本题目来源于试卷: Kinematics，类别为 IB数学
[问答题]
The acceleration, $a \mathrm{~ms}^{-2}$ of a particle moving in a vertical trajectory at time t seconds, $t \geq 0$ , is given by a(t)=-(3+v) where v is the particle's velocity in $\mathrm{ms}^{-1} $. At t=0 , the particle is at a fixed origin O and has an initial velocity of $ v_{0} \mathrm{~ms}^{-1}$ .
1. By solving an appropriate differential equation, show that the particle's velocity is given by $ v(t)=\left(v_{0}+3\right) e^{-t}-3 $.
The particle initially moves upwards until it reaches its maximum height from O , and then returns to O . Let s metres represent the particle's displacement from O , and $s_{\max } $ the maximum displacement from O .
2. 1. Show that the time T taken for the particle to reach $ s_{\max }$ satisfies the equation $ e^{-T}=\frac{3}{v_{0}+3}$ .
2. Hence, solve for T in terms of $ v_{0}$ .
3. By solving an appropriate differential equation and using the results from part (b) (i) and (ii), find an expression for $s_{\text {max }}$ in terms of $v_{0} $.
Let v(T-k) represent the particle's velocity k seconds before it reaches $ s_{\max } $, where
$v(T-k)=\left(v_{0}+3\right) e^{-(T-k)}-3$
3. By using the result from part (b) (i), show that v(T-k)=3 e^{k}-3 .
Similarly, let v(T+k) represent the particle's velocity k seconds after it reaches $ s_{\text {max }} $.
4. Deduce a similar expression for v(T+k) in terms of k .
5. Hence, show that v(T-k)+v(T+k) $\geq 0 $.
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