Verified answer

dm/dt=-cm, where m is the mass remaining at time t and c is the decay constant.

We have ∫dm/m=-∫cdt→ln(m)=-ct+k’ (k’ the arbitrary constant)

→m=ke^(-ct) (k=e^k’). Now m(0)=200→k=200→m(t)=200e^(-ct).

m(40)=100→100=200e^(-40c) →c=(ln2/40)/s. So m(60)=

200e^[{-(ln2)/40}x60)}=70.7g, so mass decayed =129.3g ((a)).

We want t for 200e^(-ct)=1→t=305.8→day 306 is the first day on which

m<1. ((b)).

let

P = amount present initially

then

dP/dt = rate of decay

dP/dt = -kP

dP/P = -kdt

integrate both sides,

ln P = -kt + c

P = e^(-kt+c)

P = Ce^(-kt)

if t = 0, then

P = Ce^0 = C;

P = C

then

a.) not sure what you mean by mass of substance decay

P = Ce^(-kt)

P = 200e^(-kt)

if t = 40 days

1/2 ( 200 ) = 200 e^(-k(40))

1/2 = e^(-40k)

ln (1/2) = -40k

k = .01732868

after 60 days, if you mean the amount of substance remaining

P = 200e^(-.01732868t)

P = 200e^[-.01732868(60)]

P = 70.71 g

the amount that decayed is

200 - 70.71 = 129.29 g

b. 1 = 200e^(-.01732868t)

ln (1/200 ) = -.01732868t

t = 305.75 = 306 days

let M = Moe^(rt)

M = Mass (g)

Mo = initial mass (g)

r = some constant

t = time (days)

We know:

100 = 200e^(40r) ie Ln(1/2) = 40r

r = -0.0173 (4dp)

we have:

M = Moe^(-0.0173t)

a) M = 200e^(-0.0173*60) = 70.8g (1dp)

b) 1 = 200e^(-0.0173t) hence t = 306.3 (1dp) hence answer is 307th day.