Question

a dump truck weighs 11.25 tons. a conveyor belt pours sand into the truck at a constant rate of 1/4 ton per minute. the dump truck weighs 18 tons when filled. at the same time the dump truck is being filled, an identical truck filled to capacity is being emptied at a rate of 1/8 ton per minute. how much sand is in each dump truck when the trucks are the same weight?

Answer 1

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:

$S = W_{1B} - W_{1A} = 18-11.25 = 6.75Tons$

Let's analyze Truck 1:

Given that the rate of change is:

$\frac{1}{4} ton/min$ 

This means that in 1min the amount of sound poured is 1/4Ton, then the time elapsed when filled using rule of three is:

$t = \frac{6.75}{ \frac{1}{4} } = 27min$

A similar analysis happens with Truck 2: 

Given that the rate of change is:

$\frac{1}{8} ton/min$ 

Then:

$t = \frac{6.75}{ \frac{1}{8}} = 54min$

Now we can find two straight lines:

Truck 1:

$\left \{ {{t=0min} \atop {S=0Ton}} \right.$

$\left \{ {{t=27min} \atop {S=6,75Ton} \right$

$S-S_{0} = m(t-t_{0})$

$S-0 = \frac{6.75-0}{27-0} (t-0)$

∴ $S = 0.25t$

Truck 2:

$\left \{ {{t=0min} \atop {S=6.75Ton}} \right$

$\left \{ {{t=54min} \atop {S=0Ton}} \right$

$S-S_{0} = m(t-t_{0})$

$S-6.75 = \frac{6.75-0}{0-54} (t-0)$

∴ $S = -0.125t+6.75$

The trucks are the same weight when they have the same amount of sand, so solving for t:

$0.25t = -0.125t+6.75$
$0.375t = 6.75$
$t = 18min$

Finally, the amount of sand in each truck:

$S = 0.25(18) = 4.5Ton$