= Solution
For a continuous <survival distribution>, $S(t)=e^{-H(t)}$, $h(t)=H'(t)$ and $f(t)=h(t)S(t)$. Applying these identities gives the <Weibull distribution>
$$
\boxed{S(t)=e^{-(\lambda t)^p},\qquad
h(t)=p\lambda^pt^{p-1},\qquad
f(t)=p\lambda^pt^{p-1}e^{-(\lambda t)^p},\quad t>0.}
$$
The <survivor function> has value one at zero; the <hazard> may diverge there when $p<1$, which does not invalidate the distribution.
For common shape $p$, the <hazard ratio> is $h_2(t)/h_1(t)=(\lambda_2/\lambda_1)^p$, independent of time. Thus the two distributions lie in the same <proportional hazards family>. They also satisfy
$$
S_2(t)=S_1\left(\frac{\lambda_2}{\lambda_1}t\right),\qquad
T_2\overset d=\frac{\lambda_1}{\lambda_2}T_1.
$$
Hence \b[the time multiplier is $\lambda_1/\lambda_2$ and the <hazard> multiplier is $(\lambda_2/\lambda_1)^p$], establishing the common <accelerated life family> as well. Here $\lambda$ is reciprocal scale, not the ordinary time-scale parameter.
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