Suppose that the position of one particle at time t is given by x1 = 5 sin(t), y1 = 2 cos(t), 0 ≤ t ≤ 2π and the position of a second particle is given by x2 = −5 + cos(t), y2 = 1 + sin(t), 0 ≤ t ≤ 2π.if so, find the collision points. (enter your answers as a comma-separated list of ordered pairs. if an answer does not exist, enter dne.)

This question involves the concept of simultaneous equations and collision points.

The collision points of both particles are "(-5, 0)".

The collision point is the point where the coordinates of both the particles are the same:

Solving the simultaneous equations above by substitution. From eqn (2):

substitute in en (1):

For  :

x₁ = 5, y₁ = 0

x₂ = -5, y₂ = 2

This does not satisfy the condition. Hence, this can not be the answer.

For  :

x₁ = -5, y₁ = 0

x₂ = -5, y₂ = 0

This pair satisfies the condition. Hence, the correct answer of the collision point is (-5, 0).

Learn more about the simultaneous equations here:

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The attached picture hows the collision point of two random simultaneous equations on graph.

t = 3π/2

Explanation:

They ask us the collision points of the two particles, at the points where their coordinates collide they must be equal. Let's write the equations of the particles

Particle 1

     x1 = 5 sin t

.     y1 = 2 cos t

Particle 2

    x2 = -5 + cos t

    y2 = 1 + sin t

At the point of collision

    x1 = x2

    y1 = y2

Let's solve

    5 sint = -5 + cos t

    2 cos t = 1 + sint

Let's solve the system of equations

    5 sin t = -5 + cos t

    sin t = -1 + 2 cost

     5 (-1 +2 cos t) = -5 + cos t

    -5 + 10 cos t = -5 + cost

     9 cost = 0

     Cost = 0

      t1 = 90º = pi / 2

      t2 = 270º = 3pi / 2

We have two times for which the equations are fulfilled. To know if both times are correct, substitute in the equations of the position of the particles

     t1 = π/2

Particle 1

     x1 = 5,   y1 = 0

Particle 2

     x2 = -5,  y2 = 1 + 1 = 2

We see that for this value the shock is not met

Let's try the other value t2 = 3π/2

Particle 1

     x1 = -5,     y1 = 0

Particle 2

    x2 = -5,     y2 = 1 -1 = 0

This point if it meets the shock of the particles,

The result is t = 3π/2

the working equation for the angular size is the product of the physical size and 360 degrees divided by the product of two pi and distance from earth. the minimum size would be 0.559 degrees while the maximum size would be 0.49 degrees.