A rocket begins its third stage of launch at a velocity of 2.25 x 10^2 m/s [fwd]. it undergoes a constant acceleration of 6.25 x 10^1 m/s(sq), while traveling 1.86km, all in the same direction. what is the rockets velocity at the end of this motion?

Question

anonymous · Answer

First you will need to solve the position formula for t.

$$ x = x_0 + vt + \frac{1}{2}at^2 $$

x_0 will be - and x will be 1.86km. Plug the value of t into the velocity formula with v_0 being 2.25e2 m/s.

$$v=v_0 +at $$

anonymous · Answer

so I would solve for t in the first equation using the quadratic formula right? and then whatever positive answer i get just plug it in to the second equation?