sabinadhindsa
Answered

Welcome to Westonci.ca, the ultimate question and answer platform. Get expert answers to your questions quickly and accurately. Explore a wealth of knowledge from professionals across various disciplines on our comprehensive Q&A platform. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.

pls help asap
for the function s(t)=-16t^2 + 104t, find the avg velocity at the time interval [1,1.01]

Sagot :

BrotherGoku

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

semsee45

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.

Thanks for using our platform. We're always here to provide accurate and up-to-date answers to all your queries. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.