What SI unit and common variables are associated with **length**?

**Length **or **distance** has SI units of **meters (m)**.

The variables d and x are used for distance, h for height, z for depth, and r for radius.

What SI unit and common variables are associated with **time**?

**Time **has SI units of **seconds (s)**.

The variable t is used for time, T is used for period and also has SI units of seconds.

What SI unit and common variables are associated with **area**?

**Area **has SI units of **meters**^{2 }**(m ^{2}).**

The variable A is used for area, S is used for surface area.

What SI unit and common variables are associated with **volume**?

**Volume **has SI units of **meters ^{3 }**

**(m**

^{3}).The variable V is used for volume.

What SI unit and common variables are associated with **velocity**?

**Velocity **or **speed **has SI units of **meters/second (m/s)**.

The variable v is used for velocity (a vector), though rarely used, s may be used for speed (magnitude only, scalar).

What SI unit and common variables are associated with **acceleration**?

**Acceleration **has SI units of **meters/second ^{2 }**

**(m/s**

^{2}).The variable a is used for acceleration (a vector).

What two characteristics are necessary to define a **vector**?

**Magnitude **and **direction**.

**magnitude** of a vector

**Magnitude **is the quantity, size, or amount and is a scalar value since it lacks direction.

**direction **of a vector

**Direction **provides spacial orientation, angle, or path.

By convention, two perpendicular directions are fixed as positive (right and up) and their opposites as negative (left and down).

What is the magnitude of the velocity vector pointing directly down at 5 m/s?

5 m/s.

Magnitude is the quantity, size, or amount.

What is the direction of the velocity vector pointing directly up at 13 m/s?

Directly upwards.

Direction is the angle, orientation, or path.

Describe the general process for **vector addition**.

Vectors are added **"tip to tail"** so that there is a continuous pathway formed from the first arrow through the last one.

The sum vector is made by connecting the original tail to the final tip.

Describe the general process for **vector subtraction**.

Vectors are subtracted in the same process as addition:** "tip to tail"**. The direction of the vector being subtracted must be reversed prior to adding.

**component vectors**

**Component vectors **represent the magnitude of a given vector along the x axis and y axis.

Generally these are referred to as, for instance, the "x component of [vector]". In the image below, vector A is shown with its x and y components as A_{x} and A_{y} respectively.

Give the formula for the x component of a force in the xy plane.

**F _{x} = F * cos θ**

In general: [vector]_{x} = [original vector] * cos θ

What is the x component of velocity for a ball thrown upwards at 4 m/s, 30º to the horizon?

(sin(30º) = .5, cos(30º) = .86)

3.4 m/s

v_{x} = 4*cos (30º) = 4 (.86) = 3.4 m/s

Give the formula for the y component of a force in the xy plane.

**F _{y} = F * sin θ**

In general: [vector]_{y} = [original vector] * sin θ

What is the y component of velocity for a ball thrown upwards at 4 m/s, 30º to the horizon?

(sin(30º) = .5, cos(30º) = .86)

2 m/s

v_{y} = 4*sin (30º) = 4 (.5) = 2 m/s

**instantaneous velocity (v _{o})**.

**Instantaneous velocity** is the direction and magnitude of the rate of change for distance per unit time.

**v _{o} = **

**Δ**

**x / t**

or: **v _{o} = p/m**

Where p is the object's momentum and m is its mass.

Give the instantaneous velocity for a 3kg block with a momentum of 12kg*m/s.

4m/s

From p = mv, 12 = 3(v), v = 4m/s.

**average velocity (v _{ave})**

**Average velocity** is the total distance and direction traveled from an initial position, divided by the total time.

**v _{ave} = Δx_{total} / Δt_{total}**

or: **v _{ave} = (v_{f} + v_{i})/2**

Note: the second equation assumes constant acceleration, which is also assumed on the AP exam unless specified otherwise.

Give the average velocity for a 1kg block after falling from rest for 2 seconds.

9.8 m/s

From: v = (1/2)at^{2}

v_{f }= (1/2) (9.8) (2)^{2} = 2 (9.8)

= 19.6 m/s and given: v_{i} = 0.

From v_{ave} = (v_{f} + v_{i})/2

v_{ave} = (19.6 + 0)/2 = 9.8 m/s

What formulas can be used to find final velocity, if given initial velocity?

**v _{f} = v_{i} + at**

or:** v _{f}^{2} = v_{i}^{2} + 2a**

**Δx**

Where:

v_{f} = final velocity in m/s

v_{i} = initial velocity in m/s

a = acceleration in m/s^{2}

Δx = displacement in m

t = time in s

Give the velocity of a 1kg block after falling for 2 seconds from rest.

19.6 m/s

From: v_{f} = v_{i} + at

v_{f }= 0 + (9.8)(2) = 2(9.8)

= 19.6 m/s

**acceleration**

**Acceleration **is the rate at which velocity changes per unit of time.

**a = Δv / ****t**

or: **a = Δx / t ^{2}**

What must have happened to the value of acceleration, if the same change in distance is now traveled in 1/2 the time?

4x the acceleration.

Since a is proportional to Δx/t^{2}, halving t will quadruple a.

What is the acceleration always associated with an object in free fall?

9.8 m/s^{2}

This is the acceleration due to gravity, commonly called g. On the AP exam, any falling object is assumed to be in free fall (no air resistance) unless told otherwise.

Note: while it's convenient to use 10 m/s^{2} for calculations, the AP exam expects you to still choose the correct answer mathematically.

How is the maximum height of an object in projectile motion found?

The maximum height of an object in projectile motion can be found by finding the point where the vertical component of velocity, v_{y}, equals zero. Then input this value into the equation with distance, to solve.

The object moves upwards until it reaches this point, and then starts moving downwards.

What is the maximum height of a ball thrown directly upwards from the ground with a velocity of 10 m/s?

The ball's maximum height is 5m.

The kinematic equation that most easily solves this question is

v_{f}^{2} = v_{0}^{2} + 2aΔx

Setting v_{0} to 10m/s, v_{f} to 0m/s, and a to -10m/s^{2} yields:

0 = 100 + 2(-10)Δx

Δx = 5m

**air resistance**

**Air resistance** is the frictional force that opposes an object moving in free fall.

**F _{air} = -kv^{2}**

The force of air resistance is proportional to the square of the velocity of the object falling. The constant k is a function of the object's shape and surface area.