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Sagot :
Using integration, it is found that the total distance traveled by P in the interval between 0 and 4 seconds was of 230.67 meters.
What is the role of derivatives in the relation between acceleration, velocity and position?
- The velocity is the derivative of the position, hence the position is the integrative of the velocity.
- The acceleration is the derivative of the velocity, hence the velocity is the integrative of the acceleration.
In this problem, the velocity is:
[tex]v(t) = 2t^2 - 9t + 4[/tex]
Then, the position will be given by:
[tex]s(t) = \int v(t) dt[/tex]
[tex]s(t) = \int (2t^2 - 9t + 4) dt[/tex]
[tex]s(t) = \frac{2t^3}{3} - \frac{9t^2}{2} + 4t + K[/tex]
Considering that when t = 0, P is 15 m from the origin O, K = 15, and:
[tex]s(t) = \frac{2t^3}{3} - \frac{9t^2}{2} + 4t + 15[/tex]
The total distance traveled by P between 0 and 4 seconds is:
[tex]D = \int_{0}^{4} s(t) dt[/tex]
[tex]D = \int_{0}^{4} \left(\frac{2t^3}{3} - \frac{9t^2}{2} + 4t + 15\right) dt[/tex]
[tex]D = \left(\frac{t^4}{6} - \frac{3t^3}{2} + 2t^2 + 15t + K\right)_{t = 0}^{t = 4}[/tex]
Then, applying the Fundamental Theorem of Calculus:
[tex]D(4) = \frac{4^4}{6} - \frac{3(4)^3}{2} + 2(4)^2 + 15(4) = 230.67[/tex]
[tex]D(0) = \frac{0^4}{6} - \frac{3(0)^3}{2} + 2(0)^2 + 15(0) = 0[/tex]
[tex]D = D(4) - D(0) = 230.67 - 0 = 230.67[/tex]
The total distance traveled by P in the interval between 0 and 4 seconds was of 230.67 meters.
You can learn more about integration at https://brainly.com/question/14800626
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