Printable View
During a recent Formula 1 race meeting, Sterling managed to complete a lap with an average speed of 150 mph. He managed to complete the first two fifths of the lap length at a speed of 123 mph and the second two fifths of the lap length at a speed of 164 mph. At what speed was the final one fifth of the lap length covered - the answer is not 176 mph?
----------------------------
suresh
James,
i want to move this one into the sticky board..
-----------------------
suresh
That's simple I guess.
The answer is 205mph.
The only trick here is, we can't directlly do (v1+v2+v3)/3 to get Vav.
Simply because distance covered with velocity v3 is different than distance covered with v1 and v2. (He's covered 2/5 of the distance with v1, 2/5 of distance with v2 and 1/5 of distance with v3).
Just use elimentery physics of v=d/t. You'll get the answer.
(Correct me if I'm wrong)
That is absolutely correct kalayama....
Great work.....
My solution are as follows..
If we take the number of miles in a lap to be x miles.
The first 2/5 are covered at 123 mph. Therefore, the time taken was (2/5) * x/123 hours.
The next 2/5 are covered at 164 mph. Therefore, the time taken was (2/5) * x/164 hours.
The last 1/5 is covered at y mph. Therefore, the time taken was (1/5) * x/y hours.
The entire trip is covered at an average of 150 mph. we can conclude that:
(2x/ 5*123) + (2x/ 5*164) + (x/ 5*y) = x / 150
Working this out we get:
(1/30) - (-2/123) - (-2/164) = 1/y
The means that y = 1 / (1/30 - 2/123 - 2/164)
Therefore y = 205 QED.
--------------------------------------------
suresh
Well. I too worked it out similarly. Didn't feel like utting the entire logic up there :)
Ans : 205.33mph
Nice contributing ramadidamu.........
But already our members post the answers...
Read all the post and then reply....
Anyway thanks for your contributing.....
-----------------------------
suresh