1 )A race car is one lap behind the lead race car when the lead car has 49 laps to go in a race. If the speed of the lead car is 69.7m/s,
what must be the average speed of the second car to catch the lead car just before the end of the race (i.e. right at the finish line)? Assume 1 lap is 1.34 km. Answer in units of m/s.
3 Answers | Add Yours
This is a distance, velocity, time problem where d = vt.
The lead car has to travel 49 laps which is a distance of 49 laps x 1340 m/lap = 65660 meters. At a speed of 69.7 m/s this will take 65660/69.7 = 942.04 seconds.
Your car has to go 50 laps or 50 x 1340 m = 67,000 m in the same 942.04 seconds.
v = d/t = 67,000/942.04s = 71.12 m/s
The car is one lap behind the leading car.
The lead car has 49 laps remaining.
Therefore the second car has 49 + 1 = 50 laps remaining.
For the lead and the second car to complete the race at the same time, the relative speed of the two cars be such that:
(Speed of lead car)/(Laps remaining for lead car) = (Speed of second car)/(Laps remaining for second car)
Substituting values of laps remaining and speed of lead car in above equation:
(69.7/49) = (Speed of second car)/50
==> Speed of second car = 50*(69.7/49) = 71.12245 m/s
Required speed of second car = 71.12245 m/s
(Please note that to solve this problem, it is not necessary to know the length of each lap.)
The distance the lead race car when the lead race car has yet to go for 49 lap is 1 lap.
Given 1 lap = 1.34 kms.
Coverting the laps:
The ahead distance of lead car is 1.3kms and it has to yet cover 49 laps = 49*1.34 kms = 65.66 kms = 65660 meters
The speed of the lead race car = 69.7m/hr.
The speed of the other race car be assumed x km/hr
The time the at which the lead car 65660m/(69.7m/s) = 65660/69.7 seconds.
So the other car whose speed is x is to overtake the first lead car exactly at the time 65660/69.7 seconds and it covers the distance gap of 1lap between it and the lead race car and the 49 lap to the target end.Or 50 lap = 50*1.34 kms =50*1.34*1000meters distance the other race car should cover to over take the lead car in 65660/69.7seconds
So the speed of the other car = x = 50*1.34*1000 / (65660/69.7) m/s= 50*1.34*1000*69.7/65660 m/s.
= 71.2244898m/s.( Purposely not rounded as othrwise overtaking point may be not behind or after the marked taget.
We’ve answered 317,372 questions. We can answer yours, too.Ask a question