Chapter 3 Lecture 2

Two Dimensional Motion Variables

Motion variables are position, displacement, velocity and acceleration.

Position

The position vector of a particle is defined to be the vector whose tail is at the origin of the coordinate plane and whose head is at the location of the particle. Position vector is customarily represented by r. If the Cartesian coordinate of the particle is (x, y), then the x-component of the position vector is equal to x and the y-component of the position vector is equal to y.

r_x = x

r_y = y

Example: A particle is located at a point whose ordered pair is (-8, 6) m.

Calculate the x and y components of its position vector.

Solution: (x, y) = (-8, 6) m; x = -8 m; y = 6 m; r_x = ? ; r_y = ?

r_x = x = -8 m

r_y = y = 6 m

Calculate the magnitude and the direction of its position vector.

Solution: x = -8 m; y = 6 m; r = ? ; θ = ?

r = √(x² + y²) = √{(-8)² + 6²} m = 10 m

θ = arctan (y/x) = arctan (6/-8) = 143°

Displacement

Displacement is defined to be change in the position vector of a particle. Graphically, a displacement vector is a vector whose tail is the initial location of the particle and whose head is at the final location of the particle. Displacement is customarily represented by Δr. If r_i is the initial position vector and r_f is the final position vector, then the displacement vector is given by:

Δr = r_f - r_i

The horizontal (vertical) component of a displacement is equal to the difference between the horizontal (vertical) components of the final and initial position vector.

Δr_x = Δx = x_f - x_i

Δr_y = Δy = y_f - y_i

Example: A particle is displaced from the point (4, -2) m to the point (8, -6) m.

Calculate the horizontal and vertical components of its displacement vectors.
Solution: x_i = 4 m; y_i = -2 m; x_f = 8 m; y_f = -6 m; Δr_x = ? ; Δr_y = ?
Δr_x = x_f - x_i = (8 - 4) m = 4 m
Δr_y = y_f - y_i = (-6 -(-2)) m = -4 m

Calculate the magnitude and direction of the displacement vector.
Solution:
Δr = √(Δx² + Δy²) = √{4² + (-4)²} m ≈ 5.66
θ = arctan (Δy/Δx) = arctan (-4/4) = -45°

Average Velocity

Average velocity is defined to be displacement per a unit time.

v_av = Δr/Δt

The horizontal (vertical) component of average velocity is equal to the horizontal (vertical) component of displacement vector per a unit time.

v_avx = Δx/Δt

v_avy = Δy/Δt

Example: A particle is displaced from the point (3, 6) m to the point (-5, 6) m in 2 seconds. Calculate the horizontal and vertical components of its average velocity.

Solution: x_i = 3 m; y_i = 6 m; x_f = -5 m; y_f = 6 m; Δt = 2 s; v_avx = ? ; v_avy = ?

v_avx = (x_f - x_i)/Δt = (-5 - 3)/2 m/s = -4 m/s.

v_avy = (y_f - y_i)/Δt = (6 - 6)/2 m/s = 0 m/s.

Instantaneous Velocity

Instantaneous velocity is velocity at a given instant of time. Or, it is average velocity evaluated at a very very small interval of time.

Average Acceleration

Average Acceleration is defined to be change in velocity per a unit time.

a_av = Δv/Δt = (v_f - v_i )/Δt

Instantaneous Acceleration

Instantaneous acceleration is acceleration at a given instant of time. It is average acceleration evaluated at a very very small interval of time.

Uniformly Accelerated Motion

Uniformly accelerated motion is motion with constant acceleration. The horizontal and vertical components of the acceleration are constant. The horizontal and vertical components of the motion are independent of each other in a sense that what goes on horizontally does not affect what goes on vertically or vice versa. Therefore, a two dimensional motion can be decomposed into a one dimensional horizontal motion and a one dimensional vertical motion. The one dimensional equations of a uniformly accelerated motion obtained earlier are applicable to these one dimensional components of a two dimensional motion. A very popular example of a uniformly accelerated two dimensional motion is gravitational motion in a curved path commonly referred as projectile motion.

Projectile Motion

Motion under gravity is a uniformly accelerated motion. The magnitude of gravitational acceleration is 9.8 m/s² and is directed downwards. Thus, the components of gravitational acceleration are:

a_x = 0 m/s²

a_y = g = -9.8 m/s²

The horizontal component of the motion is a non-accelerated motion. The equations that govern this component of the motion are equations of uniform motion. The horizontal component of the velocity remains the same since there is no acceleration in this direction. The horizontal component of the displacement is obtained as the product of the horizontal component of the velocity and time taken. (That is, equation of a uniform motion)

v_fx = v_ix

Δx = v_ixt

The vertical component of the motion is a uniformly accelerated motion with an acceleration g. The equations governing this component of this motion are the already obtained equations of a one dimensional uniformly accelerated motion.

v_fy = v_iy + gt

Δy = v_iyt + (g/2)t²

v_fy² = v_iy² + 2gΔy

Δy = (v_iy + v_fy )(t/2)

Before using these equations, an initial point and a final point should be chosen. t is the time taken for the object to go from the initial point to the final point. Time is always positive. Δx and Δy are the horizontal and vertical components of the displacement vector whose tail is at the initial point and whose head is at the final point. Appropriate signs should be used. Δx is taken to be positive if the final point is to the right of the initial point and negative if the final point is to the left of the initial point. Δy is taken to be positive if the final point is above the initial point and negative if the final point is below the initial point. v_ix and v_iy are the horizontal and vertical components of the velocity at the initial point. v_fx and v_fy are horizontal and vertical components of the velocity at the final point. A horizontal component of a velocity is positive if it points to the right and negative if it points to the left. A vertical component of a velocity is positive if it points up and negative if it points down.

Example: A ball is thrown from a 2 m table horizontally (to the right) with a velocity of 10 m/s.

Calculate the horizontal and vertical components of the initial velocity.
Solution: v_i = 10 m/s. ; Δy = -2 m; θ_i = 0° ; v_ix = ? ; v_iy = ?
v_ix = v_i cos θ_i = 10 m/s cos 0° = 10 m/s
v_iy = v_i sin θ_i = 10 m/s sin 0° = 0 m/s

Calculate the time taken to hit the ground.
Solution: t = ?
Δy = v_iyt + (g/2)t² = (g/2)t²
because v_iy = 0 m/s

Calculate the horizontal and vertical components of its velocity by the time it hits the ground.
Solution: v_fx = ? ; v_fy = ?
v_fx = v_ix = 10 m/s

How far will it fall (horizontally)?

Example: A bullet is fired from the ground with a speed of 1000 m/s making an angle of 37° with the horizontal.

Calculate the horizontal and vertical components of its initial velocity.
Solution: v_i = 1000 m/s; θ_i = 37°; v_ix = ? ; v_iy = ?
v_ix = v_i cos θ_i = (1000 cos 37°) m/s = 800 m/s
v_iy = v_i sin θ_i = (1000 sin 37°) m/s = 600 m/s

Calculate the time taken to reach the maximum height.
Solution: At the maximum height the direction of the velocity at the maximum height is horizontal which means the vertical component of the velocity at the maximum height is zero: v_fy = 0 m/s; t = ?
v_fy = v_iy + gt = 0 m/s

Calculate the maximum height reached.
Solution: Δy = ?

How far will it fall (horizontally)?
Solution: The time taken to return to the ground is twice the time taken to reach the maximum height: t = 2(61.22) s = 122.44 s; Δx = ?