TIME, SPEED AND DISTANCE - PART 2Linear RacesRaces are another variation of the concept of Relative Speed as discussed in Part 1. Few concepts are important to understand here. The concept of head start, beating by a certain distance and beating by a certain time.

Head Start - Let there be 2 runners A and B. If B gives a head start to A, it simply means, A starts ahead of B. Head start can be measured in terms of distance or time. Say, in a 100m race B gives A a head start of 10m. For solving a problem, it is equivalent to considering A running 90m and B runs 100m. Similarly, if A is given a head start say, 10 sec, then it means B starts 10 seconds after A starts.

Beating by a distance X - If A beats B by a distance of 10m in a 100m race, it means when A finishes the race, B has completed 90m.

Beating by time T - If A beats B by 10 seconds, it means B completes the 100m race 10 seconds after A.

With this understanding, let's see a few problems.

Sample Problem: In a 1200m race, P beats Q by 300m and Q beats R by 400m. Find the distance by which P beats R.

What is being tested in this question? It is testing the test takers ability to establish a ratio relationship between P, Q and R using the distance traveled by P & Q and Q & R. Note that time is constant. When P beats Q, P and Q have run for the same time - say T1. Similarly when Q beats R, Q and R have traveled for the same time say T2.

Now, Distance = Speed * Time, so if time is constant, distances must be in proportion. In other word ratio of distance covered by P and Q should be equal to the ratio of their respective speeds. Now, P beats Q by 300m, so by the time P runs 1200m, Q runs 900m. Hence ratio of their speeds is 12:9 = 4:3.

Similar arguments can be applied for Q and R too and their respective speeds can be established as 1200:800 = 3:2.

Combine the two ratios we get P:Q:R = 4:3:2 (Since Q correspnds to 3 in both the ratios they can be combined as is)

Thus P:R = 4:2 or 2:1. Hence by the time P finishes the race, R would finish 600m. Hence P beats R by 600m.

Now, try the following two easy problems to solidify the concept.

Practice Problem 1: In a 200m race A beats B by 10m or 2 seconds. Find the speeds of A and B in m/sec.

(A) 10 and 8

(B) 10/9 and 8/9

(C) 100/19 and 5

(D) 15 and 10

(E) 9/5 and 7/5

Practice Problem 2: P is 50% faster than Q. In a race, P gave Q a head start of 200m. Both finished the race simultaneously. Find the length of the race.

(A) 300m

(B) 400m

(C) 500m

(D) 600m

(E) 700m

Circular RacesCircular races are a variation of linear races with the exception that the start and the end points are the same. So, we can think of the track as a wire ring, and if we cut the ring at the starting/ending point and straighten the wire, we get the above case of linear races. This visualization helps solve some of the problems very easily. Now for circular races, it is important to understand 2 cases which can be generalized further.

Case 1: Two people running around a Circular Track of length LSame DirectionWhen 2 people a and b run around a circular track with respective speeds a and b where a>b, then

(i) Time taken to meet for the first time ever is

L/(a-b)(ii) Time taken to meet for the first time at the starting point is

LCM[L/a, L/b]Opposite DirectionWhen 2 people a and b run around a circular track with respective speeds a and b where a>b, then

(i) Time taken to meet for the first time ever is

L/(a+b)(ii) Time taken to meet for the first time at the starting point is

LCM[L/a, L/b]Case 2: Three people running around a Circular Track of length LSame DirectionWhen 3 people a, b and c run around a circular track with respective speeds a, b and c where a>b>c, then

(i) Time taken to cross/meet for the first time ever is

LCM [L/(a-b), L/(b-c)](ii) Time taken to cross/meet for the first time at the starting point is

LCM[L/a, L/b, L/c]Same DirectionFor opposite direction, there is no generalized case, and if asked, it should be done considering 2 of the 3 people at a time. The time to meet at the starting point would however remain the same as above.

Logic for the above casesAs mentioned, earlier, for the first case, time to meet at the starting point, consider cutting open the wire frame and treating the race as a linear one. The only exception being every time a person reaches the end of the race track, we'll have to imagine that he starts from the beginning of the track for the next lap. With that assumption, it simply becomes a relative speed problem where the faster person has to cross the length of the race track once to cross the slower person. Hence distance traveled = L and relative speed is a-b, so time taken is L/(a-b). For the three people scenario we take 2 people at a time and consider the LCM because a we are looking for a certain specific multiple of the time when both a crosses B and B crosses C. Note that this is the condition for all 3 people crossing for the very first time. Similar argument holds tru for the opposite direction as well.

With this understanding, give the next two problems a try.

Practice Problem 1: Peter and Paul are running in opposite directions along a circular track of 600m with initial speeds of 3m/sec and 6 m/sec. They start at the same time from the same point. Whenever they meet, they exchange their speeds and carry on in their respective directions. Find the distance between them in metres when Peter completes 17/4 rounds.

(A) 150

(B) 200

(C) 300

(D) 120

(E) 175

Practice Problem 2: P and Q are running along a circular track having started at the same time from the same point, in the same direction. How much more distance would P have traveled compared to Q by the time they meet for the 11th time? The radius of the track us 7m and the speeds of P and Q are 22 and 11 m/sec respectively.

(A) 440m

(B) 384m

(C) 524m

(D) 484m

(E) 600m

ClocksClocks involve the same concept as Circular Tracks. Only two things need to be remembered for simplifying Clock problems.

1. Minute hand covers 360 degrees in 60 minutes => minute hand covers 6 degrees/minute

2. Hour hand covers 30 degrees in 60 minutes => hour hand covers 0.5 degrees per minute

These also imply that the relative speed is 5.5 degrees per minute.

Sample Problem: Find the smaller angle between the hands of a clock when the time is 3:40 PM.

Note that, at 3 PM, the hour hand has a head start of 3 * 30 degrees = 90 degrees.

Now in 40 minutes, minutes hand covers 40*6 = 240 degrees and the hour hand covers 90+40*.5=110 degrees. Hence the smaller angle between them is 240-110 = 130 degrees.

Practice Problem - IR Two parts Analysis QuestionAfter how many minutes post 5 PM and before 6 PM will the hands of the clock be at 90 degrees to each other?

(i) First Time(A) 80/11 minutes

(B) 100/11 minutes

(C) 110/11 minutes

(D) 120/11 minutes

(E) 140/11 minutes

(i) Second Time(A) 300/11 minutes

(B) 450/11 minutes

(C) 480/11 minutes

(D) 500/11 minutes

(E) 600/11 minutes

Do let me know, if you found the post useful. If you're having any question on any of the materials covered, please post here as a reply. I shall try my best to clarify them

P.S. - I have added the answers the the questions I included in the first post

. Please refer the above post.

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