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:

$a(t) = t^2-3t+9\\\int{a(t)} \, dt=v(t) = \frac{t^3}{3}-\frac{3t^2}{2}+9t +A\\\int{v(t)} \, dt=s(t) = \frac{t^4}{12}-\frac{3t^3}{6}+\frac{9t^2}{2} +At+B$

Use the given values s(0) = 0 and s(s) = 20 to find the constants A and B:

$s(0) =0= 0-0-0 +A*0+B\\B=0\\ s(1) =20= \frac{1^4}{12}-\frac{3*1^3}{6}+\frac{9*1^2}{2} +A*1+0\\ A=15.91667\\\\s(t) = \frac{t^4}{12}-\frac{3t^3}{6}+\frac{9t^2}{2} +15.91667t$

8 0

4 years ago

May someone help me I’ll give you brainiest

Pepsi [2]

Feel like you joined the class

8 0

3 years ago

You purchase a cell phone for $125. the value of the cell phone decreases by about 20% annually. write a function at models the

djverab [1.8K]

F(x)=125-(125*0.2x)
to find the value of the phone after 3 years, you put x=3.
f(3)=125-(125*0.2(3))
f(3)=50

5 0

4 years ago