$\begin{aligned} & \begin{array}{l}\frac{d p}{d t}=k[x]^1 \\ t=0 \quad 2 \quad 1 \\ t=50 s(2-1) \quad y \quad(1-0.5) \\ \quad-\frac{1}{2} \frac{d x}{d t}=\frac{d p}{d t}=k[x]^1 \\ \quad-\frac{d x}{d t}=2 k=k[x]^1 \\ \quad 2 k=\frac{\ln 2}{50} \Rightarrow k=\frac{\ln 2}{100} \\ \text { At } \quad 50 \sec \frac{-d x}{d t}=2 k \times(1)^1=\frac{\ln 2}{50} \\ \text { At } \quad 100 \sec -\frac{1}{2} \frac{d x}{d t}=\frac{-d y}{d t} \Rightarrow \frac{d y}{d t}=\frac{\ln 2}{100} \times \frac{1}{2}\left\{\frac{-d y}{d t}=k[x]^1\right\}\end{array}\end{aligned}$