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Algebra 17 Online
OpenStudy (anonymous):

WILL MEDAL AND FAN!!! John and Mike are carpooling to see their favorite baseball team play. On their way to the game, they drove 60 miles in the first 45 minutes of the trip. Unfortunately then they ran into heavy traffic and were only able to travel at a constant speed of 20 miles per hour. Choose the correct function which describes the average speed during the entire trip as a function of time, h, in hours. Question 6 options: f(h)=60+20(h−0.75)h where h≥0.75 f(h)=20h+1.3 f(h)=60h+20 f(h)=60h+45h

OpenStudy (anonymous):

@SolomonZelman could you help? ill medal.:)

OpenStudy (anonymous):

@superdavesuper

jimthompson5910 (jim_thompson5910):

Were you able to figure out how far they traveled at any given time h ?

OpenStudy (anonymous):

well at least 45, i was thinking its option D?

jimthompson5910 (jim_thompson5910):

so they travel 60 mi in the first part of their trip how far do they travel on the second part?

OpenStudy (anonymous):

20mph

OpenStudy (anonymous):

other than that we dont know

jimthompson5910 (jim_thompson5910):

that's the speed for the second part

jimthompson5910 (jim_thompson5910):

let's say they travel 20 mph for h hrs the distance would be 20*h, no?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

so b?

jimthompson5910 (jim_thompson5910):

here's the thing though: they didn't travel a full h hours for 20 mph they spent the first 0.75 hrs going some speed (first part of the trip) then the rest of the h-0.75 hrs is them going 20 mph

jimthompson5910 (jim_thompson5910):

so they travel 60 mi on the first leg of the trip then an additional 20(h-0.75) for the second leg

jimthompson5910 (jim_thompson5910):

in total, they travel a distance of `60 + 20(h - 0.75)` miles

jimthompson5910 (jim_thompson5910):

to find the average speed over the entire trip, you divide that total distance by the total time (in this case, h) so the average speed is \(\Large \frac{60 + 20(h - 0.75)}{h}\) mph

OpenStudy (anonymous):

ohh huh i wouldnt have guessed that...

OpenStudy (anonymous):

thank you:)

jimthompson5910 (jim_thompson5910):

yeah it's probably not the most intuitive thing out there, but just remember distance = rate*time rate = distance/time "rate" is another term for "speed"

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