Answer:
a) 103.176 m / min
b) 1751.28 meters
Explanation:
Provided:
- The velocities for Emma and Lily, represented by functions E(t) and L(t) respectively:

- Notably, E ( t ) and L ( t ) are measured in m / min
- Both individuals run for a total duration of 15 minutes in the same direction along a straight track within the interval:
( 0 ≤ t ≤ 15 ) mins
- Initially, it is established that Emma is 10 meters in front of Lily at t = 0.
Task:
a) Calculate the value of
with appropriate units and clarify its interpretation
b) Determine the furthest distance between Emma and Lily during the period 0 ≤ t ≤ 15?
Solution:
- The mean value of a function f ( x ) across an interval [ a , b ] is found using calculus with the formula:

- The first question requests the calculation of Emma's average speed during the time frame ( 2 , 8 ). Hence, ( E_avg ) can be resolved using the stated formula:

- We will calculate the integral to ascertain Emma's average speed over the 2 ≤ t ≤ 8 minute duration:

- Simplify the denominator through completing the square:

- Implement the following substitution:

- Replace the variables (u) and (dt) in the E_avg formula as described above.

- Utilize the designated standard integral:

- Compute the values:

- Revert the substitutions for ( u ):
![E_a_v_g = \frac{1}{6}*[\frac{75100* arctan ( \frac{5*(16 - 7 )}{\sqrt{6797} } )}{\sqrt{6797} } - \frac{75100* arctan ( \frac{5*(4 - 7 )}{\sqrt{6797} } )}{\sqrt{6797} } ]\\\\](https://tex.z-dn.net/?f=E_a_v_g%20%3D%20%5Cfrac%7B1%7D%7B6%7D%2A%5B%5Cfrac%7B75100%2A%20arctan%20%28%20%5Cfrac%7B5%2A%2816%20-%207%20%29%7D%7B%5Csqrt%7B6797%7D%20%7D%20%20%29%7D%7B%5Csqrt%7B6797%7D%20%7D%20%20-%20%5Cfrac%7B75100%2A%20arctan%20%28%20%5Cfrac%7B5%2A%284%20-%207%20%29%7D%7B%5Csqrt%7B6797%7D%20%7D%20%20%29%7D%7B%5Csqrt%7B6797%7D%20%7D%20%5D%5C%5C%5C%5C)
- Insert the limits and deduce Emma's average speed:
![E_a_v_g = 151.82037*[arctan (0.54582 ) - arctan ( -0.18194 ) ]\\\\E_a_v_g = 103.176 \frac{m}{min}](https://tex.z-dn.net/?f=E_a_v_g%20%3D%20151.82037%2A%5Barctan%20%280.54582%20%29%20-%20arctan%20%28%20-0.18194%20%29%20%5D%5C%5C%5C%5CE_a_v_g%20%3D%20103.176%20%5Cfrac%7Bm%7D%7Bmin%7D)
Result: Emma's average velocity within the interval ( 2 ≤ t ≤ 8 ) is 103.179 meters per minute.
- The displacement S ( E ) of Emma over the duration from t = 0 to ( t ) regarded in the specified interval 0≤t≤15 is expressed by the relationship:
![S (E) = S_o + \int\limits^t_0 {E(t)} \, dt\\\\S ( E ) = 10 + \frac{75100*arctan( \frac{5*(2t - 7 )}{\sqrt{6797} }) }{\sqrt{6797} } |_0^t\\\\S ( E ) = 10 + [ \frac{75100*arctan( \frac{5*(2t - 7 )}{\sqrt{6797} }) }{\sqrt{6797} } - \frac{75100*arctan( \frac{5*(0 - 7 )}{\sqrt{6797} }) }{\sqrt{6797} } ]\\\\S ( E ) = \frac{75100*arctan( \frac{5*(2t - 7 )}{\sqrt{6797} }) }{\sqrt{6797} } + 375.71098\\](https://tex.z-dn.net/?f=S%20%28E%29%20%3D%20S_o%20%2B%20%5Cint%5Climits%5Et_0%20%7BE%28t%29%7D%20%5C%2C%20dt%5C%5C%5C%5CS%20%28%20E%20%29%20%3D%2010%20%2B%20%5Cfrac%7B75100%2Aarctan%28%20%5Cfrac%7B5%2A%282t%20-%207%20%29%7D%7B%5Csqrt%7B6797%7D%20%7D%29%20%7D%7B%5Csqrt%7B6797%7D%20%7D%20%7C_0%5Et%5C%5C%5C%5CS%20%28%20E%20%29%20%3D%2010%20%2B%20%5B%20%5Cfrac%7B75100%2Aarctan%28%20%5Cfrac%7B5%2A%282t%20-%207%20%29%7D%7B%5Csqrt%7B6797%7D%20%7D%29%20%7D%7B%5Csqrt%7B6797%7D%20%7D%20-%20%5Cfrac%7B75100%2Aarctan%28%20%5Cfrac%7B5%2A%280%20-%207%20%29%7D%7B%5Csqrt%7B6797%7D%20%7D%29%20%7D%7B%5Csqrt%7B6797%7D%20%7D%20%5D%5C%5C%5C%5CS%20%28%20E%20%29%20%3D%20%20%5Cfrac%7B75100%2Aarctan%28%20%5Cfrac%7B5%2A%282t%20-%207%20%29%7D%7B%5Csqrt%7B6797%7D%20%7D%29%20%7D%7B%5Csqrt%7B6797%7D%20%7D%20%20%2B%20375.71098%5C%5C)
- The displacement S ( L ) of Lily across the time period from t = 0 to ( t ) over the range of 0 ≤ t ≤ 15 is defined by:

Utilize integration by parts:

Perform integration by parts two additional times:

- The separation between Emma and Lily within the span of 0 < t < 15 mins can be calculated by deducting S ( L ) from S ( E ):

- The peak distance ( S ) between Emma and Lily is determined by the critical value of S ( t ) where the function reaches either a maximum or minimum.
- To find the critical points for the function S ( t ), the first derivative concerning t will be evaluated and equated to zero:
![\frac{dS}{dt} = \frac{d [ S(E) - S(L)]}{dt} \\\\\frac{dS}{dt} = E(t) - L(t) \\\\\frac{dS}{dt} = \frac{7510}{t^2 - 7t+80.22} - 12t^3*e^-^0^.^5^t = 0\\\\( 12t^5 - 84t^4 + 962.64t^3) *e^-^0^.^5^t - 7510 = 0\\\\t = 4.233 , 11.671](https://tex.z-dn.net/?f=%5Cfrac%7BdS%7D%7Bdt%7D%20%3D%20%5Cfrac%7Bd%20%5B%20S%28E%29%20-%20S%28L%29%5D%7D%7Bdt%7D%20%20%20%5C%5C%5C%5C%5Cfrac%7BdS%7D%7Bdt%7D%20%3D%20E%28t%29%20-%20L%28t%29%20%5C%5C%5C%5C%5Cfrac%7BdS%7D%7Bdt%7D%20%3D%20%5Cfrac%7B7510%7D%7Bt%5E2%20-%207t%2B80.22%7D%20%20%20-%2012t%5E3%2Ae%5E-%5E0%5E.%5E5%5Et%20%3D%200%5C%5C%5C%5C%28%2012t%5E5%20-%2084t%5E4%20%2B%20962.64t%5E3%29%20%2Ae%5E-%5E0%5E.%5E5%5Et%20-%207510%20%3D%200%5C%5C%5C%5Ct%20%3D%204.233%20%2C%2011.671)
- Each critical t value will be substituted into the displacement function S(t) for evaluation: