chapter 1

Question 1

vectors

Answer 1

quantities that incorporate both magnitude and direction

Question 2

scalars

Answer 2

quantities that only refer to magnitude

Question 3

pythagorean theorem

Answer 3

a²+ b² = c²

Question 4

let’s say we have a graph with points (3,4)…

Answer 4

to find the magnitude of this vector use the pythagorean theorem to find the the direction of this vector use trigonometry for the x and y components V_y= vsin(theta) V_x = vcos(theta) where theta is the angle formed with the x-axis

Question 5

adding vectors

Answer 5

add their x and y coordinates (X1 + y1, X2 + y2) to give new coordinate 1. take a vector (doesn’t matter which) and add its tail to the head of the other vector 2. draw a vector connecting the tail of the first vector with the head of the second vector

Question 6

multiplying vectors example

Answer 6

imagine that a ball is moving at a velocity of 10 m/s at an angle of 60 degrees to the ground. how far will it travel horizontally in 2 seconds?

1. determine the ball’s horizontal velocity (x component)

10 m/s x cos(60) = 10 m/s x 0.5 = 5 m/s in the x direction

2. solve for distance using distance = velocity x time

distance = 5 m/s x 2 s = 10 m

Question 7

displacement

Answer 7

vector quantitiy referring to change in location (distance between where something starts and where it ends up, regardless of the path it took to get there)

scalar equivalent is distance

Question 8

velocity

Answer 8

vector quanitity describing change in displacement over time with common units of m/s or miles/hour

scalar equivalent is speed

v = d/t

Question 9

acceleration

Answer 9

vector quanitity referring to change in velocity over time (m/s²)

Question 10

positive velocity and positive accelelration

Answer 10

object is moving upwards faster and faster

Question 11

positive velocity and no acceleration

Answer 11

object is moving upwards but slowing down

Question 12

positive velocity but negative acceleration

Answer 12

object is moving upwards but slowing down

Question 13

zero velocity and positive acceleration

Answer 13

object is just beginning to move upwards from rest, or is instantaneously transitioning from downwards to upwards motion

Question 14

zero velocity and zero acceleration

Answer 14

object is stationary

Question 15

zero velocity and negative acceleration

Answer 15

object is just beginning to move downwards from rest or is instantaneously transitioning from upwards to downwards motion

Question 16

negative velocity and positive acceleration

Answer 16

object is moving downwards but slowing down

Question 17

negative velocity and zero acceleration

Answer 17

object is moving downwards at a constant velocity

Question 18

negative velocity and negative acceleration

Answer 18

object is moving downwards faster and faster

Question 19

linear graph of displacement versus time

Answer 19

the slope of the line is equal to velocity

Question 20

linear graph of velocity over time

Answer 20

the slope of the line is acceleration

Question 21

area under the curve of an acceleration versus time graph is equal to the

Answer 21

change in velocity

Question 22

area under the curve of a grpah of velocity versus time is

Answer 22

displacement

Question 23

to calculate area under the curve

Answer 23

use base x height for rectangle and b x h/2 for triangle and add the pieces

Question 24

displacment vs time graph summary

Answer 24

slope = velocity

slope is positive and linear: object is moving in (+) direction at a constant rate

slope is zero: object is stationary

slope is negative: object is moving in (-) direction at a constant rate

area under the curve: not meaningful

Question 25

velocity vs time graph summary

Answer 25

slope = acceleration

slope is positive and linear: velocity is increasing (becoming more positive or less negative)

slope is zero: object is moving at a constant velocity

slope is negative: velocity is decreasing (becoming less positive and more negative)

area under the curve: displacement

Question 26

acceleration vs time graph

Answer 26

slope = no meaning

slope is positive and linear: acceleration is increasing (becoming more positive or less negative)

slope is zero: object is speeding up or slowing down at a constant rate

slope is negative: acceleration is decreasing (becoming less positive or more negative)

area under the curve: change in velocity

Question 27

kinematics equation missing acceleration

Answer 27

d = 1/2 (vi + vf)t

Question 28

kinematics equation missing displacement

Answer 28

vf = vi + at

Question 29

kinematics equation missing final velocity

Answer 29

d = vi•t + 1/2 at²

Question 30

kinematics equation missing time

Answer 30

vf² = vi² + 2ad

Question 31

free fall

Answer 31

occurs when something is falling under constant acceleration from gravity (g) -9.8 m/s² without any air resistance

the simplest case occurs when something is dropped so only motion in the y axis needs to be analyzed so vi = 0 m/s, yi = 0 m/s a = g and the convention that downward motion is negative in the y direction

Question 32

simple free fall equation without acceleration variable

Answer 32

y = (vf/2)t

Question 33

simple free fall equation without displacement variable

Answer 33

vf = gt

Question 34

simple free fall equation without final velocity variable

Answer 34

t = square root of (2y/g)

Question 35

simple free fall equation without time variable

Answer 35

v_f² = 2gy

Question 36

projectile motion (no air resistance)

Answer 36

involves horizontal motion and the initial motion in the y-direction is upward

1. break up the initial velocity into its x and y components
2. since V_x is constant, x = V_xt
3. If the projectile is launched at the same height it lands on height is given by h = h_i + v_yi²/2g, solve for t and multiply by two using tlaunch to peak = v_yi/-g
4. if the projectile is launched and there is a difference in heights, determine time from launch to peak and then from peak to ground

time from launch to peak (t = v_yi/-g) and time from peak to ground is t = square root of 2y/g then add the two times together to solve for distance travelled by the projectile