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

Answer from: 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:

Answer from: Quest

the small 0.38-kg slider is known to move from position a to position b along the vertical-plane slot. determine (a) the work done on the body by its weight and (b) the work done on the body by the spring. the distance r = 0.70 m, the spring modulus k = 154 n/m, and the unstretched length of the spring is 0.53 m

explanation:

-reverse card

Answer from: Quest

I'm not sure what your asking buddy.. but i know that you can weld two peices of steel by pure force and pressure.

Another question on Physics

Physics, 22.06.2019 06:00

Imagine that someone pushes one marble toward a motionless marble. would there still be action-reaction forces involved in the collision? how might the marbles’ motions be changed? ?

Answers: 1

Answer

Physics, 22.06.2019 17:40

The first-order reaction a→2b+c has rate constant 0.538 s−1. if the initial concentration of a is 0.867 mol l−1, what is the half-life of the reaction in seconds? remember to use correct significant figures in your answer (round your answer to the nearest hundredth). do not include units in your response.

Answers: 3

Answer

Physics, 22.06.2019 19:30

Aspacecraft that renters the earth’s atmosphere drastically slows down. the amount of kinetic energy the spacecraft has as it reenters the earths atmosphere

Answers: 2

Answer

Physics, 22.06.2019 22:30

Find v1, the speed of the fluid in the left end of the main pipe. express your answer in terms of h1, h2, g, and either a1 and a2 or γ, which is equal to a1a2.

Answers: 2

Answer

You know the right answer?

Emma and Lily jog in the same direction along a straight track. For 0≤t≤15, Emma’s velocity at time...

Questions