Using derivatives, it is found that:
i) [tex]v(t) = 3t^2 - 6t[/tex]
ii) 9 m/s.
iii) [tex]a(t) = 6t - 6[/tex]
iv) 6 m/s².
v) 1 second.
What is the role of derivatives in the relation between acceleration, velocity and position?
- The velocity is the derivative of the position.
- The acceleration is the derivative of the velocity.
In this problem, the position is:
[tex]s(t) = t^3 - 3t^2 + 8[/tex]
item i:
Velocity is the derivative of the position, hence:
[tex]v(t) = 3t^2 - 6t[/tex]
Item ii:
[tex]v(3) = 3(3)^2 - 6(3) = 27 - 18 = 9[/tex]
The speed is of 9 m/s.
Item iii:
Derivative of the velocity, hence:
[tex]a(t) = 6t - 6[/tex]
Item iv:
[tex]a(2) = 6(2) - 6 = 6[/tex]
The acceleration is of 6 m/s².
Item v:
t for which a(t) = 0, hence:
[tex]6t - 6 = 0[/tex]
[tex]6t = 6[/tex]
[tex]t = \frac{6}{6}[/tex]
[tex]t = 1[/tex]
Hence 1 second.
You can learn more about derivatives at https://brainly.com/question/14800626
