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Bob beats Judy by 10-m in a 100-m dash. Bob, claiming to give Judy an equal chance, agrees to race her again but to start from 10 m behind the starting line. Does this really give Judy an equal chance?

Bob beats Judy by 10-m in a 100-m dash. Bob, claiming to give Judy an equal chance, agrees to race her again but to start from 10 m behind the starting line. Does this really give Judy an equal chance?

Grade:upto college level

1 Answers

Deepak Patra
askIITians Faculty 471 Points
6 years ago
Given:

Length of the track, x = 100 m.
Distance by which Bob beats Judy, d = 10 m.

We assume that the average speed of the Bob and the Judy is maintained in every race.
Let us assume that the average speed of the Bob be represented by vb and the time he takes to run 100 m is represented by t .
Therefore the average speed of the bob when he runs x meters is:
v_{b} = \frac{total\ distance\ traveled}{ elapsed\ time}
v_{b} = \frac{x}{t}
Substituting the value of x, we have
v_{b} = \frac{100\ m}{t} …… (1)
Let us assume that the average speed of the Judy be represented by v1. By the time Bob run meters, the Judy could run only x-d meters.
Therefore the average speed of Judy is given as:
v_{j} = \frac{total\ distance\ traveled}{ elapsed\ time}
v_{j} = \frac{x-d}{t}
Substituting the value of x , we have
v_{j} = \frac{100\ m-10\ m}{t}
v_{j} = \frac{90\ m}{t} …… (2)
Now we consider the situation when Bob decides to start 10 meters behind the starting line. Therefore the distance traveled by the Bob now is x+10 meters.
Assuming that the Bob maintains his average speed as vb, we can calculate the time (say t’)taken by the Bob to reach the finish line as:
v_{b} = \frac{total\ distance\ traveled}{elapsed\ time}
v_{b} = \frac{x + 10\ m}{t^{'}}
Substituting the value of vb from equation (1) and the given value of x , we have
t^{'} = \frac{100\ m + 10\ m}{(100\ m/\ t)}
Judy, on the other hand has to travel distance x with the same average speed, and therefore the time (say t”) after which he will reach the finish line can be calculated as:
v_{j} = \frac{total\ distance\ traveled}{elepsed\ time}
v_{j} = \frac{x}{t''}
Substituting the value of vj from equation (2) and the given value of x, we have
t'' = \frac{100\ m}{90\ m /\ t}
t'' = 1.11t
Thus we find that t’‘ > t’.
Therefore with same average speed and under the condition that the Bob started from
10 meters behind the starting line, Bob will still manage to reach the finish line earlier than Judy.

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