Question

The mass, m grams, of a radioactive substance, present at time t days after first being observed, is given by the formula m=24e^-0.02t. Find
(i) the value of m when t=30.
(ii) the value of t when the mass is half of its value at t=0.
(iii) the rate at which the mass is decreasing when t=50.

Answer 1

Answer:

(i) The value of m when t = 30 is 13.2

(ii) The value of t when the mass is half of its value at t=0 is 34.7

(iii) The rate of the mass when t=50 is -0.18            

Step-by-step explanation:

(i) The m value when t = 30 is:

$m = 24e^{-0.02t} = 24e^{-0.02*30} = 13.2$

Then, the value of m when t = 30 is 13.2

(ii) The value of the mass when t=0 is:

$m_{0} = 24e^{-0.02t} = 24e^{-0.02*0} = 24$    

Now, the value of t is:

$ln(\frac{m_{0}/2}{24}) = -0.02t$

$t = -\frac{ln(\frac{24}{2*24})}{0.02} = 34.7$

Hence, the value of t when the mass is half of its value at t=0 is 34.7

(iii) Finally, the rate at which the mass is decreasing when t=50 is:

$\frac{dm}{dt} = \frac{d}{dt}(24e^{-0.02t}) = 24(e^{-0.02t})*(-0.02) = -0.48* (e^{-0.02*50}) = -0.18$

Therefore, the rate of the mass when t=50 is -0.18.

I hope it helps you!