91影视

It is given that,

The mass of the ice block is$M_1=1.3kg$

The mass of the ice block is$M_2=2.8kg$

The rate of change of mass,$\frac{d{M}_1}{dt}=0.2\frac{kg}{s}$

The problem is based on Newton鈥檚 second law of motion which states that the rate of change of momentum of a body is equal in both magnitude and direction of the force acting on it. Using Newton鈥檚 second law, the acceleration and from that,the rate of change of acceleration of the containers can be found.

Formulae:

Newton鈥檚 second law is,

$F_{net}=\underset{}{鈭�}Ma$ (i)

$\frac{da}{dt}=\frac{da}{dM}脳\frac{dM}{dt}$ (ii)

where,$F_{net}$ is the net force, Mis mass and a is an acceleration.

Apply Newton鈥檚 second law,

$T-M_{1}g=M_{1}a$ (iii)

$M_{2}g-T=M_{2}a$ (iv)

Solving equations (iii) and (iv),

$a=\left(\frac{M_2-M_1}{M_1+M_2}\right)g$

At t = 0 ,$M_1=1.3kg$ .The rate of change of mass,

$\frac{d{M}_1}{dt}=0.2\frac{kg}{s}$

So,its rate of change of acceleration is,

$\frac{da}{dt}=\frac{da}{d{M}_{10}}脳\frac{d{M}_1}{dt}$

Using division rule and considering$M_2$as constant,

$$\begin{gathered} \frac{da}{dt}=\left(\frac{\left\{\left(-1\right)\left(M_1+M_2\right)-\left(M_2-M_1\right)\right\}}{\left(M_1+M_2\right)}\right)\frac{d{M}_1}{dt} \\ \frac{da}{dt}=-\frac{2M_{2}g}{\left(M_1+M_2\right)}脳\frac{d{M}_1}{dt} \\ \frac{da}{dt}=\frac{-2\left(2.8\right)\left(9.8\right)\left(-0.2\right)}{{\left(2.8+1.3\right)}^2} \\ \frac{da}{dt}=0.653m/s^3 \end{gathered}$$

Therefore, the rate of change of acceleration of the containers at t = 0 is$0.653m/s^3.$

At t = 3 s

$$\begin{gathered} M_1=M_1+\left(\frac{d{M}_1}{dt}\right)t \\ {M}_1=1.3+\left(-2\right)\left(3\right) \\ =0.7kg \end{gathered}$$

The rate of change of mass,

$\frac{d{M}_1}{dt}=0.2kg/s$

So, its rate of change of acceleration is,

$$\begin{gathered} \frac{da}{dy}=\frac{da}{d{M}_1}+\frac{d{M}_1}{dt} \\ \frac{da}{dt}=\left[\frac{\left\{\left(-1\right)\left(M_1+M_2\right)-\left(M_2-M_1\right)\right\}}{{\left(M_1+M_2\right)}^2}\right]\frac{d{M}_1}{dt} \\ \frac{da}{dt}=-\frac{2M_{2}g}{{\left(M_1+M_2\right)}^2}脳\frac{d{M}_1}{dt} \\ \frac{da}{dt}=\frac{-2\left(2.8\right)\left(9.8\right)\left(-0.2\right)}{{\left(2.8+0.7\right)}^2} \\ \frac{da}{dt}=0.896m/s^2 \end{gathered}$$

Hence, the rate of change of acceleration of the containers at t = 3s is $0.896m/s^2.$

The acceleration reaches its maximum when$M_1=0$ , so,

$$\begin{gathered} M_1+\left(\frac{d{M}_1}{dt}\right)t=0 \\ 1.3-\left(0.2\right)t=0 \\ t=6.5s \end{gathered}$$

Hence, at$t=6.5s$ the acceleration reaches its maximum value.