Chapter 3

Question 1

dimension (units)

Answer 1

an abbreviation for the quantity that us being measured. for example if we are measuring a length it’ll just be L, regardless of what unit we use (metres, km, etc), and for speed L/T (length/time).
*pg40

Question 2

Kinematics

Answer 2

The description of motion in terms of an object’s position, velocity, and acceleration.
*pg42

Question 3

Displacemenr

Answer 3

The displacement of an object is its change in position. Calculated by final position - initial position. It is a vector. it gives us the net distance plus the direction. The most common symbol is ‘d’, but Δx (for horizontal) Δy (for vertical) can also be used
*pg42

Question 4

Velocity

Answer 4

How fast an object’s position changes. To calculate: divide the displacement by time. This is known as the average velocity. It is a vector
*pg44

Question 5

Speed

Answer 5

The magnitude of the velocity vector. it is a scalar quantity, no direction and can never be negative.
*pg45

Question 6

Acceleration

Answer 6

Tells us how fast an object’s velocity changes. The acceleration can be changing even if the speed is constant but if velocity is constant then acc is 0 as long as the direction isnt changing.
change in vel divided by time.
If the acc points in the same direction as the initial velocity, then the object’s speed is increasing, and vice verse.
*pg46/48

Question 7

Uniformly accelerated motion and the equations

Answer 7

This is motion in which the object's acceleration, a, is constant. 
d = 1/2(v0 + v)t
v = v0 +at
d = v0t + 1/2at^2
d = vt + 1/2at^2
v^2 = v0^2 + 2ad
*pg51

Question 8

Position vs time graph and velocity vs time graoh

Answer 8

The slope of position vs time gives you the velocity.
the slope of velocity vs time gives the acceleration. The area under this graph gives the displacement.
*pg53

Question 9

Projectile motion

Answer 9

the motion of an object experiencing only the constant, downward acceleration due to gravity. Also a case if uniformly accelerated motion. This experiences both horizontal and vertical motion. This is in a parabola shape. At the top of the parabola, the vertical velocity is 0
*pg60

Question 10

Projectile motion equations

Answer 10

displacement
horizontal: x = (v_ox)t
vertical: y = (v_oy)t + 1/2(-g)t^2
velocity
horizontal: v_x = (v_ox) (constant)
vertical: v_y = (v_oy) + (-g)t
acceleration
horizontal: a_x = 0; v_0x = v_o cos θ_o
vertical: a_y = -g ; v_0y = v_o sin θ_o
*pg61