Question #e6b1e

2 Answers
Dec 14, 2017

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.

Dec 14, 2017

#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"#