Using derivatives, it is found that:
i) 
ii) 9 m/s.
iii) 
iv) 6 m/s².
v) 1 second.
<h3>What is the role of derivatives in the relation between acceleration, velocity and position?</h3>
- The velocity is the derivative of the position.
- The acceleration is the derivative of the velocity.
In this problem, the position is:
item i:
Velocity is the <u>derivative of the position</u>, hence:
Item ii:
The speed is of 9 m/s.
Item iii:
Derivative of the velocity, hence:
Item iv:
The acceleration is of 6 m/s².
Item v:
t for which a(t) = 0, hence:
Hence 1 second.
You can learn more about derivatives at brainly.com/question/14800626
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Answer:
C. 1/(4a)
Step-by-step explanation:
We assume you're comparing the vertex form ...
y = a(x -h)^2 +k
to the form used to write the equation in terms of the focal distance p.
y = 1/(4p)(x -h)^2 +k
That comparison tells you ...
a = 1/(4p)
p = 1/(4a) . . . . . . multiply by p/a; matches choice C
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<em>Additional comment</em>
When using plain text to write a rational expression, parentheses are needed around any denominator that has is more than a single constant or variable. The order of operations requires 1/4a to be interpreted as (1/4)a. The value of p is 1/(4a).
When rational expressions are typeset, the fraction bar serves as a grouping symbol identifying the entire denominator:
A=8ft^2
A=l•w
8=l•w
l=8/w
P=2(l+w)
12=2(8/w+w)
12=2(8/w+w/1)
12=2(8/w+w^2/w)
12=16w^2/w
12=16w
w=12/16
w=3/4=0.75ft
l=8/0.75=10.67ft
