An elevator is initially moving upward at a speed of 12.00m/s. The elevator experiences a constant downward acceleration of magnitude 4.00m/s^2 for 3.00s.(a) Find the magnitude and direction of the elevators final velocity. (B) how far did it move during the 3.00s interval
Help please
(A) The final velocity is [tex]12\ m/s\ up+4\ m/s^{2}\ down*3\ s[/tex] [tex]=12\ m/s-4\ m/s^{2}*3\ s[/tex] [tex]=12\ m/s-12\ m/s[/tex] [tex]=0[/tex] So the elevator has final velocity 0 (and no direction).
(B) The distance moved is [tex]d=v_{i}t+\frac{1}{2}at^{2}[/tex] where [tex]v_{i}=[/tex] the initial velocity, [tex]a=[/tex] acceleration, and [tex]t=[/tex] time.
So we get [tex]d=12\ m/s*3\ s-\frac{1}{2}(4\ m/s^{2})(3\ s)^{2}[/tex] [tex]=36-2*9[/tex] [tex]=36-18[/tex] [tex]=18\ m[/tex] so the elevator moved 18 meters upwards.
