Physics, 22.05.2020 04:05, destinywashere101

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

Answers: 3

Show answers

Answers

answered: mcdowelle360

a) 103.176 m / min

b) 1751.28 meters

Explanation:

Given:-

- Emma's and Lily's velocities ( E(t) and L(t) ) are given as functions respectively:

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

- Where, E ( t ) and L ( t ) are given in m / min

- Both run for a total time of 15 minutes in the same direction along the straight track defined by the absolute interval:

( 0 ≤ t ≤ 15 ) mins

- It is known that Emma is 10 meters ahead of Lily at time t = 0.

Find:-

a) Find the value of $\frac{1}{6}*\int\limits^8_2 {E(t)} \, dt$ using correct units, interpret the meaning of

b) What is the maximum distance between Emma and Lily over the time interval 0 ≤ t ≤ 15?

Solution:-

- The average value of a function f ( x ) over an interval [ a , b ] is determined using calculus via the following relation:

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

- The first part of the question is asking us to determine the average velocity of Emma over the time interval of ( 2 , 8 ). Therefore, ( E_avg ) can be determined using the above relation:

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

- We will evaluate the integral formulation above to determine Emma's average velocity over the 2 ≤ t ≤ 8 minute range:

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

- Complete the square in the denominator:

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

- Use the following substitution:

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

- Substitute the relations for (u) and (dt) in the above E_avg expression.

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

- Use the following standard integral:

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

- Evaluate:

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

- Apply back substitution for ( u ):

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

- Plug in the limits and find Emma's average velocity:

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

Emma's average speed over the interval ( 2 ≤ t ≤ 8 ) is 103.179 meters per minute.

- The displacement S ( E ) of Emma from time t = 0 till time ( t ) over the absolute interval of 0≤t≤15 is given by the relation:

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

- The displacement S ( L ) of Lily from time t = 0 till time ( t ) over the absolute interval of 0 ≤ t ≤ 15 is given by the relation:

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

Apply integration by parts:

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

Re-apply integration by parts 2 more 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 distance between Emma and Lily over the time interval 0 < t < 15 mins can be determined by subtracting 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 maximum distance ( S ) between Emma and Lily is governed by the critical value of S ( t ) for which function takes either a minima or maxima.

- To determine the critical values of the function S ( t ) we will take the first derivative of the function S with respect to t and set it to zero:

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

- We will plug in each value of t and evaluate the displacement function S(t) for each critical value:

answered: Guest

the answer is a 3.8cm

explanation:

An object is located 5.0 cm from a concave mirror. the focal length is 15.0 cm. what is the image di

answered: Guest

bacteria that live on the ocean floor are sustained by phytoplankton. phytoplankton, microscopic marine photosynthetic organisms, have a vastly significant role to play not only in the marine food web of which they’re part of, but also on a more global scale. despite their infinitely small size in comparison to other marine organisms, these tiny creatures occupy an immensely important ecological niche: they are the foundation of the marine food web, and as primary producers, play key roles in supporting all other organisms in the marine environment, as well as in the regulation of the earth’s climate through the sequestration of carbon, oxygen production, and other related processes. phytoplankton account for roughly half of all global primary productivity; therefore, their significance extends far beyond the marine environment alone. there is an intriguing sense of irony in the realization that these tiny living beings that often live out their existence unnoticed and undetected by the rest of the world, have such a far-reaching impact on the lives of virtually all other living organisms on the planet, particularly on those in the marine environment. the world that we have become accustomed to has been and is continuously shaped by the workings of these miniscule yet vital “plants of the sea”.

hope this !