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:

math image

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

math image

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.)

 

Example 1: Parametric Equations

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rectangular grid
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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 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0
x(t) 0 .48 .84 1.0 .91 .60 .14 -.35 -.76 -.98 -.96 -.71 -.28
y(t) 1 .88 .54 .07 -.42 -.80 -.99 -.94 -.65 -.21 .28 .71 .96

Answer


Horizontal and Vertical Components of Velocity

The horizontal component of the velocity is written:

math image

and the vertical component is written:

math image

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

math image

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

math image

 

Example 2 

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


Answer


Example 3

If

math image

and

math image

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


Answer


Let's see how LiveMath obtained this graph.

LIVEMath


Acceleration of a Body in Curvilinear Motion

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

Horizontal component of acceleration:

math image

Vertical component of acceleration:

math image

Magnitude of acceleration:

math image

Direction of acceleration:

math image

 

Example 4

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

(i) Sketch the curve for 0 ≤ t ≤ 60.

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


Answer


Let's see how we can solve this problem in LiveMath.

LIVEMath

 

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).


Answer


Example 6

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

rocket 1

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.


Answer



Now let's return to the example we saw at the beginning of the calculus section. We have not seen how to differentiate sin x and cos x yet, but let's get LiveMath to do it for us:

LIVEMath




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