Chapter 4: Movement in 1D and 2D

Question 1

Week 2

What is displacement?

Answer 1

Displacement is a term used to describe how far and in what direction an object has moved from its initial position to its final position. It is a vector.

The displacement between two points is described as the shortest distance between two points irrespective of the path taken.

For instance, let’s assume that you want to walk to the tree nearest to your house, you walk 3 meters to the right, and 4 meters forward.

• The distance you traveled will be equal to 7 meters, because that is how much you moved in physical space to get to your destination.
• However, your displacement will be equal to 5 meters, since that is the length of shorest path between your house and the tree. Keep in mind that you moved in the direction of the tree since displacement is a vector and needs both a magnitude and a direction.

Displacement Formulae:

x = x₀ + vt
x = x₀t + 1/2 at²

Question 2

Week 2

What is velocity?

Answer 2

Velocity is the rate of change of displacement.

Velocity is a vector. It is represented using the following equation:

v = dx/dt

Question 3

Week 2

What is acceleration?

Answer 3

Acceleration is the rate of change of velocity.

Acceleration is a vector. It is represented using the following equation:

a = dv/dt

Question 4

Week 2

What condition is true for all objects traveling in projectile motion?

Answer 4

All projectile motion calculations assume that the object travels along a flat surface, which means that, in order for the projectile motion equations to work properly, we must assume that the earth is flat (which it pretty much is at small distances).

Question 5

Week 2

What trajectory do all objects moving in projectile motion have?

Answer 5

A parabolic trajectory.

Question 6

Week 2

At what angle do objects traveling in projectile motion travel the farthest?

Answer 6

At an angle of 45°

Question 7

Week 2

List all projectile motion equations.

Answer 7

All projectiles move in two dimensions at this level: vertical and horizontal.

Horizontal Motion Equations

• v_x = v₀cosθ
• Δx = v_0xt
• Δx = u²/g * sin(2θ) (only use this equation if Δy is 0)

Vertical Motion Equations

• v_y = v₀sinθ - gt
• Δy = v_0yt - 1/2gt²

Question 8

Week 2

How can we relate the projectile motion equations back to the equation of a parabola?

Answer 8

The equation of a parabola is as follows:

y = ax² + bx

If we rearrange the equation for horizontal displacement in terms of t, we obtain the following relation:

t = Δx/v_0x

If we then substitute this expression for t into our equation for horizontal displacement, we obtain the following relation:

Δy = v_0y(Δx/v_0x) - 1/2g(Δx/v_0x)²

Once simplified, we can see a relation that appears to be in the form we wanted it to be in.

Δy = (v_y/v_x)Δx - g/2(v_x)²(Δx)²

Further expansion yields the final form, which successfully forms a link between the projectile motion equations and the equation of a parabola.

Δy = -g/2v₀²cos²θ (Δx)² + tanθ (Δx)