Finding the Particular Solution for a Velocity Equation with Initial Conditions

[andrey21]

Given that v= 20√10∙ ((1+Ae^(t/√10) ÷ (1-Ae^(t/√10) find the particular solution to satisfy the initial condition v(0) = 0.From my working I have come to the answer A = -1, is this correct? If not where could I be going wrong.
My Working

0 = 20√10 ∙ ((1 + A) ÷ (1 - A))
0 = 20√10 ∙ (1 + A)
0 = 20√10 + 20√10 A
- 20√10 = 20√10 A
A = -1

Homework Statement

Homework Equations

The Attempt at a Solution

 

[HallsofIvy]

Given that v= 20√10∙ ((1+Ae^(t/√10) ÷ (1-Ae^(t/√10) find the particular solution to satisfy the initial condition v(0) = 0.


From my working I have come to the answer A = -1, is this correct? If not where could I be going wrong.

Yes, that is correct.



My Working

0 = 20√10 ∙ ((1 + A) ÷ (1 - A))
0 = 20√10 ∙ (1 + A)[/quote]
It is simpler to divide by $20\sqrt{10}[/math] and get 0= 1+ A immediately<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> 0 = 20√10 + 20√10 A<br /> - 20√10 = 20√10 A <br /> A = -1 </div> </div> </blockquote>$

 

[andrey21]

Thanks and yes I see the easier method of finding A now , oooops!

The question goes on to say:
You now have a solution for velocity, substituting this into y' = v. Use this to fing y(t) assuming that y(0) = 4000.

Heres what I know.

v= 20√10∙ ((1+Ae^(t/√10) ÷ (1-Ae^(t/√10)
y= -1/2 gt^2 + At + B
B = 4000

Where do I go from here? Simply subs A = -1 y= -1/2 gt^2 + At + B
and solve from there?