Answer
At $t=3.00 \mathrm{s},$
$$m_{1}=m_{10}+\left(d m_{1} / d t\right) t$$$$=1.30 \mathrm{kg}+(-0.200 \mathrm{kg} / \mathrm{s})(3.00 \mathrm{s})=0.700 \mathrm{kg},$$ and the rate of change of acceleration is
$$\frac{d a}{d t}=\frac{d a}{d m_{1}} \frac{d m_{1}}{d t}=-\frac{2 m_{2} g}{\left(m_{2}+m_{1}\right)^{2}} \frac{d m_{1}}{d t}$$$$=-\frac{2(2.80 \mathrm{kg})\left(9.80 \mathrm{m} / \mathrm{s}^{2}\right)}{(2.80 \mathrm{kg}+0.700 \mathrm{kg})^{2}}(-0.200 \mathrm{kg} / \mathrm{s})$$
$$=0.896 \mathrm{m} / \mathrm{s}^{3}$$
Work Step by Step
At $t=3.00 \mathrm{s},$
$$m_{1}=m_{10}+\left(d m_{1} / d t\right) t$$$$=1.30 \mathrm{kg}+(-0.200 \mathrm{kg} / \mathrm{s})(3.00 \mathrm{s})=0.700 \mathrm{kg},$$ and the rate of change of acceleration is
$$\frac{d a}{d t}=\frac{d a}{d m_{1}} \frac{d m_{1}}{d t}=-\frac{2 m_{2} g}{\left(m_{2}+m_{1}\right)^{2}} \frac{d m_{1}}{d t}$$$$=-\frac{2(2.80 \mathrm{kg})\left(9.80 \mathrm{m} / \mathrm{s}^{2}\right)}{(2.80 \mathrm{kg}+0.700 \mathrm{kg})^{2}}(-0.200 \mathrm{kg} / \mathrm{s})$$
$$=0.896 \mathrm{m} / \mathrm{s}^{3}$$
