Position Velocity Acceleration vectors - Parametric equations

Problem Statement:
The parametric equations (in m) of the trajectory of a particle are given by:

x(t) = 3t
y(t) = 4t2

  1. Write the position vector of the particle in terms of the unit vectors.
  2. Calculate the velocity vector and its magnitude (speed).
  3. Express the trajectory of the particle in the form y(x)..
  4. Calculate the unit tangent vector at each point of the trajectory.
  5. Calculate the acceleration of the particle.

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Solution:

The position vector is given by:

Where i y j are the unit vectors defining the positive direction of the x and y axes respectively. Since the position vector doesn’t have z component, the particle describes a two dimensional trajectory.


The velocity vector is given by:

And its magnitude (speed):


To express the trajectory of the particle in the form y(x) we first isolate the variable (t) in the first equation x(t) and substitute into the second:

You can find below a plot of the trajectory, which is a parabola.


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In the plot below you can see the tree vectors (r, v y a) which allow to describe the motion of a particle.

To calculate the tangent unit vector ut we have to transform the velocity vector into a parallel vector whose magnitude is one (remember that the velocity vector is always tangent to the path of motion). This is done by dividing the velocity vector by its magnitude:

This tangent unit vector depends on time because, even if its magnitude is always unity, its direction varies for a curvilinear path.


The acceleration vector is defined:

Therefore, deriving the velocity vector:

The acceleration vector doesn’t depend on time; therefore the motion of this particle is uniformly accelerated.

Do no forget to include the units in the results.

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