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2000 q10 part B (1 Viewer)
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crazylilmonkee
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[flips thru paper]
EEEEEEEEEEEEEEEEEEEEEEEEK
nope i skipped it sorry
EEEEEEEEEEEEEEEEEEEEEEEEK
nope i skipped it sorry
N
ND
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Can someone post up the question?
The first snow of the season begins to fall during the night. The depth of the snow, "h", increases at constant rate through thte night and the following day.
At 6am a snow plough begins toc lear the raod of snow. The speed, v km/h of the snow plough is inversely proportional to the depth of the snow.
(This means v= A/h where A is a constant
)
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 6am.
i) Explain carefully why, for t > T, dx/dt = k/t where k is a constant.
ii) In the period from 6am to 8am the snow plough clears 1km of road but it takes a further 3.5 hours to clear the next kilometre.
At what time did it begin snowing?
At 6am a snow plough begins toc lear the raod of snow. The speed, v km/h of the snow plough is inversely proportional to the depth of the snow.
(This means v= A/h where A is a constant
)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 6am.
i) Explain carefully why, for t > T, dx/dt = k/t where k is a constant.
ii) In the period from 6am to 8am the snow plough clears 1km of road but it takes a further 3.5 hours to clear the next kilometre.
At what time did it begin snowing?
ballerinabarbie
bundy for me
ok i'll try
i) the speed of the snow plough is v= dx/dt
and the rate=dh/dt
v=A/h, and dx/dt=A/h
The depth of the snow is h=bt (where b is a constant rate)
therefore, dx/dt=A/bt=A/b/t
dx/dt=k/t
ii) At t=T+2, 1km of road is cleared
therefore, 1=(integral from T to T+2)k/t dt
1=[klnt](from T to T+2)
1=k[ln(T+2) - lnT]
1=k ln(T+2/T)
we use 1 again, because 1km of road is cleared in the next 3.5hrs
1=(integral from T+2 to T+5.5)k/t dt
1=[kln T+5.5/T+2] (because loga/b = log a - log b)
we make this equal to k ln(T+2/T) because they are both equal to 1
so k ln (T+5.5/T+2) = k ln (T+2/T)
cancel the k and the ln
T+5.5/T+2 = T+2/T (cross multiply)
T^2 + 5.5T = T^2 + 4T + 4
1.5T = 4
t= 4/1.5 = 2hr 40min
therefore it starts snowing at 3.20am
hope that helped... its off my maths tutors notes!! the first part is probably not explained very well... but i hope the second part makes sense
i) the speed of the snow plough is v= dx/dt
and the rate=dh/dt
v=A/h, and dx/dt=A/h
The depth of the snow is h=bt (where b is a constant rate)
therefore, dx/dt=A/bt=A/b/t
dx/dt=k/t
ii) At t=T+2, 1km of road is cleared
therefore, 1=(integral from T to T+2)k/t dt
1=[klnt](from T to T+2)
1=k[ln(T+2) - lnT]
1=k ln(T+2/T)
we use 1 again, because 1km of road is cleared in the next 3.5hrs
1=(integral from T+2 to T+5.5)k/t dt
1=[kln T+5.5/T+2] (because loga/b = log a - log b)
we make this equal to k ln(T+2/T) because they are both equal to 1
so k ln (T+5.5/T+2) = k ln (T+2/T)
cancel the k and the ln
T+5.5/T+2 = T+2/T (cross multiply)
T^2 + 5.5T = T^2 + 4T + 4
1.5T = 4
t= 4/1.5 = 2hr 40min
therefore it starts snowing at 3.20am
hope that helped... its off my maths tutors notes!! the first part is probably not explained very well... but i hope the second part makes sense
Can you post that problem here? Thanks
nsw..wollongong
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here
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Sorry I was talking abt 4u nsb math problem that @Aaryan123 mentioned.