# Thread: Time-Distance Issue

1. ## Time-Distance Issue

Hello , I am having trouble with the following question:
Tom and bill agreed to race across 50-foot pool and back again.They started together but Tom finished 10feet ahead of bill.IF there rates were constant and Tom finished the race in 27 seconds , how did it take bill to finish:
The answer is supposed to be 30 but i am getting 33.75

Here is how i am doing it:
Toms velocity is: 50/27 ft/sec
Bills velocity will be : 40/27 (Because Tom finished 10ft ahead of Bill)
So Time taken for bill to finish the race will be:
S/V = 50 / (50/27) = 33.75
Could anyone tell me what i am missing or doing wrong!! Thanks

2. ## Re: Time-Distance Issue

Tom and bill agreed to race across 50-foot pool and back again.
Tom's avg speed ... $\displaystyle \frac{100 \, ft}{27 \, sec}$

Bill's ... $\displaystyle \frac{90 \, ft}{27 \, sec}$

3. ## Re: Time-Distance Issue

Tom and bill agreed to race across 50-foot pool and back again. 100 instead of 50.

4. ## Re: Time-Distance Issue

Now i get it. The time is for the two sides... Thanks.