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...

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.

Asked on by ispyy

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ndnordic's profile pic

ndnordic | High School Teacher | (Level 2) Associate Educator

Posted on

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

krishna-agrawala's profile pic

krishna-agrawala | College Teacher | (Level 3) Valedictorian

Posted on

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

Answer:

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.)

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

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.

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