Vector kinematics AI HL

I. Vector Kinematics

(a) Constant velocity

If an object moves with constant velocity ( vec{v} ), then its position at time (t) is

( vec{r} = vec{r}_0 + vec{v}t )

• (vec{r}_0) = initial position vector
• (vec{v}) = velocity vector
• (t) = time

Example: A car starts at (begin{bmatrix}2 \ 1end{bmatrix}) km with velocity (begin{bmatrix}60 \ 0end{bmatrix}) km/h.
After 2 h:

(
vec{r} = begin{bmatrix}2 \ 1end{bmatrix} + 2begin{bmatrix}60 \ 0end{bmatrix}
= begin{bmatrix}122 \ 1end{bmatrix}
)

So the car is 122 km east, 1 km north.

(b) Colliding objects

Two objects collide if their position vectors are equal at the same time.

Example:

(
vec{r}_A = begin{bmatrix}0 \ 0end{bmatrix} + tbegin{bmatrix}2 \ 1end{bmatrix}
= begin{bmatrix}2t \ tend{bmatrix}
)

(
vec{r}_B = begin{bmatrix}10 \ 2end{bmatrix} + tbegin{bmatrix}-1 \ 0end{bmatrix}
= begin{bmatrix}10-t \ 2end{bmatrix}
)

For collision, solve (vec{r}_A=vec{r}_B):
(2t=10-t Rightarrow t=tfrac{10}{3}).
At this time, (y_A=tfrac{10}{3}), (y_B=2). They are not equal → no collision.

II. Variable Velocity: Uniform Acceleration

If acceleration (vec{a}) is constant:

• Velocity: (vec{v}=vec{u}+vec{a}t)
• Position: (vec{r}=vec{r}_0+vec{u}t+tfrac{1}{2}vec{a}t^2)

Example: A particle starts at (vec{r}_0=begin{bmatrix}0\0end{bmatrix}) with initial velocity (vec{u}=begin{bmatrix}4\0end{bmatrix}) m/s and acceleration (vec{a}=begin{bmatrix}0\-9.8end{bmatrix}) m/s².

• After 2 s, velocity:
  (vec{v}=begin{bmatrix}4\0end{bmatrix}+2begin{bmatrix}0\-9.8end{bmatrix}
  = begin{bmatrix}4\-19.6end{bmatrix})
• Position:
  (vec{r}=begin{bmatrix}0\0end{bmatrix}+2begin{bmatrix}4\0end{bmatrix}
  +tfrac{1}{2}begin{bmatrix}0\-9.8end{bmatrix}(2^2)
  = begin{bmatrix}8\-19.6end{bmatrix})

So the object is 8 m across, 19.6 m down.

III. Variable Velocity in Two Dimensions

In 2D, velocity has independent components:

(
vec{v}=begin{bmatrix}v_x \ v_yend{bmatrix}, quad
vec{a}=begin{bmatrix}a_x \ a_yend{bmatrix}
)

Equations of motion:

• (v_x=u_x+a_xt,quad v_y=u_y+a_yt)
• (x=x_0+u_xt+tfrac{1}{2}a_xt^2)
• (y=y_0+u_yt+tfrac{1}{2}a_yt^2)

Projectile motion (special case)

• Horizontal: (a_x=0) → constant horizontal velocity
• Vertical: (a_y=-g) due to gravity

Example: A ball is thrown with initial velocity
(vec{u}=begin{bmatrix}10\15end{bmatrix}) m/s from the origin.

• Horizontal: (x=10t)
• Vertical: (y=15t-4.9t^2)

At (t=2) s:
(x=20,; y=30-19.6=10.4).
So the ball is at (20 m, 10.4 m).