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?
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
To find out when runner one passes runner two, we need to solve the equation for runners one and two and find the time (t) at which their displacements are equal.
The equation for runner one is:
s1 = 0.2t(t - 5.1)(t - 9.1)
The equation for runner two is:
s2 = 2.4 + 0.75t
Since both runners start at different displacements, we can set up the following equation:
0.2t(t - 5.1)(t - 9.1) = 2.4 + 0.75t
To solve this equation, we can follow these steps:
1. Distribute the multiplication on the left side:
0.2t(t^2 - 14.2t + 46.41) = 2.4 + 0.75t
2. Expand and rearrange the equation:
0.2t^3 - 2.84t^2 + 9.282t - 2.4 - 0.75t = 0
3. Combine like terms:
0.2t^3 - 2.84t^2 + 9.282t - 0.75t - 2.4 = 0
4. Simplify the equation to standard form:
0.2t^3 - 2.84t^2 + 8.532t - 2.4 = 0
Unfortunately, this equation cannot be solved algebraically using simple methods. You would need to use numerical methods, such as graphing or a numerical solver, to find the exact time at which runner one passes runner two.
Alternatively, if you are looking for an approximate solution, you can use a graphing calculator or software to graph the equations for runner one and runner two and look for the point of intersection on the graph. This point represents the time at which runner one passes runner two.