A train freight car weighs 25 tons when empty. A conveyor belt pours coal into the car at a constant rate of 1/3 ton per minute until it is full. The dump truck (Refer to problem 8) weighs 55 tons when filled. At the same time the freight car is being filled, an identical freight car filled

to capacity is being emptied at a rate of 1/6 ton per minute.

How long, in minutes, until the freight cars are the same weight?

The tucks will weigh the same in

minutes.

How much coal will be in each truck when they weigh the same?

There will be

tons of coal in each truck.

Step-by-step explanation:

Step one:

mass of Freight car=25tons

the rate of filling is 1/3 ton per minute

let the time taken to fill the freight car be x

and the total mass after x minute be y

Hence the mass as a function of time is expressed as

y=25+1/3x1

mas of dump truck=55tons

rate of being emptied= 1/6 tons per minute

let the time taken to empty the dump truck be x

and the total mass after x minute be y

Hence the mass as a function of time is expressed as

y=55-1/6x2

Step two:

equating equations 1 and 2,

1. The tucks will weigh the same in

55-1/6x=25+1/3x

collect like terms

55-25=1/3x+1/6x

30=(2+1/6)x

30=3/6x

30=1/2x

30=0.5x

x=30/0.5

x=60min

or

1 hour

The tucks will weigh the same in  60minutes(1 hour)

2. How much coal will be in each truck when they weigh the same?

For the  train freight car, the weight after 60 minutes is

y=25+1/3(60)

y=25+20

y=55tons

55-25=20tons

There will be 20tons

For the  dump truck, the weight after 60 minutes is

y=55-1/6(60)

y=55-10

y=45tons

55-45=10tons

There will be 10tons

2^y = ln 5, or   y(ln 2) = ln 5.   solving for y,   y = (ln 5) / (ln 2).   y=5

step-by-step explanation:

5 pm

step-by-step explanation: