Explore Westonci.ca, the leading Q&A site where experts provide accurate and helpful answers to all your questions. Join our Q&A platform to connect with experts dedicated to providing precise answers to your questions in different areas. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.
Sagot :
Answer:
71.8
Step-by-step explanation:
So the function s(t) describes the displament of the object over a certain time. To find velocity, first you have to know what it is. Knowing that velocity is given by the unit m/s, which is distance over time, the slope of the s(t) under a certain time interval will be your velocity since it covers a certain distance over a time interval which is given [1,1.01].
So to find the slope, you do y2-y1 / x2 - x1. Therefore, you calculate the y value for each x then you apply the slope formula
x1 = 1
x2 = 1.01
y1 = -16(1)^2 + 104(1) = -16 + 104 = 88
y2 = -16(1.01)^2 + 104(1) = 88.718
88.718 - 88 / 1.01 - 1 = 0.718/0.01 = 71.8 s(t)/s
Answer:
[tex]\overline{v}=71.84\; \sf m/s[/tex]
Step-by-step explanation:
To find the average velocity over the time interval [1, 1.01], we can use the formula for average velocity, which is the total displacement of an object divided by the total time taken for that displacement:
[tex]\overline{v}=\dfrac{s(t_2)-s(t_1)}{t_2-t_1}[/tex]
where:
- v represents the average velocity.
- s(t₁) and s(t₂) represent the positions of the object at times t₁ and t₂ respectively.
- t₁ and t₂ represent the starting and ending times of the interval over which the average velocity is being calculated.
In this case:
- t₁ = 1
- t₂ = 1.01
Substitute the values of t into the displacement function s(t) to calculate s(t₁) and s(t₂):
[tex]s(t_1)=-16(1)^2 + 104(1) \\\\ s(t_1)=-16 + 104\\\\ s(t_1)=88[/tex]
[tex]s(t_2)=-16(1.01)^2 + 104(1.01) \\\\ s(t_2)=-16(1.0201)+104(1.01)\\\\s(t_2)=-16.3216+105.04\\\\s(t_2)=88.7184[/tex]
Now, substitute the values into the average velocity equation:
[tex]\overline{v}=\dfrac{88.7184-88}{1.01-1} \\\\\\ \overline{v}=\dfrac{0.7184}{0.01} \\\\\\ \overline{v}=71.84[/tex]
Therefore, the average velocity at the time interval [1, 1.01] is 71.84 m/s, assuming that the distance is measured in meters and the time is measured in seconds.
We appreciate your time. Please revisit us for more reliable answers to any questions you may have. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Thank you for trusting Westonci.ca. Don't forget to revisit us for more accurate and insightful answers.