The basic concepts of TSD are used in solving questions based on motion in a straight line, relative motion, circular motion, problems based on trains, problem based on boats, clocks, races, etc.
THEORY OF TSD
Concept of Motion and Mathematical Representation of Motion
Motion / movement occurs when a body of any shape or size changes its position with respect to any external stationary point. Thus, when a person travels from city A to city B, we say that he has moved from city A to city B. In general, whenever a body moves with respect to a stationary point, we say that the body has undergone a displacement / motion with respect to the starting point. Thus, for motion to have occurred, there must have been some displacement with respect to a stationary point on the ground.
The mathematical model that describes motion has three variables, namely: Speed, Time and Distance. The interrelationship between these three is also the most important formula namely:
Speed × Time = Distance (Equation for the description of one motion of one body)
The above equation is the mathematical description of the movement of a body. In complex problems, we tend to get confused regarding the usage of this equation and often end up mixing up the speed, time and distance of different motions of different bodies.
It must be mentioned here that this formula is the cornerstone of the chapter Time, Speed and Distance. Besides, this formula is also the source of the various formulae applied to the problems on the applications of time, speed and distance — to trains, boats and streams, clocks and races, circular motion and straight line motion.
In the equation above, speed can be defined as the rate at which distance is covered during the motion. It is measured in terms of distance per unit time and may have any combination of units of distance and time in the numerator and the denominator respectively. (m/s, km/hour, m/min, km/min, km/day, etc.) When we say that the speed of a body is S kmph, we mean to say that the body moves with S kmph towards or away from a stationary point (as the case may be).
Time (t) is the time duration over which the movement/motion occurs/has occurred. The unit used for measuring time is synchronous with the denominator of the unit used for measuring speed. Thus, if the speed is measured in terms of km/h then time is measured in hours.
Distance (d) is the displacement of the body during the motion.
The above equation, as is self-evident, is such that the interrelationship between the three parameters defines the value of the third parameter if two of the three are known. Hence we can safely say that if we know two of the three variables describing the motion, then the motion is fully described and every aspect of it is known.
The Proportionalities Implicit in the Equation S × T = D
The above equation has three implicit proportionality dimensions each of which has its own critical bearing on the solving of time, speed and distance problems.
1. Direct proportionality between time and distance (when the speed is constant) time μ distance
Illustration
A car moves for 2 hours at a speed of 25 kmph and another car moves for 3 hours at the same speed. Find the ratio of distances covered by the two cars.
Solution: Since, the speed is constant, we can directly conclude that time μ distance.
Hence
Since, the times of travel are 2 and 3 hours respectively, the ratio of distances covered is also 2/3.
Note: This can be verified by looking at the actual distances travelled—being 50 km and 75 km in this case.
2. Direct Proportionality between speed and distance (when the time is constant) speed μ distance
(a) A body travels at S1 kmph for the first 2 hours and then travels at S2 kmph for the next two hours. Here two motions of one body are being described and between these two motions the time is constant hence speed will be proportional to the distance travelled.
(b) Two cars start simultaneously from A and B respectively towards each other with speeds of S1 kmph and S2 kmph. They meet at a point C…. Here again, the speed is directly proportional to the distance since two motions are described where the time of both the motions is the same, that is, it is evident here that the first and the second car travel for the same time.
In such a case the following ratios will be valid: S1/S2 = d1/d2.
Illustrations
(i) A car travels at 30 km/h for the first 2 hours of a journey and then travels at 40 km/h for the next 2 hours of the journey. Find the ratio of the distances travelled at the two speeds.
Solution: Since time is constant between the two motions described, we can use the proportionality between speed and distance.
Hence, d1/d2 = s1/s2 = 3/4
Alternatively, you can also think in terms of percentage as d2 will be 33.33% higher than d1 since S2 is 33.33% higher than S1 and time is constant.
(ii) Two cars leave simultaneously from points A and B on a straight line towards each other. The distance between A and B is 100 km. They meet at a point 40 km from A. Find the ratio of their speeds.
Solution: Since time is the same for both the motions described, we have ratio of speed = ratio of distance.
SA/SB = 40/60 = 2/3
(iii) Two cars move simultaneously from points A and B towards each other. The speeds of the two cars are 20 m/s and 25 m/s respectively. Find the meeting point if d(AB) = 900 km.
Solution: For the bodies to meet, the time of travel is constant (since the two cars have moved simultaneously).
Hence, speed ratio = distance ratio Æ 4/5 = distance ratio
Hence, the meeting point will be 400 km from A and 500 km from B.
3. Inverse proportionality between speed and time (when the distance is constant) Speed μ 1/time
(a) A body travels at S1 kmph for the first half of he journey and then travels at S2 kmph for the second half of the journey. Here two motions of one body are being described and between these two motions the distance travelled is constant. Hence he speed will be inversely proportional to the time travelled for.
(b) Two cars start simultaneously from A and B respectively towards each other. They meet at a point C and reach their respective destinations B and A in t1 and t2 hours respectively. Here again, the speed is inversely proportional to the time since two motions are described where the distance of both the motions is the same, that is, it is evident here that the first and the second car travel for the distance, viz., AB. In such a case, the following ratio will be valid:
S1/S2 = t2/t1 i.e. S1t1 = S2t2 = S3t3 = K
Illustrations
(i) A train meets with an accident and moves at 3/4 its original speed. Due to this, it is 20 minutes late. Find the original time for the journey beyond the point of accident.
Solution: Speed becomes 3/4 (Time becomes 4/3)
Extra time = 1/3 of normal time = 20 minutes
Normal time = 60 minutes
Alternatively, from the table on product constancy in the chapter of percentages, we get that a 25% reduction in speed leads to a 33.33% increase in time.
But, 33.33% increase in time is equal to 20 minutes increase in time.
Hence, total time (original) = 60 minutes.
(ii) A body travels half the journey at 20 kmph and the other half at 30 kmph. Find the average speed.
Solution: The short-cut process is elucidated in the chapter on ‘averages’. Answer = 24 kmph.
(iii) A man travels from his house to his office at 5 km/h and reaches his office 20 minutes late. If his speed had been 7.5 km/h, he would have reached his office 12 minutes early. Find the distance from his house to his office.
Solution: Notice that here the distance is constant. Hence, speed is inversely proportional to time.
Solving mathematically
S1/S2 = t2/(t2 + 32)
5/7.5 = t2/(t2 + 32)
5t2 + 160 = 7.5 t2
t2 = 160/2.5 = 64 minutes
Hence, the distance is given by 7.5 × 64/60 = 8 km.
Alternatively, using the Product Constancy Table from the chapter of percentages. If speed increases by 50%, then time will decrease by 33.33%.
But the decrease is equal to 32 minutes.
Hence, original time = 96 minutes and new time is 64 minutes.
Hence, the required distance = 5 × 96/60 km = 8 km.
or distance = 7.5 × 64/60 km = 8 km
[Note: The entire process can be worked out mentally while reading the problem.]
CONVERSION BETWEEN kmph to m/s
1 km/h = 1000 m/h = 1000/3600 m/s = 5/18 m/s.
Hence, to convert y km/h into m/s multiply by 5/18.
Thus, y km/h = m/s.
And vice versa : y m/s = 18 y/5 km/h. To convert from m/s to kmph, multiply by 18/5.
Relative Speed : Same Direction and Opposite Direction
Normally, when we talk about the movement of a body, we mean the movement of the body with respect to a stationary point. However, there are times when we need to determine the movement and its relationships with respect to a moving point/body. In such instances, we have to take into account the movement of the body/point with respect to which we are trying to determine relative motion.
Relative movement, therefore, can be viewed as the movement of one body relative to another
moving body. The following formulae apply for the relative speed of two independent bodies with respect to each other:
Case I: Two bodies are moving in opposite directions at speeds S1 and S2 respectively.
The relative speed is defined as S1 + S2
Case II: Two bodies are moving in the same direction.
The relative speed is defined as
(a) S1 – S2 when S1 is greater than S2.
(b) S2 – S1 when S1 is lesser than S2.
In other words, the relative speed can also be defined as the positive value of the difference between the two speeds, that is, | S1 – S2 |.
Motion in a Straight Line
Motion in a straight line is governed by the rules of relative speed enumerated above.
A. Two or more bodies starting from the same point and moving in the same direction: Their relative speed is S1 – S2.
(a) 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 every time any one of the bodies reverses directions. 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).
Distances covered in this case: For every meeting, he total distance covered by the two bodies will be 2D (where D is the distance between the extreme points). However, notice that the value of 2D would be applicable only if both the bodies reverse the direction between two meetings. In case only one body has reversed direction, the total distance would need to be calculated on a case-by-case basis. The respective coverage of the distance is in the ratio of the individual speeds.
Thus, for the 9th meeting (if both bodies have reversed direction between every 2 meetings) the total
distance covered will be 9 × 2D = 18D.
This will be useful for solving problems that require the calculation of a meeting point.
(b) In the case of the bodies continuing to move in the same direction without coming to an end point
and reversing directions: The faster body will take a lead and will keep increasing the lead and the movement of the two bodies will be governed by the relative speed equation: S1 – S2. Here again, if the two bodies start simultaneously, their movement will be governed by the direct proportionality between speed and distance.
B. Moving in the opposite direction: Their relative speed will be initially given by S1 + S2.
(a) 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 move towards each other, 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.
The position of the meeting point will be determined by the ratio of the speeds of the bodies (since the 2 movements can be described as having the time constant between them).
Distances covered in the above case: For the first meeting, the total distance covered by the two bodies will be D (the distance between the extreme points). The coverage of the distance is in the ratio of the individual speeds. Thereafter, as the bodies separate and start coming together, the combined distance to be covered is 2D. Note that if only one body is reversing direction between two meetings, this would not be the case and you will have to work it out.
Thus, for the 10th meeting (if both bodies have reverse direction between every 2 meetings) the total
distance covered will be D + 9 × 2D = 19D. This will be useful for solving of problems that require the calculation of a meeting point.
Illustrations
(i) 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.
Solution: Since time is constant, we have ratio of speeds as 3 : 2.
Also, total distance to be covered by the two together for the fourth meeting is 7D. This distance is divided in a ratio of 3 : 2 and thus we have that A will cover 4.2D and B will cover 2.8D. The fourth meeting point can then be found out by tracking either A or B’s movement. A, having moved a distance of 4.2D, will be at a point 0.2D from P. This is the required answer.
(ii) A starts walking from a place at a uniform speed of 2 km/h 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. Calculate the speed of B.
Solution: Start solving as you read the question. From the first two sentences you see that A is 1 km ahead of B when B starts moving.
This distance of 1 km is covered by B in 9/5 hours [1 hour 48 minutes = 1(4/5) = 9/5 hours].
The equation operational here (SB – SA) × T = initial distance
(SB – 2) × 9/5 = 1
Solving, we get SB = 23/9 km/h.
(b) In the case of the bodies continuing to move in the same direction without coming to an end point and reversing directions: The bodies will meet and following their meeting they will start separating and going away from each other. The relative speed will be given by S1 + S2 initially while approaching each other and, thereafter, it will be S1 + S2 while moving away from each other.
Try to visualise how two bodies separate and then come together. Also, clearly understand the three proportionalities in the equation s × t = d, since these are very important tools for problem solving.
Concept of Acceleration
Acceleration is defined as the rate of change of speed.
Acceleration can be positive (speed increases) or negative (speed decreases Æ also known as deceleration)
The unit of acceleration is speed per unit time (e.g. m/s2)
For instance, if a body has an initial speed of 5 m/s and a deceleration of 0.1 m/s2 it will take 50 seconds to come to rest.
Final speed = Initial speed + Acceleration × Time
Some more examples:
(i) Water flows into a cylindrical beaker at a constant rate. The base area of the beaker is 24 cm2. The water level rises by 10 cm every second. How quickly will the water level rise in a beaker with a base area of 30 cm2.
Solution: The flow of water in the beaker is 24 cm2 × 10 cm/s = 240 cm3/s.
If the base area is 30 cm2 then the rate of water level rise will be 240/30 = 8 cm/s.
Note: In case of confusion in such questions we are advised to use dimensional analysis to understand what to multiply and what to divide.
(ii) A 2 kilowatt heater can boil a given amount of water in 10 minutes. How long will it take for
(a) a less powerful heater of 1.2 kilowatts to boil the same amount of water?
(b) a less powerful heater of 1.2 kilowatts to boil double the amount of water?Solution:
(a) The heating required to boil the amount of water is 2 × 10 = 20 kilowatt minutes. At the rate of 1.2 kilowatt, this heat will be generated in 20/1.2 minutes = 16.66 minutes.
(b) When the water is doubled, the heating required is also doubled. Hence, heating required = 40
kilowatt minutes. At the rate of 1.2 kilowatt, this heat will be generated in 40/1.2 = 33.33 minutes.
AN APPLICATION OF ALLIGATION IN TIME, SPEED AND DISTANCE
Consider the following situation:
Suppose a car goes from A to B at an average speed of S1 and then comes back from B to A at an average speed of S2. If you had to find out the average speed of the whole journey, what would you do?
The normal short cut given for this situation gives the average speed as: (2S1S2) / (S1+S2)
However, this situation can be solved very conveniently using the process of alligation as explained
below:
Since, the two speeds are known to us, we will also know their ratio. The ratio of times for the two parts of the journey will then be the inverse ratio of the ratio of speeds. (Since the distance for the two journeys are equal). The answer will be the weighted average of the two speeds (weighted on the basis of the time travelled at each speed)
The process will become clear through an example:
A car travels at 60 km/h from Mumbai to Poona and at 120 km/h from Poona to Mumbai. What is the average speed of the car for the entire journey.
Solution
The process of alligation, will be used here to give the answer as 80.
Note here, that since the speed ratio is 1:2, the value of the time ratio used for calculating the weighted average will be 2:1.
What will happen in case the distances are not constant?
For instance, if the car goes 100 km at a speed of 66kmph and 200 km at a speed of 110 kmph, what will be the average speed?
In this case the speed ratio being 6:10 i.e. 3:5 the inverse of the speed ratio will be 5:3. This would have been the ratio to be used for the time ratio in case the distances were the same (for both the speeds). But since the distances are different, we cannot use this ratio in this form. The problem is overcome by multiplying this ratio (5:3) by the distance ratio (in this case it is 1:2) to get a value of 5:6. This is the ratio which has to be applied for the respective weights. Hence, the alligation will look like:
Solution
Thus the required answer is 90 kmph.
APPLICATIONS OF TIME, SPEED AND DISTANCE
Trains
Trains are a special case in questions related to time, speed and distance because they have their own theory and distinct situations.
The basic relation for trains problems is the same: Speed × Time = Distance.
The following things need to be kept in mind before solving questions on trains:
(a) When the train is crossing a moving object, the speed has to be taken as the relative speed of the train with respect to the object. All the rules for relative speed will apply for calculating the relative speed.
(b) The distance to be covered when crossing an object whenever a train crosses an object will be equal to: Length of train + Length of object
Thus, the following cases will yield separate equations, which will govern the crossing of the object by the train:
For each of the following situations the following notations have been used:
ST = Speed of train SO = Speed of object t = time
LT = Length of train LO = Length of object
Case I: Train crossing a stationary object without length:
ST × t = LT
Case II: Train crossing a stationary object with length:
ST × t = (LT + LO)
Case III: Train crossing a moving object without length:
• In opposite direction: (ST + SO) × t = LT
• In same direction: (ST – SO) × t = LT
Case IV: Train crossing a moving object with length:
• In opposite direction: (ST + SO) × t = (LT + LO)
• In same direction: (ST – SO) × t = (LT + LO)
[Note: In order for a train to completely cross a stationary point on the ground, the train has to traverse a distance that is equal to its entire length.
This can be visualised by remembering yourself stationary on a railway platform and being crossed by a train. You would say that the train starts crossing you when the engine of the train comes in line with you. Also, you would say that you have been crossed by the train when the end of the guard’s compartment comes in line with you. Thus, the train would have travelled its own length in crossing you].
Illustrations
(i) A train crosses a pole in 8 seconds. If the length of the train is 200 metres, find the speed of the train.
Solution: In this case, it is evident that the situation is one of the train crossing a stationary object without length. Hence, Case I is applicable here.
Thus, ST = 200/8 = 25 m/s Æ 25 × 18/5 = 90 kmph.
(ii) A train crosses a man travelling in another train in he 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 metres and that of the train in which the man is sitting is 160 metres, find the speed of the first train.
Solution: Here, the student 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. Then applying the relevant formulae after considering the directions of the movements we get the equations:
ST + SM = 25
ST – SM = 8
ST = 33/2 = 33/2 * 18/5 = 59.4 kmph
Boats and Streams
The problems of boats and streams are also dependent on the basic equation of time, speed and distance : Speed × Time = Distance.
However, as in the case of trains the adjustments to be made for solving questions on boats and streams are:
The boat has a speed of its own, which is also called the speed of the boat in still water (SB).
Another variable that is used in boats and streams problems is the speed of the stream (SS).
The speed of the movement of the boat is dependent on whether the boat is moving:
(a) In still water the speed of movement is given by Æ SB.
(b) While moving upstream (or against the flow of the water), the speed of movement is given by Æ
SU = SB – SS
(c) While moving downstream (or with the flow of the water), the speed of movement is given by Æ
SD = SB + SS
The time of movement and the distance to be covered are to be judged by the content of the problem.
Circular motion: A special case of movement is when two or more bodies are moving around a circular track.
The relative speed of two bodies moving around a circle in the same direction is taken as S1 – S2.
Also, when two bodies are moving around a circle in the opposite direction, the speed of the two
bodies is taken to be S1 + S2.
The peculiarity inherent in moving around a circle in the same direction is that when the faster body overtakes the slower body it goes ahead of it. And for every unit time that elapses, the faster body keeps increasing the distance by which the slower body is behind the faster body. However, when the distanceby which the faster body is in front of the slower body becomes equal to the circumference of the circle
around which the two bodies are moving, the faster body again comes in line with the slower body. This event is called as overlapping or lapping of the slower body by the faster body. We say that the slower body has been lapped or overlapped by the faster body.
First meeting: Three or more bodies start moving simultaneously from the same point on the circumference of the circle, in the same direction around the circle. They will first meet again in the LCM of the times that the fastest runner takes in totally overlapping each of the slower runners.
For instance, if A, B, C and D start clockwise from a point X on the circle such that A is the fastest runner then we can define TAB as the time in which A completely overlaps B, TAC as the time in which A completely overlaps C and TAD as the time in which A completely overlaps D. Then the LCM of TAB, TAC and TAD will be the time in which A, B, C and D will be together again for the first time.
First meeting at starting point: Three or more bodies start moving simultaneously from the same point
on the circumference of a circle, in the same direction around the circle. Their first meeting at the starting point will occur after a time that is got by the LCM of the times that each of the bodies takes to complete one full round.
For instance, if A, B and C start from a point X on the circle such that TA, TB and TC are the times in which A, B and C respectively cover one complete round around the circle, then they will all meet together at the starting point in the LCM of TA, TB and TC.
Clocks
Problems on clocks are based on the movement of the minute hand and that of the hour hand as well as on the relative movement between the two. In my opinion, it is best to solve problems on clocks by considering a clock to be a circular track having a circumference of 60 km and each kilometre being represented by one minute on the dial of the clock. Then, we can look at the minute hand as a runner running at the speed of 60 kmph while we can also look at the hour hand as a runner running at an average speed of 5 kmph.
Since, the minute hand and the hour hand are both moving in the same direction, the relative speed of the minute hand with respect to the hour hand is 55 kmph, that is, for every hour elapsed, the minute hand goes 55 km (minute) more than the hour hand.
(Beyond this slight adjustment, the problems of clocks require a good understanding of unitary method. This will be well illustrated through the solved example below.)
Important Information
Number of right angles formed by a clock: A clock makes 2 right angles between any 2 hours. Thus, for instance, there are 2 right angles formed between 12 to 1 or between 1 and 2 or between 2 and 3 or between 3 and 4 and so on.
However, contrary to expectations, the clock does not make 48 right angles in a day. This happens because whenever the clock passes between the time period 2–4 or between the time period 8–10 there are not 4 but only 3 right angles.
This happens because the second right angle between 2–3 (or 8–9) and the first right angle between 3–4 (or 9–10) are one and the same, occurring at 3 or 9.
Right angles are formed when the distance between the minute hand and the hour hand is equal to 15 minutes.
Exactly the same situation holds true for the formation of straight lines. There are 2 straight lines in every hour. However, the second straight line between 5–6 (or 11–12) and the first straight line between 6–7 (or 12–1) coincide with each other and are represented by the straight line formed at 6 (or 12).
Straight lines are formed when the distance between he minute hand and the hour hand is equal to either 0 minutes or 30 minutes.
Illustration
At what time between 2–3 p.m. is the first right angle in that time formed by the hands of the clock?
Solution: At 2 p.m. the minute hand can be visualised as being 10 kilometres behind the hour hand. (considering the clock dial to be a race track of circumference 60 km such that each minute represents a kilometre).
Also, the first right angle between 2–3 is formed when the minute hand is 15 kilometres ahead of the hour hand.
Thus, the minute hand has to cover 25 kilometres over the hour hand.
This can be written using the unitary method:
Distance covered by the minute hand over the hour hand.
55 kilometres in 1 hour
25 kilometres in what time?
Æ 5/11 of an hour.
Thus, the first right angle between 2–3 is formed at 5/11 hours past 2 o’clock.
This can be converted into minutes and seconds using unitary method again as:
1 hour 60 minutes
5/11 hours ? minutes
Æ 300/11 minutes = 27 (3/11) minutes
1 minute 60 seconds
3/11 minutes ? seconds Æ 180/11 seconds = 16.3636 seconds.
Hence, the required answer is: 2 : 27 : 16.36 seconds.
CONVERSION BETWEEN kmph to m/s
1 km/h = 1000 m/h = 1000/3600 m/s = 5/18 m/s.
Hence, to convert y km/h into m/s multiply by 5/18.
Thus, y km/h = m/s.
And vice versa : y m/s = 18 y/5 km/h. To convert from m/s to kmph, multiply by 18/5.
Relative Speed : Same Direction and Opposite Direction
Normally, when we talk about the movement of a body, we mean the movement of the body with respect to a stationary point. However, there are times when we need to determine the movement and its relationships with respect to a moving point/body. In such instances, we have to take into account the movement of the body/point with respect to which we are trying to determine relative motion.
Relative movement, therefore, can be viewed as the movement of one body relative to another
moving body. The following formulae apply for the relative speed of two independent bodies with respect to each other:
Case I: Two bodies are moving in opposite directions at speeds S1 and S2 respectively.
The relative speed is defined as S1 + S2
Case II: Two bodies are moving in the same direction.
The relative speed is defined as
(a) S1 – S2 when S1 is greater than S2.
(b) S2 – S1 when S1 is lesser than S2.
In other words, the relative speed can also be defined as the positive value of the difference between the two speeds, that is, | S1 – S2 |.
Motion in a Straight Line
Motion in a straight line is governed by the rules of relative speed enumerated above.
A. Two or more bodies starting from the same point and moving in the same direction: Their relative speed is S1 – S2.
(a) 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 every time any one of the bodies reverses directions. 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).
Distances covered in this case: For every meeting, he total distance covered by the two bodies will be 2D (where D is the distance between the extreme points). However, notice that the value of 2D would be applicable only if both the bodies reverse the direction between two meetings. In case only one body has reversed direction, the total distance would need to be calculated on a case-by-case basis. The respective coverage of the distance is in the ratio of the individual speeds.
Thus, for the 9th meeting (if both bodies have reversed direction between every 2 meetings) the total
distance covered will be 9 × 2D = 18D.
This will be useful for solving problems that require the calculation of a meeting point.
(b) In the case of the bodies continuing to move in the same direction without coming to an end point
and reversing directions: The faster body will take a lead and will keep increasing the lead and the movement of the two bodies will be governed by the relative speed equation: S1 – S2. Here again, if the two bodies start simultaneously, their movement will be governed by the direct proportionality between speed and distance.
B. Moving in the opposite direction: Their relative speed will be initially given by S1 + S2.
(a) 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 move towards each other, 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.
The position of the meeting point will be determined by the ratio of the speeds of the bodies (since the 2 movements can be described as having the time constant between them).
Distances covered in the above case: For the first meeting, the total distance covered by the two bodies will be D (the distance between the extreme points). The coverage of the distance is in the ratio of the individual speeds. Thereafter, as the bodies separate and start coming together, the combined distance to be covered is 2D. Note that if only one body is reversing direction between two meetings, this would not be the case and you will have to work it out.
Thus, for the 10th meeting (if both bodies have reverse direction between every 2 meetings) the total
distance covered will be D + 9 × 2D = 19D. This will be useful for solving of problems that require the calculation of a meeting point.
Illustrations
(i) 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.
Solution: Since time is constant, we have ratio of speeds as 3 : 2.
Also, total distance to be covered by the two together for the fourth meeting is 7D. This distance is divided in a ratio of 3 : 2 and thus we have that A will cover 4.2D and B will cover 2.8D. The fourth meeting point can then be found out by tracking either A or B’s movement. A, having moved a distance of 4.2D, will be at a point 0.2D from P. This is the required answer.
(ii) A starts walking from a place at a uniform speed of 2 km/h 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. Calculate the speed of B.
Solution: Start solving as you read the question. From the first two sentences you see that A is 1 km ahead of B when B starts moving.
This distance of 1 km is covered by B in 9/5 hours [1 hour 48 minutes = 1(4/5) = 9/5 hours].
The equation operational here (SB – SA) × T = initial distance
(SB – 2) × 9/5 = 1
Solving, we get SB = 23/9 km/h.
(b) In the case of the bodies continuing to move in the same direction without coming to an end point and reversing directions: The bodies will meet and following their meeting they will start separating and going away from each other. The relative speed will be given by S1 + S2 initially while approaching each other and, thereafter, it will be S1 + S2 while moving away from each other.
Try to visualise how two bodies separate and then come together. Also, clearly understand the three proportionalities in the equation s × t = d, since these are very important tools for problem solving.
Concept of Acceleration
Acceleration is defined as the rate of change of speed.
Acceleration can be positive (speed increases) or negative (speed decreases Æ also known as deceleration)
The unit of acceleration is speed per unit time (e.g. m/s2)
For instance, if a body has an initial speed of 5 m/s and a deceleration of 0.1 m/s2 it will take 50 seconds to come to rest.
Final speed = Initial speed + Acceleration × Time
Some more examples:
(i) Water flows into a cylindrical beaker at a constant rate. The base area of the beaker is 24 cm2. The water level rises by 10 cm every second. How quickly will the water level rise in a beaker with a base area of 30 cm2.
Solution: The flow of water in the beaker is 24 cm2 × 10 cm/s = 240 cm3/s.
If the base area is 30 cm2 then the rate of water level rise will be 240/30 = 8 cm/s.
Note: In case of confusion in such questions we are advised to use dimensional analysis to understand what to multiply and what to divide.
(ii) A 2 kilowatt heater can boil a given amount of water in 10 minutes. How long will it take for
(a) a less powerful heater of 1.2 kilowatts to boil the same amount of water?
(b) a less powerful heater of 1.2 kilowatts to boil double the amount of water?Solution:
(a) The heating required to boil the amount of water is 2 × 10 = 20 kilowatt minutes. At the rate of 1.2 kilowatt, this heat will be generated in 20/1.2 minutes = 16.66 minutes.
(b) When the water is doubled, the heating required is also doubled. Hence, heating required = 40
kilowatt minutes. At the rate of 1.2 kilowatt, this heat will be generated in 40/1.2 = 33.33 minutes.
AN APPLICATION OF ALLIGATION IN TIME, SPEED AND DISTANCE
Consider the following situation:
Suppose a car goes from A to B at an average speed of S1 and then comes back from B to A at an average speed of S2. If you had to find out the average speed of the whole journey, what would you do?
The normal short cut given for this situation gives the average speed as: (2S1S2) / (S1+S2)
However, this situation can be solved very conveniently using the process of alligation as explained
below:
Since, the two speeds are known to us, we will also know their ratio. The ratio of times for the two parts of the journey will then be the inverse ratio of the ratio of speeds. (Since the distance for the two journeys are equal). The answer will be the weighted average of the two speeds (weighted on the basis of the time travelled at each speed)
The process will become clear through an example:
A car travels at 60 km/h from Mumbai to Poona and at 120 km/h from Poona to Mumbai. What is the average speed of the car for the entire journey.
Solution
Note here, that since the speed ratio is 1:2, the value of the time ratio used for calculating the weighted average will be 2:1.
What will happen in case the distances are not constant?
For instance, if the car goes 100 km at a speed of 66kmph and 200 km at a speed of 110 kmph, what will be the average speed?
In this case the speed ratio being 6:10 i.e. 3:5 the inverse of the speed ratio will be 5:3. This would have been the ratio to be used for the time ratio in case the distances were the same (for both the speeds). But since the distances are different, we cannot use this ratio in this form. The problem is overcome by multiplying this ratio (5:3) by the distance ratio (in this case it is 1:2) to get a value of 5:6. This is the ratio which has to be applied for the respective weights. Hence, the alligation will look like:
Solution
APPLICATIONS OF TIME, SPEED AND DISTANCE
Trains
Trains are a special case in questions related to time, speed and distance because they have their own theory and distinct situations.
The basic relation for trains problems is the same: Speed × Time = Distance.
The following things need to be kept in mind before solving questions on trains:
(a) When the train is crossing a moving object, the speed has to be taken as the relative speed of the train with respect to the object. All the rules for relative speed will apply for calculating the relative speed.
(b) The distance to be covered when crossing an object whenever a train crosses an object will be equal to: Length of train + Length of object
Thus, the following cases will yield separate equations, which will govern the crossing of the object by the train:
For each of the following situations the following notations have been used:
ST = Speed of train SO = Speed of object t = time
LT = Length of train LO = Length of object
Case I: Train crossing a stationary object without length:
ST × t = LT
Case II: Train crossing a stationary object with length:
ST × t = (LT + LO)
Case III: Train crossing a moving object without length:
• In opposite direction: (ST + SO) × t = LT
• In same direction: (ST – SO) × t = LT
Case IV: Train crossing a moving object with length:
• In opposite direction: (ST + SO) × t = (LT + LO)
• In same direction: (ST – SO) × t = (LT + LO)
[Note: In order for a train to completely cross a stationary point on the ground, the train has to traverse a distance that is equal to its entire length.
This can be visualised by remembering yourself stationary on a railway platform and being crossed by a train. You would say that the train starts crossing you when the engine of the train comes in line with you. Also, you would say that you have been crossed by the train when the end of the guard’s compartment comes in line with you. Thus, the train would have travelled its own length in crossing you].
Illustrations
(i) A train crosses a pole in 8 seconds. If the length of the train is 200 metres, find the speed of the train.
Solution: In this case, it is evident that the situation is one of the train crossing a stationary object without length. Hence, Case I is applicable here.
Thus, ST = 200/8 = 25 m/s Æ 25 × 18/5 = 90 kmph.
(ii) A train crosses a man travelling in another train in he 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 metres and that of the train in which the man is sitting is 160 metres, find the speed of the first train.
Solution: Here, the student 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. Then applying the relevant formulae after considering the directions of the movements we get the equations:
ST + SM = 25
ST – SM = 8
ST = 33/2 = 33/2 * 18/5 = 59.4 kmph
Boats and Streams
The problems of boats and streams are also dependent on the basic equation of time, speed and distance : Speed × Time = Distance.
However, as in the case of trains the adjustments to be made for solving questions on boats and streams are:
The boat has a speed of its own, which is also called the speed of the boat in still water (SB).
Another variable that is used in boats and streams problems is the speed of the stream (SS).
The speed of the movement of the boat is dependent on whether the boat is moving:
(a) In still water the speed of movement is given by Æ SB.
(b) While moving upstream (or against the flow of the water), the speed of movement is given by Æ
SU = SB – SS
(c) While moving downstream (or with the flow of the water), the speed of movement is given by Æ
SD = SB + SS
The time of movement and the distance to be covered are to be judged by the content of the problem.
Circular motion: A special case of movement is when two or more bodies are moving around a circular track.
The relative speed of two bodies moving around a circle in the same direction is taken as S1 – S2.
Also, when two bodies are moving around a circle in the opposite direction, the speed of the two
bodies is taken to be S1 + S2.
The peculiarity inherent in moving around a circle in the same direction is that when the faster body overtakes the slower body it goes ahead of it. And for every unit time that elapses, the faster body keeps increasing the distance by which the slower body is behind the faster body. However, when the distanceby which the faster body is in front of the slower body becomes equal to the circumference of the circle
around which the two bodies are moving, the faster body again comes in line with the slower body. This event is called as overlapping or lapping of the slower body by the faster body. We say that the slower body has been lapped or overlapped by the faster body.
First meeting: Three or more bodies start moving simultaneously from the same point on the circumference of the circle, in the same direction around the circle. They will first meet again in the LCM of the times that the fastest runner takes in totally overlapping each of the slower runners.
For instance, if A, B, C and D start clockwise from a point X on the circle such that A is the fastest runner then we can define TAB as the time in which A completely overlaps B, TAC as the time in which A completely overlaps C and TAD as the time in which A completely overlaps D. Then the LCM of TAB, TAC and TAD will be the time in which A, B, C and D will be together again for the first time.
First meeting at starting point: Three or more bodies start moving simultaneously from the same point
on the circumference of a circle, in the same direction around the circle. Their first meeting at the starting point will occur after a time that is got by the LCM of the times that each of the bodies takes to complete one full round.
For instance, if A, B and C start from a point X on the circle such that TA, TB and TC are the times in which A, B and C respectively cover one complete round around the circle, then they will all meet together at the starting point in the LCM of TA, TB and TC.
Clocks
Problems on clocks are based on the movement of the minute hand and that of the hour hand as well as on the relative movement between the two. In my opinion, it is best to solve problems on clocks by considering a clock to be a circular track having a circumference of 60 km and each kilometre being represented by one minute on the dial of the clock. Then, we can look at the minute hand as a runner running at the speed of 60 kmph while we can also look at the hour hand as a runner running at an average speed of 5 kmph.
Since, the minute hand and the hour hand are both moving in the same direction, the relative speed of the minute hand with respect to the hour hand is 55 kmph, that is, for every hour elapsed, the minute hand goes 55 km (minute) more than the hour hand.
(Beyond this slight adjustment, the problems of clocks require a good understanding of unitary method. This will be well illustrated through the solved example below.)
Important Information
Number of right angles formed by a clock: A clock makes 2 right angles between any 2 hours. Thus, for instance, there are 2 right angles formed between 12 to 1 or between 1 and 2 or between 2 and 3 or between 3 and 4 and so on.
However, contrary to expectations, the clock does not make 48 right angles in a day. This happens because whenever the clock passes between the time period 2–4 or between the time period 8–10 there are not 4 but only 3 right angles.
This happens because the second right angle between 2–3 (or 8–9) and the first right angle between 3–4 (or 9–10) are one and the same, occurring at 3 or 9.
Right angles are formed when the distance between the minute hand and the hour hand is equal to 15 minutes.
Exactly the same situation holds true for the formation of straight lines. There are 2 straight lines in every hour. However, the second straight line between 5–6 (or 11–12) and the first straight line between 6–7 (or 12–1) coincide with each other and are represented by the straight line formed at 6 (or 12).
Straight lines are formed when the distance between he minute hand and the hour hand is equal to either 0 minutes or 30 minutes.
Illustration
At what time between 2–3 p.m. is the first right angle in that time formed by the hands of the clock?
Solution: At 2 p.m. the minute hand can be visualised as being 10 kilometres behind the hour hand. (considering the clock dial to be a race track of circumference 60 km such that each minute represents a kilometre).
Also, the first right angle between 2–3 is formed when the minute hand is 15 kilometres ahead of the hour hand.
Thus, the minute hand has to cover 25 kilometres over the hour hand.
This can be written using the unitary method:
Distance covered by the minute hand over the hour hand.
55 kilometres in 1 hour
25 kilometres in what time?
Æ 5/11 of an hour.
Thus, the first right angle between 2–3 is formed at 5/11 hours past 2 o’clock.
This can be converted into minutes and seconds using unitary method again as:
1 hour 60 minutes
5/11 hours ? minutes
Æ 300/11 minutes = 27 (3/11) minutes
1 minute 60 seconds
3/11 minutes ? seconds Æ 180/11 seconds = 16.3636 seconds.
Hence, the required answer is: 2 : 27 : 16.36 seconds.
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