ii). "In the period from 6 am to 8 am the snow plough clears 1 km of road, but
it takes a further 3·5 hours to clear the next kilometre."
therefore it takes 2 hrs to clear first Km, 3.5 hours to clear 2nd Km
Now lets express how much snow is cleared in terms of how long it takes to clear it:
it took 1.5 hours more to clear the snow that built up in the 2nd km, expressing in terms of time: it took 1.5 hours longer to clear 3.5 hours worth of snow
therefore to get the rate of how much 'hours' worth of snow is cleared in 1 hour is:
3.5/1.5 = 2& 1/3
therefore since it took 2 hours to clear teh first km, 2x2&1/3 hours worth of snow must have been cleared, that is 4&2/3 hours
so snow must have started falling 4&2/3 hours before 8am, when it finished teh first kM
THEREFOE IT BEGAN TO SNOW AT 3:20AM
soz couldnt find a way to do it using the result proved in i), but i will continue to try to work it out using i).
EDIT:
found out how to use the result derived in i)., which is how i think examiners would have wanted the solution:
from teh Q: "Let x km be the distance the snow plough has cleared and let t be the time
in hours from the beginning of the snowfall. Let t = T correspond to 6 am."
therefore 8am corresponds to t=(T+2), while the 2nd km is finished clearing at t-(T+5.5)
dx/dt = k/t
integrating WRT t, from t= T, to t= (T+2), which will give us the distance the plough has cleared (1km)
let log = log base e
that is 1=log(T+2)-log(T) = log ((T+2)/T)
however, 1 km was also cleared from t=(T+2) to t=(T+5.5)
that is (integrating WRT t, from t= (T+2), to t= (T+5.5):
1=log(T+5.5)-log(T+2)= log((T+5.5)/(T+2))
since bothe quations = 1, they equate:
log((T+2)/T)=log((T+5.5)/(T+2))
(T+2)/T = (T+5.5)/(T+2)
(cross multiply)
(T+2)(T+2)=T(T+5.5)
T^2 +4T +4 = T^2 +5.5T
4=1.5T
T= 4/1.5 = 2&2/3
we need when t=0, which is 2&2/3 hours before 6AM
therefore t=0 corresponds to 3:20 am
EDIT 2:
save your money, but if your really keen on giving it away, give it to charity!!!
