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
4n 5 + 3.5n 5 - 2.1n 5 = n 5 (4 + 3.5 - 2.1) = n 5 * 5.4 = 5.4n 5
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Answer:
y = 3x -8
Step-by-step explanation:
We assume you want the tangent to the parabola y = x² -3x +1 at the given point. The slope is found using the derivative of the function at that point.
y' = 2x -3
At x=3, the slope is ...
y' = 2(3) -3 = 3
The equation of the line through point (3, 1) with a slope of 3 is ...
y -1 = 3(x -3) . . . . use the point-slope form of the equation for a line
y = 3x -9 +1 . . . . . eliminate parentheses, add 1
y = 3x -8