# Mathematical problem is classical question

1. Feb 14, 2009

### dE_logics

From the formula 2as = v2 - u2

I made v the subject...and it becomes -

(2as+u2)1/2 = v

Problem is real world value of 2as is negative...as a result it makes a complex answer...but actually its not; instead the value of v should also come negative.

Here a, u are negative...and a force applies on the body in the same direction as u, which is negative, as a result a negative a cause of that very force.

So the v should also be real and negative...but its coming as complex

Say...can I take the a and u to be positive initially and add the negative sign to the result...sounds ok to me.

2. Feb 14, 2009

### Phrak

$$v=\pm \sqrt{2as+u^2}$$

3. Feb 14, 2009

### atyy

How about the sign of s?

4. Feb 15, 2009

### dE_logics

Yeah its too negative.

But that way the value that comes by is positive, but it should be negative.

Last edited: Feb 15, 2009
5. Feb 15, 2009

### davieddy

See Phrak's response

6. Feb 15, 2009

### ZapperZ

Staff Emeritus
Show this explicitly in a specific problem and we will show you where you forgot another sign somewhere. Both Phrak and atyy have given you sufficient hints.

Zz.

7. Feb 15, 2009

### Staff: Mentor

When 2as is negative that means the body is slowing down. But a negative value of 2as cannot have a magnitude greater than u²--once the body slows to zero it must reverse direction.

8. Feb 16, 2009

### dE_logics

It is accelerating, but all the coordinates are negative, I mean...the u, a and s are all negative.

9. Feb 16, 2009

### dE_logics

Ok then...both negative and positive results will be valid?...I mean...that's true when a value is squared.

10. Feb 16, 2009

### Staff: Mentor

So then what is the issue? 2as is positive, thus v² = u² + 2as is positive. No complex answers required.

(Realize that this equation only gives you the magnitude of v; the sign is up to you.)