A particle moves along a straight line. Its positions at time $t \ge 0$ ($t$ in seconds) is given in meters. The acceleration function (in $m/s^2$) of the particle is $a(t) =t^2+ 4t+ 6$ and $s(0) = 0$, $s(1) = 20$. Find the displacement $s(t)$.
I am confused on how to do this. Any help thank you.
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$\begingroup$$$a(t) = \frac{d^2s}{dt^2} = t^2+4t+6$$Integrating once with respect to $t$ you will get $$v(t) = \frac{t^3}{3}+2t^2+6t+c_1$$
Integrating second time with respect to $t$ you will get $$s(t) = \frac{t^4}{12}+\frac{2t^3}{3}+3t^2+c_1t+c_2$$
What we need now is to solve the initial value problem. The clues are on $s(0)$ and $s(1)$. You have two unknowns and two equations so this is solvable.
$\endgroup$ 5 $\begingroup$$$a=\frac{dv}{dt}$$This can be written as $$\int_0^Tdv=\int_0^T a(t)dt$$Integrating, you get $$v(T)-v(0)=\int_0^T a(t)dt$$Do the integration of the right hand side. Then repeat the same procedure for the displacement. Note that you will get a formula that includes integrating twice the acceleration, plus two integration constants: $v(0)$ and $s(0)$. $s(0)$ it is given explicitly to you. $v(0)$ can be obtained from $s(1)$ (just plug in $t=1$ in your displacement equation)
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