El desplazamiento (en metros) de una partícula que se mueve
en línea recta esta dado por s = 12 – 8t + 18, donde t se
mide en segundos.
a) Encuentre la velocidad promedio en cada intervalo de
tiempo:
i) [3, 4] ii) [3.5, 4]
iii) [4, 5] iv) [4, 4.5]
b) Halle la velocidad instantánea cuando t = 4.
c) Dibuje la grafica de s como función de t y trace las rectas
secantes cuyas pendientes son las velocidades promedio
en el inciso a) y la recta tangente cuya pendiente es la
velocidad instantánea en el inciso b).

Answer:

c

Step-by-step explanation:

hize la misma pregunta que vos

The average velocities for the given time intervals are,

(i) -8 m/s,

(ii) -12 m/s,

(iii) 4 m/s, and

(iv) 4 m/s. The instantaneous velocity at t = 4 is -8 m/s. The attached graph of s as a function of t is a straight line with slope -8, showing the secant and tangent lines.

a) To find the average velocity in each time interval,

calculate the change in displacement (Δs) divided by the change in time (Δt) within each interval.

s = 12 - 8t + 18

i) [3, 4]:

Δs = s(4) - s(3) = (12 - 8(4) + 18) - (12 - 8(3) + 18) = -8

Δt = 4 - 3 = 1

Average velocity = Δs / Δt = -8 / 1 = -8 m/s

ii) [3.5, 4]:

Δs = s(4) - s(3.5) = (12 - 8(4) + 18) - (12 - 8(3.5) + 18) = -6

Δt = 4 - 3.5 = 0.5

Average velocity = Δs / Δt = -6 / 0.5 = -12 m/s

iii) [4, 5]:

Δs = s(5) - s(4) = (12 - 8(5) + 18) - (12 - 8(4) + 18) = 4

Δt = 5 - 4 = 1

Average velocity = Δs / Δt = 4 / 1 = 4 m/s

iv) [4, 4.5]:

Δs = s(4.5) - s(4) = (12 - 8(4.5) + 18) - (12 - 8(4) + 18) = 2

Δt = 4.5 - 4 = 0.5

Average velocity = Δs / Δt = 2 / 0.5 = 4 m/s

b) To find the instantaneous velocity when t = 4, we'll find the derivative of s with respect to t and then substitute t = 4.

s = 12 - 8t + 18

ds/dt = -8

Instantaneous velocity at t = 4 is equal to the derivative at t = 4:

Instantaneous velocity at t = 4: ds/dt (t=4) = -8 m/s

c) To sketch the graph of s as a function of t and draw the secant lines (average velocities) and the tangent line (instantaneous velocity):

Attached graph.

The graph of s as a function of t is a straight line with a slope of -8 and a y-intercept of 30. At t = 4, the instantaneous velocity is -8 m/s.

Secant lines:

Draw a straight line connecting the points (3, 10) and (4, 2) for the interval [3, 4].

Draw a straight line connecting the points (3.5, 4) and (4, 2) for the interval [3.5, 4].

Draw a straight line connecting the points (4, 2) and (5, 6) for the interval [4, 5].

Draw a straight line connecting the points (4, 2) and (4.5, 4) for the interval [4, 4.5].

Tangent line:

At t = 4, draw a straight line with a slope of -8 and passing through the point (4, 2).

The attached graph will show the secant lines with different slopes as the average velocities in each interval and the tangent line with a slope of -8 as the instantaneous velocity at t = 4.

learn more about average velocity here

brainly.com/question/32663715

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The above question is incomplete , the complete question is:

The displacement (in meters) of a moving particle

in a straight line is given by s = 12 – 8t + 18, where t is

measured in seconds.

a) Find the average velocity in each time interval:

i) [3, 4] ii) [3.5, 4]

iii) [4, 5] iv) [4, 4.5]

b) Find the instantaneous velocity when t = 4.

c) Plot s as a function of t and draw the secant lines whose slopes are the average velocity in part (a) and the tangent line whose slope is the

instantaneous velocity in part (b).

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