Question

a particle is moving with the given data. find the position of the particle, ( s(t) ).

( a(t)=t^{2}-4 t+5, quad s(0)=0, quad s(1)=20 )

( s(t)= )

Explanation:

Step1: Integrate acceleration to get velocity

Since \(a(t)=t^{2}-4t + 5\), and \(v(t)=\int a(t)dt\).
Using the power rule \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n
eq - 1)\), we have \(v(t)=\int(t^{2}-4t + 5)dt=\frac{t^{3}}{3}-2t^{2}+5t + C\).

Step2: Integrate velocity to get position

Since \(s(t)=\int v(t)dt\), then \(s(t)=\int(\frac{t^{3}}{3}-2t^{2}+5t + C)dt=\frac{t^{4}}{12}-\frac{2t^{3}}{3}+\frac{5t^{2}}{2}+Ct+D\).

Step3: Use initial conditions

Given \(s(0) = 0\), substitute \(t = 0\) into \(s(t)\):
\(s(0)=\frac{0^{4}}{12}-\frac{2\times0^{3}}{3}+\frac{5\times0^{2}}{2}+C\times0+D=0\), so \(D = 0\).
Given \(s(1)=20\), substitute \(t = 1\) and \(D = 0\) into \(s(t)\):
\(s(1)=\frac{1^{4}}{12}-\frac{2\times1^{3}}{3}+\frac{5\times1^{2}}{2}+C\times1=\frac{1}{12}-\frac{2}{3}+\frac{5}{2}+C\).
Simplify \(\frac{1}{12}-\frac{8}{12}+\frac{30}{12}+C=\frac{1 - 8+30}{12}+C=\frac{23}{12}+C\).
Since \(s(1) = 20=\frac{240}{12}\), then \(C=\frac{240}{12}-\frac{23}{12}=\frac{217}{12}\).

Answer:

\(s(t)=\frac{t^{4}}{12}-\frac{2t^{3}}{3}+\frac{5t^{2}}{2}+\frac{217t}{12}\)