# 3. Curvilinear Motion

by M. Bourne

In the last Chapter (in The Derivative as an Instantaneous Rate of Change), we found out how to find the velocity from the displacement function using:

v=(ds)/(dt)

and the acceleration from the velocity function (or displacement function), using:

a=(dv)/dt=(d^2s)/(dt^2)

These formulae are only appropriate for rectilinear motion (i.e. velocity and acceleration in a straight line). This is inadequate for most real situations, so we introduce here the concept of curvilinear motion, where an object is moving in a plane along a specified curved path.

We generally express the x and y components of the motion as functions of time. This form is called parametric form. (See another example using parametric form in Lissajous Figures.)

### Need Graph Paper?

Draw the curve

y(t) = cos t,
x(t) = sin t

for t = 0 to in 0.5 intervals.

First, we need to set up a table of values which we obtain by substituting various values of t:

 t 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 x(t) 0 0.48 0.84 1 0.91 0.6 0.14 -0.35 -0.76 -0.98 -0.96 -0.71 -0.28 y(t) 1 0.88 0.54 0.07 -0.42 -0.8 -0.99 -0.94 -0.65 -0.21 0.28 0.71 0.96

## Horizontal and Vertical Components of Velocity

The horizontal component of the velocity is written:

v_x=(dx)/(dt)

and the vertical component is written:

v_y=(dy)/(dt)

We want to find the magnitude of the resultant velocity v once we know the horizontal and vertical components. We use:

v=sqrt((v_x)^2+(v_y)^2

The direction θ that the object is moving in, is found using:

tan\ theta_v=(v_y)/(v_x

### Example 2

If x = 5t3 and y = 4t2 at time t, find the magnitude and direction of the velocity when t = 10.

### Example 3

If

x=(20t)/(2t+1)

and

y=0.1(t^2+t)

at time t, find the magnitude and direction of the velocity when t = 2. Plot the curve.

## Acceleration of a Body in Curvilinear Motion

The expressions for acceleration are very similar to those for velocity:

Horizontal component of acceleration:

a_x=(dv_x)/(dt)

Vertical component of acceleration:

a_y=(dv_y)/(dt)

Magnitude of acceleration:

a=sqrt((a_x)^2+(a_y)^2)

Direction of acceleration:

tan\ theta_a=(a_y)/(a_x)

### Example 4

A car on a test track goes into a turn described by x = 20 + 0.2t3, y = 20t − 2t2, where x and y are measured in metres and t in seconds.

(i) Sketch the curve for 0 ≤ t ≤ 8.

(ii) Find the acceleration of the car at t = 3.0 seconds.

## What if x and y are NOT given as functions of t?

### Example 5

A particle moves along the path y = x2 + 4x + 2 where units are in centimetres. If the horizontal velocity vx is constant at 3 cm s-1, find the magnitude and direction of the velocity of the particle at the point (-1, -1).

### Example 6

A rocket follows a path given by (distances in km):

y=x-x^3/90

If the horizontal velocity is given by V(x) = x, find the magnitude and direction of the velocity when the rocket hits the ground (assume level terrain) if time is in minutes.

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