Emma and Lily jog in the same direction along a straight track. For 0≤t≤15, Emma’s velocity at time t is given by E(t)=7510t2−7t

Question

2 months ago

4

Emma and Lily jog in the same direction along a straight track. For 0≤t≤15, Emma’s velocity at time t is given by E(t)=7510t2−7t

+80.22 and Lily’s velocity at time t is given by L(t)=12t3e−0.5t. Both E(t) and L(t) are positive for 0≤t≤15 and are measured in meters per minute, and t is measured in minutes. Emma is 10 meters ahead of Lily at time t=0, and Emma remains ahead of Lily for 0

Physics

1 answer:

Keith_Richards [3.2K] · Answer 1 · 2026-02-16T13:22:37+03:00

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:

$E(t) = \frac{7510}{t^2-7t + 80.22} \\\\L ( t ) = 12t^3*e^-^0^.^5^t$

- 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 $\frac{1}{6}*\int\limits^8_2 {E(t)} \, dt$ 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:

$f_a_v_g = \frac{1}{b-a}\int\limits^a_b {f(x)} \, dx$

- 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:

$E_a_v_g = \frac{1}{8 - 2}*\int\limits^8_2 {E(t)} \, dt\\\\E_a_v_g = \frac{1}{6}*\int\limits^8_2 {E(t)} \, dt\\$

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

$E_a_v_g = \frac{1}{6}*\int\limits^8_2 {\frac{7510}{t^2 - 7t + 80.22} } \, dt\\\\E_a_v_g = \frac{1}{6}*37550\int\limits^8_2 {\frac{1}{50t^2 - 350t + 4011} } \, dt\\\\$

- Simplify the denominator through completing the square:

$E_a_v_g = \frac{1}{6}*37550\int\limits^8_2 {\frac{1}{(5\sqrt{2}*t - \frac{35}{\sqrt{2} })^2 + \frac{6797}{2} } } \, dt\\\\$

- Implement the following substitution:

$u = \frac{5*(2t - 7 )}{\sqrt{6797} } \\\\\frac{du}{dt} = \frac{10}{\sqrt{6797} } \\\\dt = \frac{\sqrt{6797}}{10}.du$

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

$E_a_v_g = \frac{1}{6}*37550\int {\frac{\sqrt{6797} }{5*(6797u^2 + 67997) } } \, du\\\\E_a_v_g = \frac{1}{6}*37550*\frac{1}{5\sqrt{6797}} \int {\frac{1 }{(u^2 + 1) } } \, du$

- Utilize the designated standard integral:

$arctan(u) = \int {\frac{1}{u^2 + 1} } \, du$

- Compute the values:

$E_a_v_g = \frac{1}{6}*37550*\frac{1}{5\sqrt{6797}}* arctan ( u ) |$

- Revert the substitutions for ( u ):

$E_a_v_g = \frac{1}{6}*[\frac{75100* arctan ( \frac{5*(16 - 7 )}{\sqrt{6797} } )}{\sqrt{6797} } - \frac{75100* arctan ( \frac{5*(4 - 7 )}{\sqrt{6797} } )}{\sqrt{6797} } ]\\\\$

- Insert the limits and deduce Emma's average speed:

$E_a_v_g = 151.82037*[arctan (0.54582 ) - arctan ( -0.18194 ) ]\\\\E_a_v_g = 103.176 \frac{m}{min}$

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:

$S (E) = S_o + \int\limits^t_0 {E(t)} \, dt\\\\S ( E ) = 10 + \frac{75100*arctan( \frac{5*(2t - 7 )}{\sqrt{6797} }) }{\sqrt{6797} } |_0^t\\\\S ( E ) = 10 + [ \frac{75100*arctan( \frac{5*(2t - 7 )}{\sqrt{6797} }) }{\sqrt{6797} } - \frac{75100*arctan( \frac{5*(0 - 7 )}{\sqrt{6797} }) }{\sqrt{6797} } ]\\\\S ( E ) = \frac{75100*arctan( \frac{5*(2t - 7 )}{\sqrt{6797} }) }{\sqrt{6797} } + 375.71098\\$

- 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:

$S (L) = \int\limits^t_0 {L(t)} \, dt\\\\S (L) = \int\limits^t_0 ({12t^3 *e^-^0^.^5^t } )\, .dt\\$

Utilize integration by parts:

$S ( L ) = 24t^3*e^-^0^.^5^t - 64*\int\limits^t_0 ({e^-^0^.^5^t*t^2} \,) dt\\$

Perform integration by parts two additional times:

$S ( L ) = -24t^3*e^-^0^.^5^t + 64*[ -2t^2*e^-^0^.^5^t - 2\int\limits^t_0 ({e^-^0^.^5^t*t} \,) dt ]\\$ $S ( L ) = -24t^3*e^-^0^.^5^t + 64*[ -2t^2*e^-^0^.^5^t - 2*( -2t*e^-^0^.^5^t - (4e^-^0^.^5^t - 4 ) ]\\\\$

$S ( L ) = e^-^0^.^5^t* ( -24t^3 -128t^2+ 256t + 512) - 512 \\$

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

$S = S ( E ) - S ( L )\\\\S = \frac{75100*arctan( \frac{5*(2t - 7 )}{\sqrt{6797} }) }{\sqrt{6797} } - e^-^0^.^5^t* ( -24t^3 -128t^2+ 256t + 512) + 887.71098\\$

- 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:

$\frac{dS}{dt} = \frac{d [ S(E) - S(L)]}{dt} \\\\\frac{dS}{dt} = E(t) - L(t) \\\\\frac{dS}{dt} = \frac{7510}{t^2 - 7t+80.22} - 12t^3*e^-^0^.^5^t = 0\\\\( 12t^5 - 84t^4 + 962.64t^3) *e^-^0^.^5^t - 7510 = 0\\\\t = 4.233 , 11.671$

- Each critical t value will be substituted into the displacement function S(t) for evaluation:

Download docx