Grade 12th passMechanics

Bob beats Judy by-10 m in a l00m 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?

faraz

8 Years agoGrade 12th pass

1 Answer

Arun

8 Years ago

Dear student

…… (2) $v_{j} = \frac{90\ m}{t}$ $v_{j} = \frac{100\ m-10\ m}{t}$ Substituting the value of x , we have $v_{j} = \frac{x-d}{t}$ $v_{j} = \frac{total\ distance\ traveled}{ elapsed\ time}$ . By the time Bob run meters, the Judy could run only x-d meters.Therefore the average speed of Judy is given as:₁ …… (1)Let us assume that the average speed of the Judy be represented by v $v_{b} = \frac{100\ m}{t}$ Substituting the value of x, we have $v_{b} = \frac{x}{t}$ $v_{b} = \frac{total\ distance\ traveled}{ elapsed\ time}$ 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:_bGiven: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 v

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 v_b, 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 v_b 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 v_j 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.

Regards

Arun (askIITians forum expert)