# I need guidance

1. Apr 13, 2004

### franz32

I need guidance (updated)

The website has changed a lot! Very beautiful and attactive! I like it.

Well, I need help. This is on integral calculus . How do I find an equation involving the velocity "v" and the distance "s" given that the acceleration "a" is 800; and that v = 20 when s = 1?

Last edited: Jun 19, 2004
2. Apr 13, 2004

### philosophking

Velocity(v) is the integral of the acceleration function. Position(s) is the integral of the velocity function. The variable that you're integrating with respect to is t. I hope this helps you, it's pretty easy from there =P

3. Apr 13, 2004

s(t) = position
s'(t) = v(t) = velocity
s''(t) = v'(t) = a(t) = acceleration

(EDIT: Sorry...See below...I made a mistake).

Last edited: Apr 13, 2004
4. Apr 13, 2004

### philosophking

Yes, you have to say v(t) = 20 and s(t) = 1

5. Apr 13, 2004

### philosophking

After trying this problem for quite some time, I have realized that in its present wording, it is impossible to solve :)

6. Apr 14, 2004

### HallsofIvy

Right. There will be two unknowns and only one additional condition.

Since the acceleration is a constant 800, the velocity at any time t is
v(t)= 800t+ v0 where v0 is the (unkown) velocity at time t=0. The position is s(t)= 400t2+ v0t+ s0 where s0 is the (unknown) position at time t= 0.

Knowing that v(t)= 20 and s(t)= 1 for some t allows us to reduce to only one unknown but not get rid of both. If we knew what that "t" was, then we could answer this question.

7. Apr 15, 2004

### Nexus[Free-DC]

Wouldn't the sensible thing be to add a disclaimer like, "...assuming the starting point to be zero, we find that..."

8. Apr 27, 2004

### franz32

Thank you for all of ur helps. =)

9. Jun 19, 2004

### franz32

Oh... I think there's another way one can solve it:

a = dv/dt = dv/ds X ds/dt = v X dv/ds