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"