Answer
$$\frac{d a}{d t}=0.653 \mathrm{\ m} / \mathrm{s}^{3}$$
Work Step by Step
At $t=0,\quad m_{10}=1.30 \mathrm{kg} .$
With $d m_{1} / d t=-0.200 \mathrm{kg} / \mathrm{s}$ , we find the rate of change of acceleration to be
$$\frac{d a}{d t}=\frac{d a}{d m_{1}} \frac{d m_{1}}{d t}=-\frac{2 m_{2} g}{\left(m_{2}+m_{10}\right)^{2}} \frac{d m_{1}}{d t}$$$$=-\frac{2(2.80 \mathrm{kg})\left(9.80 \mathrm{m} / \mathrm{s}^{2}\right)}{(2.80 \mathrm{kg}+1.30 \mathrm{kg})^{2}}(-0.200 \mathrm{kg} / \mathrm{s})$$$$=0.653 \mathrm{\ m} / \mathrm{s}^{3}$$
