of the train.
------> St X t = Lt -----> St X 8 = 200 --------> St = 200/8 --------> St = 25 m/s or 90kmph

Example 2 :- A Train travelling at 20 m/s crosses a platform in 30 seconds and a man standing on the platform in 18 seconds. what is the length of the platform
A) 240 meters
B) 360 meters
C) 420 meters
D) 600 meters
E) Cannot be determined

Solution :- Check both the statements one by one

I - Train travelling at 20 m/s crosses a platform in 30 seconds. Here train is crossing a stationary object with length --------------> St X t = Lt + Lo ------> 20 X 30 = Lt + Lo ------> 600 = Lt + Lo ----> 600 is the sum of train length and platform length

II - Train travelling at 20 m/s crosses a man standing on the platform in 18 seconds. Here train is crossing a stationary object without length --------------> St X t = Lt ------> 20 X 18 = Lt ------> 360 = Lt ----> train’s length is 360.

We will put this value in equation one 600 = Lt + Lo ----> 600 = 360 + Lo Lo = 240 so platform’s length will be 240. Choice A.



Train crosses a Moving object without length.

In same direction (St - So) X t = Lt

Here the train has to cover its own length to completely cross any moving object that have a negligible length - a running person, a motorist etc – so here the distance covered( d) by the train while crossing will be equivalent to its own length
So the equation will be ( speed of train – speed of object) X time to cross object = length of train

In Opposite direction (St + So) X t = Lt

Here the train has to cover its own length to completely cross any moving object that have a negligible length - a running person, a motorist , bullet of a weapon, arrow etc. – so here the distance covered( d) by the train while crossing will be equivalent to its own length
So the equation will be ( speed of train + speed of object) X time to cross object = length of train

NOTE :- The Concept below ,presented in italic style, is beyond the scope of GMAT. It is presented only for the purpose of fulfillment of topic and for them who are curious to know about it.
Readers can avoid it out rightly if they don’t find the information illuminating, as the chance of GMAT testing this concept in the exam is extremely rare, infact virtually zero.


Train crosses a Moving object with length.


In same direction (St - So) X t = Lt + Lo

Here the train has to cover its own length and the length of moving object to completely cross the object that have a certain length and that moving in the same direction – Train passing another(slower) train – so here the distance covered( d) by the train while crossing will be equivalent to its own length + the length of moving object
So the equation will be (speed of train – speed of object) X time to cross object = length of train + length of object

In Opposite direction (St + So) X t = Lt + Lo

Here the train has to cover its own length and the length of moving object to completely cross the object that have a certain length and that moving in the opposite direction – Train passing another(slower) train – so here the distance covered( d) by the train while crossing will be equivalent to its own length + the length of moving object
So the equation will be (speed of train – speed of object) X time to cross object = length of train + length of object


Here is a bit complex question
A Train crosses a man travelling in another train in the opposite direction in 8 seconds. However, the train requires 25 seconds to cross the same man if the trains are travelling in the same direction. If the length of the first train is 200 mtrs and that of the train in which the man is sitting is 160 mtrs, find the speed of the first train.
Note :- Here we should understand that the situation is one of the train crossing a moving object without length. Thus the length of the man’s train is useless or redundant data and given intentionally in order to confuse test taker.
A Train crosses a man travelling in another train in the opposite direction in 8 seconds.----> (St + Sm) X t = Lt ------------------------> (St + Sm) X 8 = 200 ------> St + Sm = 200/8 = 25 Note :- Sm stands for speed of a man’s train.
the train requires 25 seconds to cross the same man if the trains are travelling in the same direction ----> (St - Sm) X t = Lt ------------------------> (St - Sm) X 25 = 200 ------> St - Sm = 200/25 = 8
We have two unknowns and two equations. So solving them we get St = 16.5 m/s or 59.4kmph



Boats and Streams

The problems with boats and streams are also based on the basic equation Distance = Speed X Time
Following variables are generally used in these problems.
SB = Speed of Boat
SS = Speed of stream
The speed of the movement of the boat is dependent on how the boat is moving
1) In still water = SB
2) Moving Upstream = SB – SS
3) Moving Downstream = SB + SS


Average Speed.


We all know the general formula to calculate the average speed

Total Distance / Total Time


Example :- If Joe covered first 100km of his trip at 50kmph and rest 320 km at 80kmph. What was his average speed throughout the trip
-------> It took Joe 2 hours to cover first 100kms and 4 hours to cover rest 320 kms. Total time = 6 hours
Avg speed = (Total Distance / Total Time) -------> 420 / 6 -----> 70 kmph

When the distance covered for two journeys is same and we know the indivisual speeds of those journeys then average speed is given by

Average Speed = 2S1S2/(S1 + S2)


Example :- A car travels at 60 kmph from Mumbai to Pune and at 120kmph from Pune to Mumbai. What is the average speed of the car for the entire journey.
2S1S2/(S1 + S2) -------> 2 X 60 X 120 / 60 + 120 --------> 14400/180 --------> 80 kmph

On the GMAT, within time constraints, divisions like 14400/180 may take much of valuable time

Here is another way to calculate average speed (when distance is same)


Speeds are 60 and 120.
Their ratio will be ½
sum of numerator and denominator of ratio is 1 + 2 = 3
Difference in speeds is 120 – 60 = 60.
Divide this difference by the sum of numerator and denominator i.e. by 3 ---------> 60/3 = 20.
Now our Average speed will be 20 X 1 parts away from lower speed 60 + 20 X 1 = 80

See below examples for better understanding of this concept

Speed1 Speed2 Ratio of speeds Sum of ratio elements difference between speeds Division difference/sum Average speed
40 60 2/3 5 20 (20/5) = 4 40 + 4 X 2 = 48
45 105 3/7 10 60 (60/10) = 6 45 + 6 X 3 = 63
66 110 3/5 8 44 (44/8)=5.5 66 + 3 X 5.5 = 82.5



Here is the GMAT type example

Example 3 :- The Sinhagad Express left Pune for Mumbai at noon sharp. Two hours later, the Deccan Queen started from Pune in the same direction. The Deccan Queen passed the Sinhagad Express at 8 P.M.
Find the average speed of the two trains over the journey if the sum of their average speeds is 70 kmph
a) 34.28 kmph
b) 35 kmph
c) 50 kmph
d) 12 kmph
e) 16 kmph

Speed Sinhagad Express = x
Time Sinhagad Express = 8
Speed Deccan Queen = 70 – x (Sum of their speeds is 70)
Time Deccan Queen = 6
We know that Deccan queen started 2 hours later, Hence till the time Deccan starts, Sinhagad would have travelled for 2 hours and have covered 2x distance.
Deccan has covered this 2x distance by ((70 – x) – x) speed i.e. by 70 – 2x in 6 hours
as per the formula distance = speed x time ----> 2x = (70-2x)6 -----> 2x = 420 – 12x -----> 14x = 420 ----> x= 30
Note :- If you understood the principle Inverse Proportionality between Speed and Time (when Distance is constant) well, you can notice that since the time ratio is 4:3, speed ratio must be 3:4; Since sum of their speeds is 70, the lower speed must be 30 and higher speed must be 40. From here you can directly apply any of the formula to calculate average speed.


So
Speed Sinhagad = 30, Time Sinhagad = 8, Distance Sinhagad = 240

Speed Deccan = 40, Time Deccan = 6, Distance Deccan = 240

Their Speed Ratio = ¾, We know they travelled for equal distance so their time ratio will be 4/3

Average Speed = Total Distance / Total Time = 480 / 14 = 34.28 kmph
Average Speed = 2S1S2/(S1 + S2) -------> 2 X 30 X 40 / 30 + 40 --------> 2400/70 --------> 34.28 kmph
Average Speed = (40-30)/7 ---------> 30 + 3 X 1.42 -------> 30 + 4.26 ------> 34.26 kmph



Courtesy for the Information
1) Prof. Dr. R. D. Sharma - Author of CBSE Math Books
2) Mr. Arun Sharma - Alumnus IIM Bangalore

Special Thanks to
Mr. Mike McGarry – Magoosh GMAT Instructor[/size][/color]


Practice Problems.

Regards,

Narenn

1. Walking at ¾ of his normal speed, Mike is 16 minutes late in reaching his office. The usual time taken by him to cover the distance between his home and his office is
a. 48 minutes
b. 60 minutes
c. 42 minutes
d. 62 minutes
e. 66 minutes


2. Two trains for Mumbai leave Delhi at 6 am and 6.45 am and travel at 100 kmph and 136 kmph respectively. How many kilometers from Delhi will the two trains be together.
a. 262.4 km
b. 260 km
c. 283.33 km
d. 275 km
e. None of these


3. Ron walks to a viewpoint and returns to the starting point by his car and thus takes a total time of 6 hours 45 minutes. He would have gained 2 hours by driving both ways. How long would it have taken for him to walk both ways.
a. 8 h 45 min
b. 7 h 45 min
c. 5 h 30 min
d. 6 h 45 min
e. None of these


4. Two trains of length 100 m and 250 m run on parallel tracks. When they run in the same direction, they take 70 sec to cross each other and when they run in opposite directions, they take 10 sec to cross each other. The speed of the faster train is
a. 5 m/s
b. 15 m/s
c. 20 m/s
d. 25 m/s
e. 35 m/s


5. A man walking at a constant rate of 4 miles per hour is passed by a woman traveling in the same direction along the same path at a constant rate of 20 miles per hour. The woman stops to wait for the man 5 minutes after passing him, while the man continues to walk at his constant rate. How many minutes must the woman wait until the man catches up?
a. 16 mins
b. 20 mins
c. 24 mins
d. 25 mins
e. 28 mins


6. A dog is passed by a train in 8 seconds. Find the length of the train if its speed is 36 kmph
a. 70 mtrs Note the formula to convert the speed from kmph to m/s and vice versa
b. 80 mtrs Y kmph = 5Y/18 m/s and Y m/s = 18Y/5 kmph
c. 85 mtrs You can remember it as KM = 5/18 and MK = 18/5
d. 90 mtrs
e. 60 mtrs


7. A Train requires 7 seconds to pass a pole while it requires 25 seconds to cross a stationary train which is 378 mtrs long. Find the speed of the train.
a. 75.6 kmph
b. 75.4 kmph
c. 76.2 kmph
d. 21 kmph
e. 20 kmph


** Happy Solving **
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New post 08 Apr 2013, 08:21
Amazing effort. Thanks for this great work , and looking forward to seeing the remaining part soon :)
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New post 13 Apr 2013, 15:27
thanks for all these great articles ... awaiting your valuable additions to this article as promised ... thanks in advance :)
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TheNona wrote:
thanks for all these great articles ... awaiting your valuable additions to this article as promised ... thanks in advance :)


Yes! Sure.
Actually this permutations and combinations article took a lot more time than was expected, but surely will complete the remaining part of TSD in next 2 days.

Thanks,

Narenn
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New post 25 Aug 2013, 20:06
Where can I get the answers / explanations for the 7 practice problems of Time speed and distance mentioned here.
Appreciate your help.
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New post 25 Aug 2013, 22:43
Narenn wrote:
karannanda wrote:
Where can I get the answers / explanations for the 7 practice problems of Time speed and distance mentioned here.
Appreciate your help.


I shall post the OAs and Explanations soon. Till that time, you can post your answers here.





Sure.
My Answers below:
1. A (48 mins)
2. C (283.33 KMs)
3. A (8h 45min)
4. C (20 m/s)
5. B (20 mins)
6. B (80 m)
7. D (21 kmph)
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New post 27 Aug 2013, 12:07
Narenn wrote:
1. Walking at ¾ of his normal speed, Mike is 16 minutes late in reaching his office. The usual time taken by him to cover the distance between his home and his office is
a. 48 minutes
b. 60 minutes
c. 42 minutes
d. 62 minutes
e. 66 minutes


Correct Answer is A

REMEMBER the Inverse Proportionality between Speed and Time (when Distance is constant)

If the distance to be covered is constant, then time will vary inversely as speed i.e. as speed increases, time decreases and vice versa.
\(\frac{Speed1}{Speed2} = \frac{time2}{time1}\)

So in our case when mike started walking at \(\frac{3}{4}\) of his normal speed, he must have taken \(\frac{4}{3}\) of his usual time to complete the journey.

That means increased time, which is equivalent to 16 minutes, is 1/3 of usual time. So the usual time must be 48 minutes.
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New post 27 Aug 2013, 12:27
Narenn wrote:
2. Two trains for Mumbai leave Delhi at 6 am and 6.45 am and travel at 100 kmph and 136 kmph respectively. How many kilometers from Delhi will the two trains be together.
a. 262.4 km
b. 260 km
c. 283.33 km
d. 275 km
e. None of these


Correct Answer is C

Train 1 (that starts at 6:00 AM)
Train 2 (that starts at 6:45AM)

Train 1 has a 45 mins headstart. In 45 mins it travels \(100 * \frac{45}{60} = 75 kms\)

Relative speed of two trains (as they are traveling in the same direction) = 136 - 100 = 36 kmph

Train 2 will cover the headstart distance that Train 1 got in \(\frac{75}{36} = 2.083 hours\)

SO when the two trains meet, distance covered will be equivalent to that covered by Train 2 in 2.083 hours. which is 136 * 2.083 = 283.33 kms
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Hi Narenn,

Please post OE...
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Narenn wrote:

Two Important Concepts Circular Motion and Clocks will be added in the same article in next couple of days


Hi Narenn,

Please post the Circular motion and clocks theory please :)
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Re: Time, Speed, and Distance Simplified [#permalink]

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New post 18 Nov 2013, 10:15
Gnpth wrote:
Narenn wrote:

Two Important Concepts Circular Motion and Clocks will be added in the same article in next couple of days


Hi Narenn,

Please post the Circular motion and clocks theory please :)


Thanks for hitting the nail on my head - I almost forgot the posting of remaining part. I will do that on this weekend, positively. :)
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Re: Time, Speed, and Distance Simplified [#permalink]

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New post 04 Nov 2014, 13:58
Narenn wrote:
Gnpth wrote:
Narenn wrote:

Two Important Concepts Circular Motion and Clocks will be added in the same article in next couple of days


Hi Narenn,

Please post the Circular motion and clocks theory please :)


Thanks for hitting the nail on my head - I almost forgot the posting of remaining part. I will do that on this weekend, positively. :)




Hi Narenn,
Is there a link for the Circular motion and clocks theory and some related questions that i can refer to.
Thanks a lot.
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Re: Time, Speed, and Distance Simplified [#permalink]

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New post 29 Dec 2015, 10:07
Hi, are the answers for these 7 questions posted anywhere?
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Re: Time, Speed, and Distance Simplified [#permalink]

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New post 14 Mar 2016, 22:15
can i get the explanations please
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Re: Time, Speed, and Distance Simplified [#permalink]

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New post 14 Mar 2016, 22:32
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Re: Time, Speed, and Distance Simplified [#permalink]

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New post 20 Mar 2016, 02:31
Bunuel wrote:
gps5441 wrote:
can i get the explanations please


There are explanations in the topic. Any specific one you want to discuss?


Hi
Have been able to solve all except for the 3rd question.
can you please explain it to me
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Time, Speed, and Distance Simplified [#permalink]

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New post 20 Mar 2016, 02:44
gps5441 wrote:
Bunuel wrote:
gps5441 wrote:
can i get the explanations please


There are explanations in the topic. Any specific one you want to discuss?


Hi
Have been able to solve all except for the 3rd question.
can you please explain it to me


Ron walks to a viewpoint and returns to the starting point by his car and thus takes a total time of 6 hours 45 minutes. He would have gained 2 hours by driving both ways. How long would it have taken for him to walk both ways.

A. 8 h 45 min
B. 7 h 45 min
C. 6 h 45 min
D. 5 h 30 min
E. None of these

1. (Walking to to a viewpoint) + (Driving back) = (6 hours 45 minutes)

2. (Driving to a viewpoint) + (Driving back) = (6 hours 45 minutes - 2 hours) = (4 hours 45 minutes), therefore (one way driving) = (4 hours 45 minutes)/2 = (2 hours 22.5 minutes).

3. From 1. (one way driving) = (6 hours 45 minutes) - (2 hours 22.5 minutes) = (4 hours 22.5 minutes).

4. (Walking to to a viewpoint) + (Walking back) = (4 hours 22.5 minutes) + (4 hours 22.5 minutes) = (8 hours 45 minutes).

Answer: A.

Hope it's clear.
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Re: Time, Speed, and Distance Simplified [#permalink]

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New post 07 Apr 2016, 18:17
Narenn wrote:

2. Relative Speed of bodies moving in opposite direction.

In the case of the bodies moving to and fro between two points A and B starting from opposite ends of the path, The two bodies will meet at a point in between A and B, then move apart away from each other. The faster body will reach its extreme point first followed by the slower body reaching its extreme point next. Relative speed will change every time ; one of the bodies reverses direction.
In this case, the position of the meeting point will be determined by the ratio of speeds of the bodies – since the time travelled of both the bodies is same. (Remember (S1 / S2) = (d1 / d2))
In this case, for the first meeting, the total distance covered by the two bodies will be d (d = distance between the extreme points). The respective coverage of the distance will in the ratio of the individual speeds. Thereafter, as the bodies separate and start coming together, the combined distance to be covered will be 2d. Thus for the 9th meeting the total distance covered will be d + 8 X 2d = 17d

NOTE :- TRY THIS CLASSIC EXAMPLE. IT CAN BE SOLVED USING ABOVE CONCEPT IN A SHORT TIME.
m03-09-ps-trains-71044.html



Hi Narenn,
Thank you for this informative post. I have some questions concerning relative speed. I followed your suggestion and tried the question given in the link above; however, I cannot arrive at the correct answer using the method provided.


---
The question reads: Two trains continuously travel between Washington DC and Baltimore, which is 120 miles away. They start simultaneously, train A at Washington and train B at Baltimore, and run at 30 and 90 mph respectively. The station turnaround times are negligible. What is the distance between the point where the trains meet for the first time and the point where they meet for the second time?

Using the approach you suggested, I would solve the question in the following way: Sa/Sb=1/3 (Sa is the speed of train A, Sb speed of train B). So, Da/Db=1/3. For the first meeting, train A covered the distance: Da=(1/4)*120=30 miles.
The combined distance covered by the two train: D+2D=3D=3*120=360 miles.
Hence, for the second meeting, train A covered the distance: Da=(1/4)*3D=(1/4)*360=90 miles.
So, the distance between the the meeting: 90-30=60 miles.

However, the correct answer is 30 miles. What am I missing here? Please help.
---


As I tried to figure it out, I think the problem lies in the word "meet". If I am correct, your approach applies in the case the two bodies meet 'face to face', which means they are in opposite directions by the time they're about to meet. However, in the problem discussed above, the second meeting happens when the two trains 'cross' each other (the trains are in the same direction when they 'meet' the second time). In this case (the crossing case), the total distance covered by the two trains for the second meeting is not equal 2D (the faster train doesn't need to turn around at extreme point B to 'meet' the slower train; they 'cross' by the time the faster train reaches point B).

So, my questions:
1. Is my reasoning correct?
2. How should I interpret the word "meet" in this type of problem?

Thank you very much!
Nhi
Re: Time, Speed, and Distance Simplified   [#permalink] 07 Apr 2016, 18:17
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