We know these variables:
W1B = Weight of the truck 1 before starting the filling.
W1A = Weight of the truck 1 when filled.
R1F = Rate of change (Truck 1 being filled)
W2B = Weight of the truck 2 before starting the filling.
W2A = Weight of the truck 2 when filled.
R2F = Rate of change (Truck 2 being filled)
Given that a conveyor belt pours sand into the truck and stops when filled, then the amount of sand, when the truck is filled to capacity, is:
[tex]S = W_{1B} - W_{1A} = 18-11.25 = 6.75Tons[/tex]
Let's analyze Truck 1:
Given that the rate of change is:
[tex] \frac{1}{4} ton/min[/tex]
This means that in 1min the amount of sound poured is 1/4Ton, then the time elapsed when filled using rule of three is:
[tex]t = \frac{6.75}{ \frac{1}{4} } = 27min[/tex]
A similar analysis happens with Truck 2:
Given that the rate of change is:
[tex] \frac{1}{8} ton/min[/tex]
Then:
[tex]t = \frac{6.75}{ \frac{1}{8}} = 54min[/tex]
Now we can find two straight lines:
Truck 1:
[tex]\left \{ {{t=0min} \atop {S=0Ton}} \right. [/tex]
[tex] \left \{ {{t=27min} \atop {S=6,75Ton} \right[/tex]
[tex]S-S_{0} = m(t-t_{0})[/tex]
[tex]S-0 = \frac{6.75-0}{27-0} (t-0)[/tex]
∴ [tex]S = 0.25t[/tex]
Truck 2:
[tex] \left \{ {{t=0min} \atop {S=6.75Ton}} \right[/tex]
[tex] \left \{ {{t=54min} \atop {S=0Ton}} \right[/tex]
[tex]S-S_{0} = m(t-t_{0})[/tex]
[tex]S-6.75 = \frac{6.75-0}{0-54} (t-0)[/tex]
∴ [tex]S = -0.125t+6.75[/tex]
The trucks are the same weight when they have the same amount of sand, so solving for t:
[tex]0.25t = -0.125t+6.75[/tex]
[tex]0.375t = 6.75[/tex]
[tex]t = 18min[/tex]
Finally, the amount of sand in each truck:
[tex]S = 0.25(18) = 4.5Ton [/tex]
