Question

Question 13 Water is poured into a container that has a leak. The mass m of the water is given as a function of time t by m = 5.10.8 - 2.7t + 18, with t > 0, m in grams, and t in seconds. (a) At what time is the water mass greatest, and (b) what is that greatest mass? What is the rate of mass change at (c) t = 2.2 s and (d) t = 5.3 s? (a) Number: Unit: (b) Number: Unit: (c) Number: Unit: (d) Number: Unit:

Answer 1

The mass as a function of time is given as

Part A) The mass flow becomes zero at the maximum mass in the container i.e.

d (5.170.8 – 2.7t +18) dt dm dt dm dt - 4.08-0.2 - 2.7

As per condition

dm dt = 0 4.08-0.2 – 2.7 = 0 2-02 2.7 4.08 4.08 2.7 +0.2 t 4.08 2.7 t = 7.88 s.

Part B) Maximum Mass in the container

m = 5.1 x (3.88 s)0.8 – 2.7 x 3.88 s + 18 m = 23.32 g

Part C) Mass flow rate at t = 2.2 s

$\\ \frac{\mathrm{d} m}{\mathrm{d} t} = 4.08\times (2.2 \ \rm s)^{-0.2} -2.7 \\ \\ \frac{\mathrm{d} m}{\mathrm{d} t}= 0.785 \ \rm g/s$

Part D) Mass flow rate at t = 5.3 s

$\\ \frac{\mathrm{d} m}{\mathrm{d} t} = 4.08\times (5.3 \ \rm s)^{-0.2} -2.7 \\ \\ \frac{\mathrm{d} m}{\mathrm{d} t}= 0.223 \ \rm g/s$