Figure 5-47 shows Atwood\'s machine, in which two containers are connected by a

Question

Figure 5-47 shows Atwood's machine, in which two containers are connected by a cord (of negligible mass) passing over a frictionless pulley (also of negligible mass). At time t = 0, container 1 has mass 1.30 kg and container 2 has mass 2.80 kg, but container 1 is losing mass (through a leak) at the constant rate of 0.200 kg/s. At what rate is the acceleration magnitude of the containers changing at (a) t = 0 and (b) t = 3.00 s? (c) When does the acceleration reach its maximum value?

I am not sure how to even start this question. All i can find is the acceleration (3.58 m/s^2). I do not think that it is helpful though. Below are the answers, I just need to know how to get them.
Answer:

(a) 0.653 m/s3; (b) 0.896 m/s3; (c) 6.50 s

Explanation / Answer

m_1 = (1.3 - 0.2*t) kg m_2 = 2.8 kg Acceleration a = g * ((m_1 - m_2) / (m_1 + m_2 )) a(t) = g*((1.3 -0.2*t - 2.8)/(1.3 - 0.2*t + 2.8) a(t) = g*(-1.5-0.2*t)/(4.1- 0.2*t)..............(1) differentiating the above equation with respect to 't' The rate of change of acceleration d(a(t)) / dt = g*(-1.12) / (4.1- 0.2*t)^2 (a) At time t = 0 s d(a(t)) / dt = g*(-1.12) / (4.1- 0.2*0.0 s)^2 d(a(t)) / dt = g*(-1.12) / (4.1- 0.2*0.0 s)^2 d(a) / dt   = 0.653 m /s^3 (b) At time t = 3 s d(a) / dt    =  9.8*(-1.12) / (4.1- 0.2*3)^2 d(a) / dt     = 0.896 m/s^2 (3) maximum acceleation occurs at m_1 = 0 at that a = g is the maximum value

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