Motion in Two Dimension

Motion in two dimensions deals with the study of objects that move in a plane. In such motion, describing the position of a particle using a single coordinate is not sufficient. To completely analyze its motion, quantities like position, velocity, and acceleration, along with their directions, must be considered. These quantities are vector in nature and are conveniently studied using vector algebra and calculus.

Real-Life 2D Motion

Imagine a ball being thrown at an angle θ to the horizontal. The ball moves forward while simultaneously rising and falling under gravity. Its motion is not along a straight line both horizontal and vertical displacements change with time independently.

To analyse such motion completely, we break it into two perpendicular components:

• Horizontal motion is governed by constant velocity if air resistance is neglected.
• Vertical motion is governed by acceleration due to gravity.

This approach helps us calculate quantities like position, velocity, acceleration, and trajectory using vector methods and kinematic equations.

Motion in a Plane

When a particle moves in a plane (xy-plane), its motion is called motion in a plane. To completely describe this motion, the following vector quantities are required.

1. Position
2. Velocity
3. Acceleration

Position Vector

The position vector of a particle is defined as the vector joining the origin of the coordinate system to the position of the particle.

\vec{r} = x\hat{i} + y\hat{j}

Where:

• x and y are the coordinates of the particle.
• \hat{i} \text{ and } \hat{j} → unit vector along the x-axis

Velocity

Velocity is defined as the rate of change of position with respect to time. The velocity of a particle can be described in two ways - average velocity and instantaneous velocity.

Instantaneous Velocity

When velocity changes continuously, we define velocity at a particular instant called instantaneous velocity.

\vec{v} = \lim_{\Delta t \to 0}\frac{\Delta \vec{r}}{\Delta t} = \frac{d\vec{r}}{dt}

Velocity is component form

\vec{v} = \frac{dx}{dt}\hat{i} + \frac{dy}{dt}\hat{j}

Instantaneous velocity gives the speed and direction of motion at a given instant.

Average Velocity

The average velocity is the ratio of total displacement over total time. Suppose a particle goes from \vec{r}  to  \vec{r'}  in a total time of \Delta t

The velocity is given by, 

\vec{v} = \frac{\vec{r} - \vec{r'}}{\Delta t}

Acceleration

The acceleration of a body moving in a plane is given by the rate of change in its velocity. Similar to velocity, there can be two cases here too - average acceleration and instantaneous acceleration. Average acceleration is given by the ratio of the net change in the velocity of the object with the total time taken. Let initial and final velocities be denoted by  \vec{v}_{i} and \vec{v_{f}}

\vec{v} = \frac{\vec{v_{f}} - \vec{v_{i}}}{\Delta t}

Instantaneous acceleration is used when the acceleration of the body is changing with time. 

\vec{a} = \lim_{\Delta t \to 0}\frac{\Delta \vec{v}}{\Delta t}\\ = \vec{a} = \frac{dv}{dt}

This can also be decomposed into its components. 

a = \frac{dv_{x}}{dt}\hat{i} + \frac{dv_{y}}{dt}\hat{j} \\ = a = \frac{d^2x}{dt^2}\hat{i} + \frac{d^2y}{dt^2}\hat{j}

The magnitude of acceleration can also be calculated using the components of velocity, 

a = \sqrt{a_x^2 + a_y^2}

And the direction is given by angle\theta,

\theta = tan^{-1}(\frac{a_y}{a_x})

Kinematic Relations in Two Dimensions

When a particle moves in a plane with constant acceleration its motion can be studied by resolving it into two mutually perpendicular directions usually along the x-axis and y-axis. Each direction behaves like an independent one-dimensional motion. The standard equations of kinematics are applied separately along each axis.

Motion along x-direction

v_x = u_x + a_x t

x = u_x t + \frac{1}{2} a_x t^2

v_x^2 = u_x^2 + 2 a_x x

Motion along y-direction

v_y = u_y + a_y t

y = u_y t + \frac{1}{2} a_y t^2

v_y^2 = u_y^2 + 2 a_y y

Resultant Velocity

The actual velocity of the particle at any instant is obtained by combining the x and y components

v = \sqrt{v_x^2 + v_y^2}

Direction of Motion

The direction of velocity with respect to the x-axis is given by

\theta = \tan^{-1}\!\left(\frac{v_y}{v_x}\right)

Comparison of 1D, 2D and 3D Motion

Sample Problems

Question 1: Find the velocity at t = 2, for the particle which is moving in a plane and whose position is given r = t²i + t²j.

Answer: 

Given: the initial and final position vectors, 

r = t²i + t²j

The position vector changes with time. The velocity in this case is given by the formula, 

 v = \frac{dx}{dt}\hat{i} + \frac{dy}{dt}\hat{j}

Here x(t) = t² and y(t) = t²

Plugging these values into the equation, 

 v = \frac{dx}{dt}\hat{i} + \frac{dy}{dt}\hat{j} \\ v= \frac{d}{dt}(t^2)\hat{i} + \frac{d}{dt}(t^2)\hat{j} \\ v = 2t \hat{i} + 2t\hat{j}

At t = 2, 

v = 2t \hat{i} + 2t\hat{j} \\ = v = 2(2)\hat{i} + 2(2)\hat{j} \\ = v = 4\hat{i} + 4\hat{j}

Question 2: Find the velocity at t = 0, for the particle which is moving in a plane and whose position is given below, r = (t+2)i + (4t²+2)j.

Answer: 

Given: the initial and final position vectors, 

r = (t+2)i + (4t²+2)j

The position vector changes with time. The velocity in this case is given by the formula, 

 v = \frac{dx}{dt}\hat{i} + \frac{dy}{dt}\hat{j}

Here x(t) = t+2 and y(t) = 4t²+2

Plugging these values into the equation, 

 v = \frac{dx}{dt}\hat{i} + \frac{dy}{dt}\hat{j} \\ v= \frac{d}{dt}(t+2)\hat{i} + \frac{d}{dt}(4t^2+2)\hat{j} \\ v = \hat{i} + 8t\hat{j}

at t = 0, 

v = \hat{i} + 8t\hat{j} \\ = v = 4\hat{i} + 8(0)\hat{j} \\ = v = 4\hat{i}

Question 3: Find the average acceleration between t = 0 and t = 3, for the particle which is moving in a plane and whose position is given below, v = 3ti + 3t³j.

Answer: 

Given: velocity as a function of time. 

v = 3ti + 3t³j

The velocity vector changes with time. The average acceleration is given by the formula, 

\vec{a} = \frac{\vec{v_f} - \vec{v_i}}{\Delta t}

At t = 0 

v_i = 0i + 0j 

At t = 3 

v_f = 9i + 81j 

\Delta t = 3

Plugging the values into the above equation, 

\vec{a} = \frac{\vec{v_f} - \vec{v_i}}{\Delta t}\\ = \vec{a} = \frac{9\hat{i} + 81\hat{j}}{3} \\ = \vec{a} = 3\hat{i} + 27\hat{j}

Question 4: Find the average acceleration between t = 0 and t = 2, for the particle which is moving in a plane and whose position is given below, v = ti + 3tj.

Answer: 

Given: velocity as a function of time. 

v = ti + 3tj

The velocity vector changes with time. The average acceleration is given by the formula, 

\vec{a} = \frac{\vec{v_f} - \vec{v_i}}{\Delta t}

At t = 0 

v_i = 0i + 0j 

At t = 2 

v_f = 2i + 6j 

\Delta t = 2

Plugging the values into the above equation, 

\vec{a} = \frac{\vec{v_f} - \vec{v_i}}{\Delta t}\\ = \vec{a} = \frac{2\hat{i} + 6\hat{j}}{2} \\ = \vec{a} = \hat{i} + 3\hat{j}

Question 5: Find the instantaneous acceleration at t = 1, for the particle which is moving in a plane and whose position is given r = ti + tj.

Answer: 

Given: the initial and final position vectors, 

r = ti + tj

The position vector changes with time. The acceleration in terms of position given by, 

a = \frac{d^2x}{dt^2}\hat{i} + \frac{d^2y}{dt^2}\hat{j}

x(t) = t and y(t) = t

Plugging the values into the equation, 

a = \frac{d^2x}{dt^2}\hat{i} + \frac{d^2y}{dt^2}\hat{j} \\ a = \frac{d^2(t)}{dt^2}\hat{i} + \frac{d^2(t)}{dt^2}\hat{j} \\ a = 0\hat{i} + 0\hat{j}

Question 6: A ball is thrown from the ground with a speed of 25 m/s at an angle of 37° to the horizontal then calculate Time of flight, Maximum height and Horizontal range.

Solution:

Given

u = 25 \,\text{m/s}

\sin 37^\circ = 0.6 \quad

\cos 37^\circ = 0.8

g = 10 \,\text{m/s}^2

Components of Velocity

u_x = u \cos\theta = 25 \times 0.8

= 20 \,\text{m/s}

u_y = u \sin\theta = 25 \times 0.6

= 15 \,\text{m/s}

Time of Flight (T) = \frac{2 u_y}{g}

= \frac{2 \times 15}{10}

= 3 \,\text{s}

Maximum Height (H):

= \frac{u_y^2}{2 g}

= \frac{15^2}{2 \times 10}

= \frac{225}{20}

= 11.25 \,\text{m}

Horizontal Range (R):

u_x \times T

= 20 \times 3

= 60 \,\text{m}

Projectile motion 2D Case

Projectile motion is a special case of 2D motion where an object is projected into the air and moves under the influence of gravity only. Its path is parabolic, meaning it rises, reaches a peak, and then falls.

Equations of Motion

x = u \cos\theta \, t

y = u \sin\theta \, t - \frac{1}{2} g t^2

Maximum Height (H)

H = \frac{(u \sin\theta)^2}{2g}

Time of Flight (T)

T = \frac{2u \sin\theta}{g}

Horizontal Range (R)

R = \frac{u^2 \sin 2\theta}{g}

Unsolved Problems

Question 1: Find the instantaneous acceleration at t = 1, for the particle which is moving in a plane and whose position is given r = 3ti + t³j.

Question 2: A body is moving in a 2D plane, its acceleration along the x-axis is 4m/s² and along the y-axis, it is 3m/s². Find the magnitude and the angle of the acceleration on the x-axis.

Question 3: A particle moves in a plane such that its position is given by r = (2t)\,i + (3t)\,j where t is in seconds. Find the velocity of the particle.

Question 4: A particle has velocity components v_x = 5 m/s, v_y = 12 m/s. Find the magnitude and direction of the velocity.

Question 5: A projectile is thrown with an initial speed of 25 m/s at an angle of 37° to the horizontal. Find time of flight and Maximum height (\sin 37^\circ = 0.6,\ \cos 37^\circ = 0.8,\ g = 10\,\text{m/s}^2)

Question 6: A projectile has a time of flight of 6 s and a horizontal range of 90 m.Find Initial speed of projection Angle of projection (g = 10 m/s2).

Question 7: A projectile is projected from the ground such that its maximum height is 20 m and horizontal range is 80 m. Find the initial speed and angle of projection (g = 10 m/s²).