I'm trying to solve these accelerated motion equations for velocity but I'm getting strange solutions with lower final velocities when using higher initial velocities, which doesn't make sense to me since acceleration is positive. I can't figure out what the problem is:
s=1/2*at²+v0t+s0
[x=( -b ± sqrt(b² - 4ac) ) /2a] 0=(1/2*a)t²+(v0)t+(s0-s) a-> (1/2*a) b-> (v0) c-> (s0-s) x-> t
t=( -(v0) ± sqrt((v0)² - 4*(1/2*a)(s0-s)) ) /2(1/2*a) t=( -v0 ± sqrt(v0² - 2a*(s0-s)) ) /a t=( -v0 ± sqrt(v0² - (2as0-2as) ) ) /a t=( -v0 ± sqrt(v0² - 2as0 + 2as ) ) /a
Case 1: Given: a=10 s0=0 s=10 V0=10
Since v=at, t=v/a, therefore I take it that: t=( -v0 ± sqrt(v0² - 2as0 + 2as ) ) /a transforms to v = -v0 ± sqrt(v0² - 2as0 + 2as ) to solve for velocity.
v = -10 ± sqrt(100 + 200) = -10 ± sqrt(300) = {-10+17.32=7.32 , -10-17.32=-27.32}
Since given acceleration and initial velocity are positive, I take it that the right answer is: 7.32 (Plus time definitely should not be negative). Although it doesn't make sense that final velocity is lower than initial velocity. My gut says the right answer should be -27.32 with the sign inverted. But I can't just take that for granted because something feels wrong about all this.
Case 2: Given: a=10 s0=0 s=10 V0=20
Since v=at, t=v/a, therefore I take it that: t=( -v0 ± sqrt(v0² - 2as0 + 2as ) ) /a transforms to v = -v0 ± sqrt(v0² - 2as0 + 2as ) to solve for velocity.
v = -20 ± sqrt(400 + 200) = -20 ± sqrt(600) = {-20+24.49=4.49 , -20-24.49=-44.49}
Again it doesn't make sense on its own. But compared to case 1 it even makes less sense to get a lower final velocity for a higher initial velocity than in case 1.
I'm sure I'm doing something terribly stupid, that's why I tried to be as clear as possible in my explanations. Please help me out.
The flaw in your calculations is that you used the relation $v=at$ erroneously. This relation is correct only when the initial velocity is zero. If we have an initial velocity $v_0$, the formula becomes $v=at +v_0$. As a result, in your examples you get that the term $-v_0$ is canceled out and the final velocity is directly given by the radical $\sqrt{v_0^2 - 2as_0 + 2as }$. The final velocities are then $17.32$ and $24.49$, respectively.