Asked by alicia
Function for runner one: s=0.2t(t-5.1)(t-9.1)
Function for runner two:s=2.4+0.75t
's' is for displacement and 't' is for time (where one minute is t = 1)
At the start of a race, runner one begins with a displacement of 0m. However, at the start of the same race, runner two starts with a displacement of 2.4m
By the beginning of the first minute of the race, runner one has a displacement of 6.64m. At this same time, runner two has a displacement of 3.15m.
When does runner one pass runner two?
Function for runner two:s=2.4+0.75t
's' is for displacement and 't' is for time (where one minute is t = 1)
At the start of a race, runner one begins with a displacement of 0m. However, at the start of the same race, runner two starts with a displacement of 2.4m
By the beginning of the first minute of the race, runner one has a displacement of 6.64m. At this same time, runner two has a displacement of 3.15m.
When does runner one pass runner two?
Answers
Answered by
Steve
yadda yadda yadda. The equations say it all. The words are redundant. If runner 1 passes runner 2, their distances are equal:
0.2t(t-5.1)(t-9.1) = 2.4 + 0.75t
t ≈ 10.1
Check out the graphs at
http://www.wolframalpha.com/input/?i=0.2t%28t-5.1%29%28t-9.1%29+%3D+2.4+%2B+0.75t
0.2t(t-5.1)(t-9.1) = 2.4 + 0.75t
t ≈ 10.1
Check out the graphs at
http://www.wolframalpha.com/input/?i=0.2t%28t-5.1%29%28t-9.1%29+%3D+2.4+%2B+0.75t
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