PHYS 200 Unit 2: Kinematics

Question	Answer
position	location of object at particular time
point of reference	used to define the position of a body precisely
distance	magnitude of displacement between two positions
distance traveled	total length of path traveled between two positions
displacement	-change in position of object - ∆x= x_f - x_0
average velocity	-displacement divided by elapsed time -v = ∆x/∆t -vector
instantaneous velocity	velocity at specific instant
instantaneous speed	-distance traveled divided by elapsed time -scalar
average acceleration	-rate at which velocity changes -a = ∆v/∆t -vector
instantaneous acceleration	acceleration at specific instant in time
deceleration	-acceleration in direction opposite to direction of velocity -has opposite sign as velocity
negative acceleration	-acceleration in negative direction
Kinematic Equations Summary	-x = x_0 + vt -v = (v_0 + v)/2 -v = v_0 + at -x = x_0 + v_0 t + 1/2 at^2 -v^2 = v_0^2 + 2a(x - x_0)
free-fall	state of movement that results from gravitational force only
acceleration due to gravity (on Earth)	g = 9.80 m/s^2
Kinematic Equations in Free-Fall	-v = v_0 - gt -y = y_0 + v_0 t - 1/2 gt^2 -v^2 = v_0^2 - 2g(y - y_0)

Resource summary

	Created by Adriana Vincelli-Joma over 4 years ago