(a) a=τa˙+mF⇒dtdv=τdtda+mF⇒∫dtdvdt=τ∫dtdadt+m1∫Fdt.
[v(t0+ϵ)−v(t0−ϵ)]=τ[a(t0+ϵ)−a(t0−ϵ)]+m2ϵFave , where Fave is the average force during the interval. But υ is continuous, so as long as F is not a delta function, we are left (in the limit ϵ→0) with [a(t0+ϵ)−a(t0−ϵ)]=0 . Thus a, too, is continuous. qed
(b) (i) a=τa˙=τdtda⇒ada=τ1dt⇒∫ada=τ1∫dt⇒lna=τt+ constant ⇒a(t)=Aet/τ, where A is a constant.
(ii) a=τa˙+mF⇒τdtda=a−mF⇒a−F/mda=τ1dt⇒ln(a−F/m)=τt+ constant ⇒a−mF=Bet/τ⇒a(t)=mF+Bet/τ , where B is some other constant.
(iii) Same as (i): a(t)=Cet/τ , where C is a third constant.
(c) At t=0,A=F/m+B; at t=T,F/m+BeT/τ=CeT/τ⇒C=(F/m)e−T/τ+B . So
a(t)=⎩⎪⎪⎨⎪⎪⎧[(F/m)+B]et/τ,[(F/m)+Bet/τ],[(F/m)e−T/τ+B]et/τ,t≤00≤t≤Tt≥T
To eliminate the runaway in region (iii), we’d need B=−(F/m)e−T/τ ; to avoid preacceleration in region (i), we’d need B = –(F/m). Obviously, we cannot do both at once.
(d) If we choose to eliminate the runaway, then
a(t)=⎩⎪⎪⎨⎪⎪⎧(F/m)[1−e−T/τ]et/τ,(F/m)[1−e(t−T)/τ],0,t≤00≤t≤Tt≥T.
(i) v=(F/m)[1−e−T/τ]∫et/τdt=(Fτ/m)[1−e−T/τ]et/τ+D , where D is a constant determined by the condition v(−∞)=0⇒D=0.
(ii) v=(F/m)[t−τe(t−T)/τ]+E , where E is a constant determined by the continuity of υ at t = 0: (Fτ/m)[1−e−T/τ]=(F/m)[−τe−T/τ]+E⇒E=(Fτ/m).
(iii) v is a constant determined by the continuity of v at t=T:v=(F/m)[T+τ−τ]=(F/m)T.
v(t)=⎩⎪⎪⎨⎪⎪⎧(Fτ/m)[1−e−T/τ]et/τ,(F/m)[t+τ−τe(t−T)/τ],(F/m)T,t≤00≤t≤T;t≥T.