“(1) A toy truck travels at a rate of 0.5t ft/sec after t seconds of travel (so after 6 seconds the truck is traveling 3 ft/sec). “
why is it going suddenly slower, you said it was going at .5 ft/sec.
(1) A toy truck travels at a rate of 0.5t ft/sec after t seconds of travel (so after 6 seconds the truck is traveling 3 ft/sec).
a) Graph the speed of the truck vs. time for 0<t<10.
b) Find the distance the truck has traveled in the first t seconds for 0<t<10. Call it T(t).
c) Find T'(5)
(2) A micorgram of Magnesium 28 contains approximately 3*10^16 atoms and this isotope has a half life of 21 hours.
a) Express the number of atoms of Magnesium 28 as a function of t, time in hours.
b) Approximately how many atoms of Magnesium 28 decay in the one second after 5 hours have passed? (Hint: what is the matematical relation for the decay as related to the amount?)
3 answers
This is really a physics problem poorly expressed. I suspect it is accelerating.
v = Vo + a t
Vo = 0
v = a t = .5 t
v versus t is a straight line through the origin with slope 1/2
T(t) = (1/2) a t^2
= .25 t^2
quadratic with slope zero at origin and going through (10, 25)
I assume T' is the time derivative of T, or the speed
T'(5) = v(5) = 2.5 ft/s
v = Vo + a t
Vo = 0
v = a t = .5 t
v versus t is a straight line through the origin with slope 1/2
T(t) = (1/2) a t^2
= .25 t^2
quadratic with slope zero at origin and going through (10, 25)
I assume T' is the time derivative of T, or the speed
T'(5) = v(5) = 2.5 ft/s
starts with 3 * 10^16 atoms
Meaning of half life
for an exponential decay in time, some constant k
mass = initial mass e^-kt
or
M = Mo e^-kt
now when is m = .5 Mo ?
.5 = e^-k T where T is half life
ln .5 = -k T
ln .5 = -k (21 hr)
-.6931 = - 21*k
k = .033
So for this stuff
M = Mo e^-.033 t where t is in hours
here Mo is 3*10^16 so make your graph
Part b
we want d M/dt in seconds when t = 5 hr
dM / dt = -(3*10^16).033 e^- (.033*5)
you can do that
then that is number lost per hour
divide by 3600 to get per second.
Meaning of half life
for an exponential decay in time, some constant k
mass = initial mass e^-kt
or
M = Mo e^-kt
now when is m = .5 Mo ?
.5 = e^-k T where T is half life
ln .5 = -k T
ln .5 = -k (21 hr)
-.6931 = - 21*k
k = .033
So for this stuff
M = Mo e^-.033 t where t is in hours
here Mo is 3*10^16 so make your graph
Part b
we want d M/dt in seconds when t = 5 hr
dM / dt = -(3*10^16).033 e^- (.033*5)
you can do that
then that is number lost per hour
divide by 3600 to get per second.
0 answers