Question

Question #e6b1e

Answer 1

See below.

Explanation:

In order to solve this, we must find a function for the volume.

You are given:

`(dV)/(dt)=-8(25-t)`

This is the derivative of some volume function, it is the function itself we require, so we must integrate this to get back the original function.

`-8(25-t)=-200+8t`

`int(-200+8t) dt=-200t+4t^2+c`

We now need to find the value of `c`.

We are told that at the start the bucket contained 2.5 litres, this is equivalent to 2500 `cm^3`. This is at time `t=0`

`:.`

`2500=-200(0)+(0)^2+c`

So `c=2500`

Our equation is now:

`V=-200t+8t^2+2500`

We need to find the value of `t` when the bucket is empty so:

`-200t+8t^2+2500=0`

`t=25` and `t=25`

I won't include the steps to solve this, as I'm sure you already know them.

So the time for the bucket to empty is:

`25color(white)()` seconds.

Answer 2

`v=4t^2-200t+2500`
`t=25" s"`

Explanation:

It is given that

`(dv)/(dt)=-8(25-t)`

Where `v` is volume function.
Integrating both sides with respect to time `t`

`int(dv)/(dt)dt=int(-8(25-t))dt`
`=>v=int(-200+8t)dt`
`=>v=-200t+4t^2+C` ......(1)
where `C` is constant of integration.

To find the value of `C` we make use initial condition. Given that at in the beginning bucket contained 2.5 litres of water. At `t=0`
`v=2.5l= 2500 cm^3`.
Inserting these values in (1) we get

`2500=-200xx0+(0)^2+C`
`=>C=2500`

Now (1) becomes

`v=-200t+4t^2+2500`

rewriting it as

`v=4t^2-200t+2500` .....(2)

When the bucket is empty `v=0`. Imposing the condition

`0=4t^2-200t+2500`
`=>t^2-50t+625=0`

This quadratic is a perfect square.

`=>(t-25)^2=0`

Both roots are equal. Therefore, we have

`t=25" s"`