So I'm answering this question on use of calculus in applied maths:
If $a= 4s+2$ and initially $s=0$ and $v=1$ m/s, find $v$ when $s=3$.
So far I've done:
$$v\frac{dv}{ds} = 4s + 2$$
which means
$$ \int{v} dv = \int (4s+2) ds $$
which I've solved to get
$$\frac{v^2}{2} = 2s^2 +2s +C$$
Then I used the information given that when $s=0$, $v=1$ to get $C= \frac{1}{2}$
I've solved $\frac{v^2}{2} = 2s^2 +2s + \frac{1}{2}$ to get $v^2 = 49$.
The answers says the answer is $7$ - what I can't work out is how we know it's +7 and not -7.
Many TIA
As per my opinion question is about the magnitude of the velocity not about the direction. So your answer should be 7 ( m/s or km/hr what ever). But when you are asked about the direction then you say if it is +7 then it is going towards the positive direction and when -7 it is going backwards.