Motion in two dimensions: position, velocity and acceleration vectors

Last week the motion was along a line and direction was just a sign. Now we are in the plane: position, velocity and acceleration are real vectors. The good news: there is no new physics. You write the equations of week 3 separately for $x$ and $y$. You combine the components with the vector tools of week 2. This page is the groundwork for next week’s projectile topic.

Golden rules:

1. Every vector equation is two scalar equations: one for $x$, one for $y$.
2. The $x$ and $y$ motions are independent of each other. The only thing that links them is time $t$.
3. Derivatives are taken component by component. $\vec v = \tfrac{dx}{dt}\,\hat i + \tfrac{dy}{dt}\,\hat j$.
4. If speed is asked, take the magnitude; if velocity is asked, give a vector.
5. There can be acceleration even at constant speed. If the direction changes, the velocity vector is changing.

Position and displacement vector

The position of the particle is the position vector drawn from the origin to it:

$\vec r = x\,\hat i + y\,\hat j$

The displacement from time $t_i$ to time $t_f$ (displacement vector):

$\Delta\vec r = \vec r_f - \vec r_i = \Delta x\,\hat i + \Delta y\,\hat j$

The magnitude of the displacement is the straight-line distance between the two points. The distance traveled (along the curve) is greater than or equal to it.

Average and instantaneous velocity

$\bar{\vec v} = \frac{\Delta\vec r}{\Delta t} \qquad \vec v = \frac{d\vec r}{dt} = \frac{dx}{dt}\,\hat i + \frac{dy}{dt}\,\hat j = v_x\,\hat i + v_y\,\hat j$

The average velocity is in the same direction as $\Delta\vec r$. The instantaneous velocity is tangent to the path at every point.

Average and instantaneous acceleration

$\bar{\vec a} = \frac{\Delta\vec v}{\Delta t} \qquad \vec a = \frac{d\vec v}{dt} = \frac{dv_x}{dt}\,\hat i + \frac{dv_y}{dt}\,\hat j$

Acceleration is nonzero in three cases: if the magnitude of the velocity changes, if its direction changes, or both. A car going at a constant 60 km/h around a curve is accelerating.

PITFALL · the difference of speeds. The magnitude of the average acceleration is not $\dfrac{|\vec v_2| - |\vec v_1|}{\Delta t}$ but $\dfrac{|\vec v_2 - \vec v_1|}{\Delta t}$. First subtract the vectors component by component, then take the magnitude. For an object turning at constant speed, the first formula gives zero; the second gives the correct nonzero value.

Example 1. $\vec r(t) = 2t^2\,\hat i + (5t - t^2)\,\hat j$ (m, s). What are the position, velocity, speed and acceleration at $t = 3.0$ s?

r(3)  = 2(9) î + (15 - 9) ĵ        = 18 î + 6 ĵ  m
v(t)  = 4t î + (5 - 2t) ĵ
v(3)  = 12 î - 1 ĵ  m/s
speed = √(144 + 1) = 12.0 m/s
direction = tan⁻¹(-1/12) = -4.8°  (quadrant IV, vx > 0, correct)
a(t)  = 4 î - 2 ĵ  m/s²   (constant)

Here the acceleration came out constant; that is why the constant-acceleration equations below apply to this motion too. If $\vec r(t)$ had a term like $t^3$, the acceleration would depend on time and the constant-acceleration equations could not be used. In that case only differentiation and integration work.

Constant-acceleration 2D motion

If $\vec a$ is constant, the equations of week 3 hold as vectors:

$\vec v = \vec v_0 + \vec a\,t \qquad \vec r = \vec r_0 + \vec v_0\,t + \tfrac12\vec a\,t^2$

Split into components:

Example 2. A particle leaves the origin with $\vec v_0 = (3.0\,\hat i - 2.0\,\hat j)$ m/s, and its acceleration is $\vec a = (0.50\,\hat i + 1.0\,\hat j)\,\text{m/s}^2$. What are the velocity and position at $t = 4.0$ s?

vx = 3.0 + 0.50(4.0) = 5.0 m/s
vy = -2.0 + 1.0(4.0) = 2.0 m/s       →  v = 5.0 î + 2.0 ĵ m/s,  |v| = 5.39 m/s
x  = 3.0(4.0) + ½(0.50)(16) = 12 + 4.0 = 16 m
y  = -2.0(4.0) + ½(1.0)(16) = -8.0 + 8.0 = 0
r  = 16 î  m

Note: at $t = 4.0$ s the object is on the $x$ axis again. $y$ first went negative ($v_{0y} < 0$), then the acceleration brought it back. You see this by looking at the $y$ equation alone; $x$ does not interfere at all.

Maximum $x$ or $y$

A component reaches its largest (or smallest) value at the moment the velocity of that component is zero: $v_x = 0$ for $x_{\max}$, $v_y = 0$ for $y_{\max}$. Find this moment $t$ and put it into the other equations.

TIP · which component? If the question asks “When is the particle farthest east”, it is $v_x = 0$. If it asks “Highest point”, it is $v_y = 0$. In that case the other component is not zero, so the object does not stop.

Independence of motions

Galileo’s observation: an object with a horizontal velocity and an object released at the same time hit the ground at the same time. The vertical motion is not affected by the horizontal velocity. Its mathematical counterpart is that the $x$ and $y$ equations can be solved separately.

Problem-solving method:

1. Choose an axis and an origin, and set the positive directions.
2. Resolve $\vec v_0$ and $\vec a$ into components.
3. Find the shared unknown (usually $t$) from one axis.
4. Put that $t$ into the equation of the other axis.
5. Combine the result as a vector: magnitude and direction.

Common mistakes