Modeling with First Order DE’s — Question 4

Question 4 – Solution

Strategy. Balance forces in each phase and carry velocity continuously across a coefficient change; integrate velocity to obtain displacement.

Step 1: Model and match the phases. Newton’s law gives $2v'=20-bv$, with forces in newtons. Thus $v'=10-v$ before opening and $v'=10-5v$ afterward. The initial value is $v(0)=0$, and no impulse means $v(4^-)=v(4^+)=v_4$.

Step 2: Solve both initial-value problems. The solutions are $$\boxed{v(t)= \begin{cases} 10(1-e^{-t}),&0\le t\le 4,\\ 2+(v_4-2)e^{-5(t-4)},&4\le t\le 6, \end{cases}\quad v_4=10(1-e^{-4})\approx 9.817\text{ m/s}.}$$ Each expression satisfies its phase equation, and both give $v_4$ at the switch. The phase terminal speeds are $10$ and $2$ m/s, respectively.

Step 3: Integrate the distance. Velocity stays positive, so downward displacement equals distance traveled: $$\boxed{d=10(3+e^{-4})+4+ \frac{8-10e^{-4}}5(1-e^{-10})\approx 35.75\text{ m}}.$$ The first term integrates the pre-opening velocity over $[0,4]$; the remaining terms integrate the post-opening velocity over $[4,6]$.

Step 4: Examine the acceleration jump. The one-sided accelerations are $$v'(4^-)=10e^{-4}\approx 0.183\text{ m/s}^2,\qquad v'(4^+)=-40+50e^{-4}\approx-39.08\text{ m/s}^2.$$ Changing a finite force changes acceleration without an instantaneous velocity jump. The velocity is continuous and piecewise differentiable, but is not differentiable at the idealized switch. Each classical phase equation applies away from $t=4$; the switch is handled by continuity, not by requiring a single derivative there.