A race car driver drove, on a 4 km long race course, at an average speed of 120 km/h for the first 2 km. How fast does he have to go the second 2 km to average 240 km/h for the entire course?
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480
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240
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360
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120
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none of the above
To average 240 km/h over 4 km, the total time allowed is 4/240 = 1/60 hours (1 minute). The driver already spent 2/120 = 1/60 hours on the first 2 km. Thus, he has 0 time left for the second half, requiring infinite speed. 'None of the above' is correct.
To solve this question, we can use the formula for average speed:
Average Speed = Total Distance / Total Time
Let's assume the driver took t1 hours to complete the first 2 km at a speed of 120 km/h. Therefore, the time taken for the first 2 km is t1 hours.
Given that the driver has to average 240 km/h for the entire 4 km race course, we can calculate the total time taken to complete the race course using the formula:
Total Time = Total Distance / Average Speed
Total Time = 4 km / 240 km/h = 1/60 hours
Now, let's find the time taken to complete the second 2 km of the race course.
Time taken for the second 2 km = Total Time - Time taken for the first 2 km
Time taken for the second 2 km = (1/60) - t1
Now, we can calculate the speed needed to average 240 km/h for the entire course:
Speed for the second 2 km = Distance / Time taken for the second 2 km
Speed for the second 2 km = 2 km / ((1/60) - t1)
To average 240 km/h for the entire course, the speed for the second 2 km must be:
240 km/h = 2 km / ((1/60) - t1)
Simplifying this equation:
240 km/h = 2 km / ((1/60) - t1) 240 km/h = 2 km / (60 - 60t1) 240 km/h = 2 km / (60 - 60t1) 240 km/h = 1 / (30 - 30t1) 30 - 30t1 = 1/240 -30t1 = 1/240 - 30 t1 = (1/240 - 30) / -30
By calculating the value of t1, we can determine the speed for the second 2 km. However, this calculation results in a negative value for t1, which is not possible. Therefore, there is no valid speed for the second 2 km that would allow the driver to average 240 km/h for the entire course.
Hence, the correct answer is E) none of the above.