Introduction to 2D Motion

Motion in Two Dimensions

In two-dimensional motion, an object moves simultaneously along two perpendicular directions — usually horizontal (x) and vertical (y). Examples include projectile motion, circular motion, and motion of a car turning on a curved road.

The motion is described using position vectors and vector components of displacement, velocity, and acceleration.

Let a particle’s position at any time $t$ be given by: $$\vec{r}(t) = x(t)\hat{i} + y(t)\hat{j}$$

• The velocity is the time derivative: $$\vec{v}(t) = \frac{d\vec{r}}{dt} = \frac{dx}{dt}\hat{i} + \frac{dy}{dt}\hat{j} = v_x\hat{i} + v_y\hat{j}$$

• The acceleration is the derivative of velocity: $$\vec{a}(t) = \frac{d\vec{v}}{dt} = a_x\hat{i} + a_y\hat{j}$$

Visual Example

Below is a trajectory of a particle in 2D motion with velocity vectors:

• Blue curve: path of the object
• Red arrows: instantaneous velocity vectors

This clearly shows how motion in a plane is the result of two independent 1D motions along x and y axes.

• Horizontal motion: uniform
• Vertical motion: uniformly accelerated (due to gravity)

Example 1:

A particle’s position is $\vec{r}(t) = 4t\hat{i} + (5t - 4.9t^2)\hat{j}$. Find velocity and acceleration at $t = 1$ s.

Solution:

• $\vec{v}(t) = \frac{d\vec{r}}{dt} = 4\hat{i} + (5 - 9.8t)\hat{j}$

• At $t=1$: $\vec{v} = 4\hat{i} - 4.8\hat{j}$ m/s

• $\vec{a}(t) = \frac{d\vec{v}}{dt} = -9.8\hat{j}$ m/s² (constant)

  Example 2:

  A ball is thrown with an initial speed of 20 m/s at an angle of 30° above the horizontal from ground level. Assuming no air resistance, find: (a) the time of flight, (b) the maximum height reached, and (c) the horizontal range of the projectile.

  Solution Let's break the problem into steps using the components of motion:

1. Resolve initial velocity into components: The initial speed is 20 m/s at 30°. The horizontal component is $v_{0x} = 20\cos 30° \approx 17.32$ m/s, and the vertical component is $v_{0y} = 20\sin 30° = 10$ m/s.
2. Time of flight (total time in air): The projectile lands at the same vertical level it was launched, so we can use the vertical motion. The vertical displacement after time $T$ is zero (returning to ground): $y - y_0 = v_{0y} T + \frac{1}{2} a_y T^2 = 0$. Here $a_y = -g = -9.8~m/s^2$. Plugging values: $10 T - \frac{1}{2}(9.8)T^2 = 0$. This simplifies to $T(10 - 4.9T) = 0$, yielding a nonzero solution $T = \frac{10}{4.9} \approx 2.04$ s. So the ball is in the air for about 2.04 seconds.
3. Maximum height: At the peak, the vertical velocity becomes zero. We can find the peak height using $v_y^2 = v_{0y}^2 + 2a_y (y - y_0)$. At the top, $v_y = 0$, so $0 = (10)^2 + 2(-9.8)(H - 0)$, where $H$ is the max height above the launch point. Solving, $2(9.8)H = 100$, so $H = \frac{100}{19.6} \approx 5.10$ m. (Alternatively, use $H = \frac{v_{0y}^2}{2g}$.)
4. Horizontal range: The range $R$ is the horizontal distance traveled in the time of flight. Since $v_x$ is constant, $R = v_{0x} \times T$. We have $v_{0x} \approx 17.32$ m/s and $T \approx 2.04$ s, so $R \approx 17.32 \times 2.04 \approx 35.3$ m.

Answer: (a) $\displaystyle 2.04~\text{s}$, (b) $\displaystyle 5.1~\text{m}$, (c) $\displaystyle 35.3~\text{m}$. These results align with the expected behavior: a modest throw stays aloft for about 2 seconds, reaching about 5 m high and covering ~35 m horizontally.