BOTH DOUBTS ARE EQUALLY IMPORTANT
DOUBT 1: I discovered that if a vehicle travels with a speed of 60 km/hr in first trip and speed of 20km/hr in second trip. Both the trips lasted for 20 minutes . D1=20km d2=20/3km
Avg speed per trip=40 km/hr
ALSO , Arithmetic mean of 60 and 20 = avg of 2 speeds = (60 km/hr + 20km/hr)/2= sum of observation / no of observation=40km/hr
So avg speed=Arithmetic mean (only when t1=t2)
Is there any intutive explanation why avg speed = AM ? other than doing a sum and then at the last finding out oh yes Avg speed per trip= AM
DOUBT 2: If i take the case of first trip only suppose trip1 was for 20 minutes not 1 hr then the car covered 20km. So the driver says we covered 20 km in 20 minutes . From that info can we say the car was travelling 60km/hr ; though the car has not travlled for 1 hr it has just travlled for 20 minutes covering 20 km , thn how can we say first trip car was travelling for 60 km/hr .
NOTE: IS it because distance and time are directly prorportional . If d1/t1=20 km/20min (from what we travelled) now if we want to measure in terms of 60 minutes that means t1 needs to be tripled so t1 becomes 60 minutes and as d1 is proportion to t1; d1 also triples so d2/t2=60km/60minutes so we can say THE FIRST TRIP WAS TRAVELLING AT A SPEED OF 60km/hr
@Dr.Peterson @JeffM
DOUBT 1: I discovered that if a vehicle travels with a speed of 60 km/hr in first trip and speed of 20km/hr in second trip. Both the trips lasted for 20 minutes . D1=20km d2=20/3km
Avg speed per trip=40 km/hr
ALSO , Arithmetic mean of 60 and 20 = avg of 2 speeds = (60 km/hr + 20km/hr)/2= sum of observation / no of observation=40km/hr
So avg speed=Arithmetic mean (only when t1=t2)
Is there any intutive explanation why avg speed = AM ? other than doing a sum and then at the last finding out oh yes Avg speed per trip= AM
DOUBT 2: If i take the case of first trip only suppose trip1 was for 20 minutes not 1 hr then the car covered 20km. So the driver says we covered 20 km in 20 minutes . From that info can we say the car was travelling 60km/hr ; though the car has not travlled for 1 hr it has just travlled for 20 minutes covering 20 km , thn how can we say first trip car was travelling for 60 km/hr .
NOTE: IS it because distance and time are directly prorportional . If d1/t1=20 km/20min (from what we travelled) now if we want to measure in terms of 60 minutes that means t1 needs to be tripled so t1 becomes 60 minutes and as d1 is proportion to t1; d1 also triples so d2/t2=60km/60minutes so we can say THE FIRST TRIP WAS TRAVELLING AT A SPEED OF 60km/hr
@Dr.Peterson @JeffM