# Planar Motion

## Position and Trajectory

Suppose a particle moves in a plane. Then its position can be expressed as

where *x*, *y* are the Cartesian coordinates and **i**, **j** are the unit vectors along the *x*- and *y*- coordinate axes, respectively.

As the coordinates *x*, *y* depend on time, we can write the position vector in parametric form:

This vector equation defines the trajectory of the particle.

## Velocity and Speed

The velocity \(\mathbf{v}\) for a particle is defined as the time derivative of the particle's position vector, that is

In coordinate form, it is given by

where \({v_x} = \frac{{dx}}{{dt}},\) \({v_y} = \frac{{dy}}{{dt}}\) are components of the velocity vector.

The particle's speed \(v\) is a scalar quantity. It is defined as the magnitude of the velocity:

## Acceleration

The acceleration \(\mathbf{a}\) for a particle is defined as the time derivative of the particle's velocity vector:

In case when \(\mathbf{v}\) is written in coordinate form, the acceleration is given by

where \({a_x} = \frac{{d{v_x}}}{{dt}} = \frac{{{d^2}x}}{{d{t^2}}},\) \({a_y} = \frac{{d{v_y}}}{{dt}} = \frac{{{d^2}y}}{{d{t^2}}}\) are components of the acceleration vector.

The magnitude of the acceleration is computed as

## Motion with Constant Acceleration

Suppose a particle or object moves with a constant acceleration \(\mathbf{a}:\)

Then the velocity at time \(t\) is written as

where \({\mathbf{v}_0}\) is the initial velocity vector at \(t = 0.\)

The position at time \(t\) is described by the equation

where \({\mathbf{r}_0}\) is the initial position at \(t = 0.\)

## Free Fall Motion

Free fall is motion of a body with a constant acceleration caused by gravity. Such an object moves with the downward acceleration

where \(g = 9.8\,\frac{\text{m}}{\text{s}^2}\) is the acceleration due to gravity, \(\mathbf{j}\) is a unit vector pointing vertically upward.

The vertical velocity \(v\left( t \right)\) and the position (or altitude) \(y\left( t \right)\) during a free fall are given by the equations

where \({v_0}\) is the initial velocity.

## Solved Problems

### Example 1.

A particle is moving in \(xy-\)plane according to the equations \[x = t, y = {t^3},\] where \(x,y\) are measured in meters. Find the particle's velocity and speed at \(t = 1\,\text{s}.\)

Solution.

From the definition of velocity,

where

Hence, the velocity vector at \(t = 1\) is given by

The speed at this time is equal to

### Example 2.

Assume that an object moves on a trajectory according to the equations \[x\left( t \right) = t + \cos t, y\left( t \right) = t - \sin t.\] Find the magnitude of the acceleration vector.

Solution.

We differentiate \(x\left( t \right)\) and \(y\left( t \right)\) twice to find the acceleration of the object:

Now we can compute the magnitude of the acceleration vector:

### Example 3.

A particle moves along the hyperbola \[y = \frac{{12}}{x},\] so that the \(x-\)coordinate is increasing at a constant rate of \(3\) meters per second. Find the speed of the particle when it is at the point \(\left( {3,4} \right).\)

Solution.

First, we determine the velocity vector of the particle using the formula

The derivative \(\frac{{dx}}{{dt}}\) is known:

Compute the derivative \(\frac{{dy}}{{dt}}\) by the chain rule:

At the given point we have

Now we can find the particle's speed:

### Example 4.

A particle is moving along the curve given by the parametric equations \[x = 1 + t, y = 1 - t.\] Determine the \(xy-\)equation of the trajectory and speed along it.

Solution.

We solve the first equation for \(t\) and plug it into the second equation:

Hence, the particle's trajectory is the straight line \(y = 2 - x.\) Find the velocity vector:

Then

The speed is equal to

The answer is

### Example 5.

A particle moves on a trajectory described by the parametric equations \[x\left( t \right) = {2^t}, y\left( t \right) = {8^t},\] where \(t \ge 0.\) What is the shape of the trajectory?

Solution.

From the first equation, we find:

Substitute this into the second equation to obtain the \(xy-\)equation of the curve:

As \(t \ge 0,\) then \(x \ge {2^0} = 1.\)

Hence, the given curve is the cubic parabola \(y = {x^3}\) where \(x \ge 1.\)

### Example 6.

A particle's position is given by the equations \[x\left( t \right) = \frac{1}{2} - {t^2}, y\left( t \right) = \sqrt 2 t,\] where \(x,y\) are measured in meters. Determine the particle's distance \(d\) from the origin as a function of time \(t.\)

Solution.

The distance from the origin is defined as

Substituting \(x\left( t \right),\) \(y\left( t \right),\) we obtain: