contestada

Consider an athlete running a 40-m dash. The position of the athlete is given by d(t)=t^2/6+4t where d is the position in meters and t is the time elapsed, measured in seconds. Compute the average velocity of the runner over the intervals: (a) [1.95, 2.05 (b) [1.995, 2.005 (c) [1.9995, 2.0005] (d) 12, 2.00001] Then, using these values, determine the instantaneous velocity of the runner at 2 seconds.

Respuesta :

jolis1796

Answer:

a), b) and c) v avg = 4.66 m/s

v = 4.66 m/s

Step-by-step explanation:

Given

We use the formulas

d(t)=(t²/6)+4*t

v avg = Δd/Δt

a) d(1.95) = ((1.95)²/6)+4*(1.95) = 8.43 m

d(2.05) = ((2.05)²/6)+4*(2.05) = 8.90 m

v avg = (8.90-8.43)m/(2.05-1.95)s = 4.66 m/s

b) d(1.995) = ((1.995)²/6)+4*(1.995) = 8.643 m

d(2.005) = ((2.005)²/6)+4*(2.005) = 8.690 m

v avg = (8.690-8.643)m/(2.005-1.995)s = 4.66 m/s

c) d(1.9995) = ((1.9995)²/6)+4*(1.9995) = 8.6643 m

d(2.0005) = ((2.0005)²/6)+4*(2.0005) = 8.6690 m

v avg = (8.6690-8.6643)m/(2.0005-1.9995)s = 4.66 m/s

d) d(2) = ((2)²/6)+4*(2) = 8.6666 m

d(2.00001) = ((2.00001)²/6)+4*(2.00001) = 8.6667 m

v avg = (8.6667-8.6666)m/(2.00001-2)s = 4.66 m/s

The instantaneous velocity of the runner at 2 seconds is obtained as follows

v(t) = d(d(t))/dt =d(t²/6)+4*t)/dt = (t/3) + 4

then

v(2) = (2/3) + 4 = 4.66 m/s

lhmarianateixeira

In this exercise we have to calculate the athlete's position, thus we find that:

A) [tex]4.66 m/s[/tex]

B)[tex]4.66 m/s[/tex]
C)[tex]4.66 m/s[/tex]
D)[tex]4.66 m/s[/tex]

E)[tex]4.66 m/s[/tex]

Knowing that the formula needed for the calculation is:

[tex]d(t)=(t^2/6)+4*t\\v_{avg} = \Delta d/\Delta t[/tex]

Then using the formula and just putting the known values ​​we find that;

a) [tex]d(1.95) = ((1.95)^2/6)+4*(1.95) = 8.43 m\\d(2.05) = ((2.05)^2/6)+4*(2.05) = 8.90 m\\v_{avg} = (8.90-8.43)m/(2.05-1.95)s = 4.66 m/s[/tex]

b) [tex]d(1.995) = ((1.995)^2/6)+4*(1.995) = 8.643 m\\d(2.005) = ((2.005)^2/6)+4*(2.005) = 8.690 m\\v_{avg} = (8.690-8.643)m/(2.005-1.995)s = 4.66 m/s[/tex]

c) [tex]d(1.9995) = ((1.9995)^2/6)+4*(1.9995) = 8.6643 m\\d(2.0005) = ((2.0005)^2/6)+4*(2.0005) = 8.6690 m\\v_{ avg} = (8.6690-8.6643)m/(2.0005-1.9995)s = 4.66 m/s[/tex]

d)[tex]d(2) = ((2)^2/6)+4*(2) = 8.6666 m\\d(2.00001) = ((2.00001)^2/6)+4*(2.00001) = 8.6667 m\\v_{ avg} = (8.6667-8.6666)m/(2.00001-2)s = 4.66 m/s[/tex]

E)The instantaneous velocity of the runner at 2 seconds is obtained as follows, will be :

[tex]v(t) = d(d(t))/dt =d(t^2/6)+4*t)/dt = (t/3) + 4\\v(2) = (2/3) + 4 = 4.66 m/s[/tex]

See more about velocity at brainly.com/question/862972

Otras preguntas