Respuesta :
Answer:
[tex] s(t) = \frac{t^4}{12}-\frac{3t^3}{6}+\frac{9t^2}{2} +15.91667t[/tex]
Step-by-step explanation:
Integrating the expression that describes the acceleration of the particle gives us an expression for its velocity. Integrating the expression for its velocity, gives us the expression for its position:
[tex]a(t) = t^2-3t+9\\\int{a(t)} \, dt=v(t) = \frac{t^3}{3}-\frac{3t^2}{2}+9t +A\\\int{v(t)} \, dt=s(t) = \frac{t^4}{12}-\frac{3t^3}{6}+\frac{9t^2}{2} +At+B[/tex]
Use the given values s(0) = 0 and s(s) = 20 to find the constants A and B:
[tex]s(0) =0= 0-0-0 +A*0+B\\B=0\\ s(1) =20= \frac{1^4}{12}-\frac{3*1^3}{6}+\frac{9*1^2}{2} +A*1+0\\ A=15.91667\\\\s(t) = \frac{t^4}{12}-\frac{3t^3}{6}+\frac{9t^2}{2} +15.91667t[/tex]
Using integration, it is found that the position of the particle is given by:
[tex]s(t) = \frac{t^4}{12} - \frac{t^3}{2} + \frac{9t^2}{2} + 15.92t[/tex]
The acceleration is given by:
[tex]a(t) = t^2 - 3t + 9[/tex]
The velocity is the integral of the acceleration, thus:
[tex]v(t) = \int a(t) dt[/tex]
[tex]v(t) = \int (t^2 - 3t + 9) dt[/tex]
[tex]v(t) = \frac{t^3}{3} - \frac{3t^2}{2} + 9t + k_1[/tex]
The position is the integral of the velocity, thus:
[tex]s(t) = \int v(t) dt[/tex]
[tex]s(t) = \int \left(\frac{t^3}{3} - \frac{3t^2}{2} + 9t + k_1\right) dt[/tex]
[tex]s(t) = \frac{t^4}{12} - \frac{t^3}{2} + \frac{9t^2}{2} + k_1t + k_2[/tex]
Since s(0) = 0, [tex]k_2 = 0[/tex].
Since s(1) = 20:
[tex]\frac{1}{12} - \frac{1}{2} + \frac{9}{2} + k_1 = 20[/tex]
[tex]\frac{1 - 6 + 54 + 12k_1}{12} = 20[/tex]
[tex]49 + 12k_1 = 240[/tex]
[tex]k_1 = \frac{240 - 49}{12}[/tex]
[tex]k_1 = 15.92[/tex]
Thus, the equation is:
[tex]s(t) = \frac{t^4}{12} - \frac{t^3}{2} + \frac{9t^2}{2} + 15.92t[/tex]
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