Paincake said:
I've seen online a kinematic equation for final velocity, which is
Vf² = Vi² + 2(a)(d),
but my teacher has not introduced it. Where does this equation come from?
That's not the equation you want to use for this problem. Try to find a formula that has
initial velocity (not final velocity), distance, acceleration, and time.
But I'll tell you where that equation comes from anyway, if you're curious. It comes from combining two of the other kinematics equations for uniform acceleration.
Start with
x = x_0 +v_0 t + \frac{1}{2}at^2
Now modify some variables, using
d = x - x0. And instead of calling the initial velocty "
v0", let's call it "
vi" instead. So now we have,
d = v_i t + \frac{1}{2}at^2
Don't forget about that equation, we'll come back to it in a second. But first let's look at a different kinematics equation for uniform acceleration:
v_f = v_i + at
Rearranging that equation, we have
t = \frac{v_f - v_i}{a}
Now let's substitute that into the modified first equation above.
d = v_i \frac{v_f - v_i}{a} + \frac{1}{2} a \left( \frac{v_f - v_i}{a} \right)^2
Expanding a little gives us
d = \frac{v_i v_f - v_i^2}{a} +\frac{1}{2}a \left( \frac{v_f^2 - 2v_f v_i + v_i^2}{a^2} \right)
= \frac{v_i v_f - v_i^2}{a} + \frac{v_f^2 - 2v_f v_i + v_i^2}{2a}
Multiplying both sides of the equation by 2
a gives,
2ad = 2v_i v_f - 2v_i^2 + v_f^2 - 2 v_i v_f + v_i^2.
And simplifying the right side of the equation produces
2ad = v_f^2 - v_i^2.
Adding
vi2 to both sides gives
v_i^2 + 2ad = v_f^2.