Question

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.)

Answer 1

Answer:

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

Answer 2

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:

$x_1=x_2\\y_1=y_2\\\\5\ Sin\ (t) = -5+Cos\ (t)------ eqn(1)\\2\ Cos\ (t)=1+Sin\ (t)-------- eqn(2)$

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

$Sin\ (t)=2\ Cos\ (t)-1$

substitute in en (1):

$5(2\ Cos\ (t)-1)=-5+Cos\ (t)\\9Cos\ (t)=0\\t = Cos^{-1}(0)\ \ \ ; \ \ \ 0\leq t\leq 2\pi\\\\t = \frac{\pi}{2}\ \ (OR) t = \frac{3\pi}{2}$

For  $t=\frac{\pi}{2}$:

x₁ = 5, y₁ = 0

x₂ = -5, y₂ = 2

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

For  $t=\frac{3\pi}{2}$:

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:

https://brainly.com/question/16763389?referrer=searchResults

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