Respuesta :
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
[tex]\bar{v} = \frac{1}{t_{2}-t_{1}}[58t-1.86t^{2}]_{t1}^{t2}[/tex]
Evaluate the results (Use v to denote the average velocity).
i. t = [1,2]
v = (108.56 - 56.14)/1 = 52.42
Answer: 52.42 m/s
ii. t=[1,1.5]
v = (82.815 - 56.14)/0.5 = 53.35
Answer: 53.35 m/s
iii. t=[1,1.1]
v = (61.5494 - 56.14)/0.1 = 54.094
Answer: 54.094 m/s
iv. t=[1,1.01]
v = (56.6826 - 56.14)/0.01 = 54.26
Answer: 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
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
[tex]\bar{v} = \frac{1}{t_{2}-t_{1}}[58t-1.86t^{2}]_{t1}^{t2}[/tex]
Evaluate the results (Use v to denote the average velocity).
i. t = [1,2]
v = (108.56 - 56.14)/1 = 52.42
Answer: 52.42 m/s
ii. t=[1,1.5]
v = (82.815 - 56.14)/0.5 = 53.35
Answer: 53.35 m/s
iii. t=[1,1.1]
v = (61.5494 - 56.14)/0.1 = 54.094
Answer: 54.094 m/s
iv. t=[1,1.01]
v = (56.6826 - 56.14)/0.01 = 54.26
Answer: 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
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, [tex]y = 58t-1.86t^2[/tex].
- t is time in seconds.
The average velocity is given by the formula:
[tex]\rm \bar{v} = \dfrac{Distance\;at \; time \;t_2-Distance\;at \; time \;t_1}{t_2-t_1}[/tex]
So, at the time interval, t = [1,2], the average velocity will be:
[tex]\rm \bar{v} = \dfrac{(58\times 2-1.86\times 2^2)-(58\times 1-1.86\times 1^2)}{2-1}[/tex]
[tex]\rm \bar{v} = \dfrac{(108.56)-(56.14)}{1}=52.42\;m/sec[/tex]
Now, at the time interval, t = [1,1.5], the average velocity will be:
[tex]\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}[/tex]
[tex]\rm \bar{v} = \dfrac{(82.815)-(56.14)}{0.5}=53.35\;m/sec[/tex]
Now, at the time interval, t = [1,1.1], the average velocity will be:
[tex]\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}[/tex]
[tex]\rm \bar{v} = \dfrac{(61.5494)-(56.14)}{0.1}=54.09\;m/sec[/tex]
Now, at the time interval, t = [1,1.01], the average velocity will be:
[tex]\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}[/tex]
[tex]\rm \bar{v} = \dfrac{(56.682614)-(56.14)}{0.01}=54.26\;m/sec[/tex]
For more information, refer the link given below:
https://brainly.com/question/18153640
