Finding v(t) from v=-4s² with s=2 at t=0

[Lancelot59]

I'm given this:
$$v=-4s^{2}$$
s=2
t=0

and I need to find v(t)

I used this differential and got t like so:
$$dt=\frac{v}{ds}$$

$$dt=\frac{ds}{v}$$
$$dt=\frac{ds}{-4s^{2}}$$
$$t=-\frac{1}{4s}+c$$

I plugged in 0 for t and 2 for s. I got -1/8 for c. Then isolating s I got
$$s=-\frac{1}{4t}+2$$
I stuck that into v=-4s^2 and got this:
$$v=-\frac{1}{4t^2}-\frac{4}{t}-16$$
However according to the online system I'm wrong.

 

[tiny-tim]

Hi Lancelot59!

Your brackets are wrong …

it's s = 1/(4t + constant) ​

 

[Lancelot59]

How does that work? I thought constants were always added to the end.

 

[tiny-tim]

Yes, which is where you correctly put it in your line t = 1/4s + C.

But it doesn't stay at the end when you up-end it …

1/4s = t - C,

4s = 1/(t - C)

 

[Lancelot59]

I see, so now I get c=1/8.
I still end up with the same expression for S.

Nevermind, I made an algebra mistake. I have the answer now.

Thanks.