The student left the house 10 minutes before the start of the lesson and walked at a speed of 4 km / h.

| December 13, 2020 | education

The student left the house 10 minutes before the start of the lesson and walked at a speed of 4 km / h. After 2 minutes he noticed that he was late and went 1 km / h faster. After another 2 minutes, he again increased the speed by 1 km / h. So he increased his speed every 2 minutes and ran to school just in time for the beginning of the lesson. How fast should he have walked in order to arrive at school on time without changing speed?

To solve it, it is logical to find the distance to the school and divide by the time remaining before the call.
2 minutes = 1/30 hours
In the first two minutes he covered 4 * 1/30 meters, for the 2nd 5 * 1/30 meters,
for the 3rd 2 sec. 6 * 1/30 etc 5 times
Total 4 + 5 + 6 + 7 + 8/30 = 1 km.
1 km / (1/6 hour) = 6 km / h-response



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