Mathematics, 30.01.2020 06:59 CateL1515

If an arrow is shot upward on mars with a speed of 58 m/s, its height in meters t seconds later is given by y = 58t − 1.86t2. (round your answers to two decimal places.) (a) find the average speed over the given time intervals.
i. t= [1, 2]
ii. t=[1, 1.5]
iii. [1, 1.1]
iv. [1, 1.01]

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Answers

The distance is given by
y = 58t - 1.86t²

The velocity is the derivative of y with respect to t. It is
v = 58 - 3.72t

In an interval [t₁, t₂], the average velocity is
$\bar{v} = \frac{1}{t_{2}-t_{1}}[58t-1.86t^{2}]_{t1}^{t2}$

Evaluate the results (Use v to denote the average velocity).
i. t = [1,2]
v = (108.56 - 56.14)/1 = 52.42
52.42 m/s

ii. t=[1,1.5]
v = (82.815 - 56.14)/0.5 = 53.35
53.35 m/s

iii. t=[1,1.1]
v = (61.5494 - 56.14)/0.1 = 54.094
54.094 m/s

iv. t=[1,1.01]
v = (56.6826 - 56.14)/0.01 = 54.26
54.26 m/s

Note:
As the time interval approaches zero, we obtain the derivative of y at t=1, which is 58-3.72 = 54.28 m/s

Answer from: taridunkley724

The average velocity at different time interval can be evaluated by using average velocity formula. So, at time interval [1,2] the average velocity is 52.42m/sec, at time interval [1,1.5] the average velocity is 53.35m/sec, at time interval [1,1.1] the average velocity is 54.09m/sec, and at time interval [1,1.01] the average velocity is 54.26m/sec.

Given :

Arrow is shot upward on Mars with a speed of 58 m/s.Height of the arrow shot upwards is, y = 58t-1.86t^2

.t is time in seconds.

The average velocity is given by the formula:

$\rm \bar{v} = \dfrac{Distance\;at \; time \;t_2-Distance\;at \; time \;t_1}{t_2-t_1}$

So, at the time interval, t = [1,2], the average velocity will be:

$\rm \bar{v} = \dfrac{(58\times 2-1.86\times 2^2)-(58\times 1-1.86\times 1^2)}{2-1}$

$\rm \bar{v} = \dfrac{(108.56)-(56.14)}{1}=52.42\;m/sec$

Now, at the time interval, t = [1,1.5], the average velocity will be:

$\rm \bar{v} = \dfrac{(58\times 1.5-1.86\times (1.5)^2)-(58\times 1-1.86\times 1^2)}{1.5-1}$

$\rm \bar{v} = \dfrac{(82.815)-(56.14)}{0.5}=53.35\;m/sec$

Now, at the time interval, t = [1,1.1], the average velocity will be:

$\rm \bar{v} = \dfrac{(58\times 1.1-1.86\times (1.1)^2)-(58\times 1-1.86\times 1^2)}{1.1-1}$

$\rm \bar{v} = \dfrac{(61.5494)-(56.14)}{0.1}=54.09\;m/sec$

Now, at the time interval, t = [1,1.01], the average velocity will be:

$\rm \bar{v} = \dfrac{(58\times 1.01-1.86\times (1.01)^2)-(58\times 1-1.86\times 1^2)}{1.01-1}$

$\rm \bar{v} = \dfrac{(56.682614)-(56.14)}{0.01}=54.26\;m/sec$

For more information, refer the link given below:

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Answer from: Quest

step-by-step explanation:

Answer from: Quest

the system of inequity is

$y\leq -x+4$

$3y> x$

step-by-step explanation:

from the given graph it is noticed that we have two related lines. one is solid and another is dashed.

if a line passing through two points, then the equation of line is

$y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$

the solid line passing through the points (0,4) and (4,0).

$y-4=\frac{0-4}{4-0}(x-0)$

$y=-x+4$

the related equation of solid line is $y=-x+4$ .

similarly, find the related equation of dashed line. the dashed liune passing through the points (0,0) and (3,1).

$y-0=\frac{0-1}{3-0}(x-0)$

$3y=x$

the related equation of dashed line is $3y=x$

from the given graph it is noticed that the point (0,1) lies on the shaded region, therefore the point (0,1) must be satisfied by the related equations.

check the related equations by the point (0,1).

$1=-0+4$

$1=4$

this statement is true if the sign of inequality is < instead of equality. since it is an solid line therefore the sign of inequality will be $\leq$ . the required inequality is

$y\leq -x+4$

similarly, check the second related equation.

$3(1)=0$

it is a dashed line, so this statement is true if the sign of inequality is > . the required inequality is

$3y> x$

15 points write a system of inequalities to represent the shaded portion of the graph

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