# Question #e6b1e

##### 2 Answers

See below.

#### Explanation:

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

You are given:

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.

We now need to find the value of

We are told that at the start the bucket contained 2.5 litres, this is equivalent to 2500

So

Our equation is now:

We need to find the value of

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:

#### Explanation:

It is given that

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

Where

Integrating both sides with respect to time

#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

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

#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"#