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Mathematics 15 Online
OpenStudy (anonymous):

REWARD AND MEDAL 2.) Grace leaves home at 8am TEN minutes later, Will notices Grace's lunch and begins bicycling after her. If Grace walks at 5km/h and Will cycles at 15 km/h, how long will it take him to catch up with her??????

OpenStudy (jdoe0001):

hmm, I see, she left at 8am, and forgot her lunch anyhow so how long do you think Will would have traveled in distance when he catches up with her? would you say by that time they'd have traveled the same distance?

jimthompson5910 (jim_thompson5910):

10 minutes = 10/60 = 1/6 hours let t = time elapsed (in hours) from when Grace leaves home If Grace walks at 5 km/hr, then she will walk a total of 5t kilometers when she walks for t hours. Will starts ten minutes or 1/6 hour later, so he will bike for t - 1/6 hours. This means he travels 15(t -1/6) km

OpenStudy (anonymous):

idk

OpenStudy (jdoe0001):

say, you leave home at 8am, and then your brother catches up with you by the time your brother, also from home, catches up with you, you have already walked 2 miles how long has he walked too?

OpenStudy (anonymous):

2 miles?

OpenStudy (jdoe0001):

if you had walked 2 miles, and your brother walked 3 miles so he really passed you by, by 1 mile if you had walked 2 miles, and your brother walked 1 mile so he's still 1 mile behind you if you had walked 2 miles, and your brother walked 1 1/2 miles he hasn't gotten to you, he's still behind

OpenStudy (jdoe0001):

however if you walked 2 miles, and he walked 2 miles, both from home has he caught up with you? or pass you by or still behind you?

OpenStudy (anonymous):

you aren't making sense.. can u just tell me what goes in the chart then tell me the equation??

OpenStudy (jdoe0001):

lol

OpenStudy (jdoe0001):

well, that's the idea behind it

OpenStudy (anonymous):

can u tell me itd be helping me

OpenStudy (jdoe0001):

anyhow, jim_thompson5910 already typed it in

jimthompson5910 (jim_thompson5910):

grace walks 5t km will bikes 5(t-1/6) km the two will meet when the two distances are the same, so you need to solve 5t = 5(t-1/6)

jimthompson5910 (jim_thompson5910):

sorry typo

jimthompson5910 (jim_thompson5910):

will bikes 15(t-1/6), so the equation should be 5t = 15(t - 1/6)

OpenStudy (anonymous):

how do u solve with the 1/6??

jimthompson5910 (jim_thompson5910):

5t = 15(t - 1/6) 5t = 15t - 15(1/6) ... distribute 5t = 15t - 15/6 ... multiply 5t = 15t - 5/2 ... reduce I'll let you finish up

OpenStudy (anonymous):

subtract 5t next?

jimthompson5910 (jim_thompson5910):

15t actually

OpenStudy (anonymous):

5t=-5/2

OpenStudy (anonymous):

???? is that right

OpenStudy (anonymous):

T=-1/2 answer so -1/2 hours

jimthompson5910 (jim_thompson5910):

5t - 15t = ???

OpenStudy (anonymous):

i didn't put that anywhere

jimthompson5910 (jim_thompson5910):

what is 5t - 15t

OpenStudy (anonymous):

ys

jimthompson5910 (jim_thompson5910):

simplify 5t - 15t for me

OpenStudy (anonymous):

-10t

jimthompson5910 (jim_thompson5910):

so we now have -10t = -5/2 t = ??

OpenStudy (anonymous):

x=-1/4

jimthompson5910 (jim_thompson5910):

it's positive though

jimthompson5910 (jim_thompson5910):

t = 1/4

jimthompson5910 (jim_thompson5910):

so it takes 1/4 of an hour or 15 minutes for Will to catch up to Grace

OpenStudy (anonymous):

how long would that be??

OpenStudy (anonymous):

oh ok 15 or 1/4 of an hour. THANKS SO MUCH im posting another question sticka round

OpenStudy (anonymous):

3.) a jet took one hour longer flying to Lincoln from Adams at 800 km/h than to return at 1200 km/h find the distance from Lincoln to Adams????

OpenStudy (anonymous):

@jim_thompson5910

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