Final answer:
The position of a particle moving in a circle of radius A can be represented by x(t) = Acos(ωt+ϕ). The speed of the particle is constant A. The acceleration of the particle is given by x''(t) = -ω²x(t). The position and velocity of another particle can be determined using the given position equation z(t) = (D²/E)-2Dt+Et².
Step-by-step explanation:
(a) To show that the particle moves in a circle of radius A, we can rewrite the position vector as r(t) = Acos(ωt+ϕ)î + Asin(ωt+ϕ)ĵ, where î and ĵ are unit vectors in the x and y directions respectively. The equation of the position vector represents a circle in the xy-plane with center (0,0) and radius A.
(b) Taking the derivative of the position vector, we find r'(t) = -Aωsin(ωt+ϕ)î + Aωcos(ωt+ϕ)ĵ. The magnitude of the velocity vector is |r'(t)| = Aω, which is constant. Therefore, the speed of the particle is a constant A.
(c) Differentiating the velocity vector, we get r''(t) = -Aω²cos(ωt+ϕ)î - Aω²sin(ωt+ϕ)ĵ. Since cos²(ωt+ϕ) + sin²(ωt+ϕ) = 1, we have cos²(ωt+ϕ) = 1 - sin²(ωt+ϕ). Substituting this into the equation for r''(t), we get r''(t) = -Aω²(1 - sin²(ωt+ϕ))î - Aω²sin(ωt+ϕ)ĵ. Simplifying further, we find r''(t) = -Aω²î - Aω²sin(ωt+ϕ)ĵ = -Aω²r(t). Thus, we have shown that x''(t) = -ω²x(t), irrespective of the values of A and ϕ.
(d) The given position equation z(t) = (D²/E)-2Dt+Et² represents a quadratic function. Taking the derivative of z(t), we get z'(t) = -2D + 2Et.
(e) Differentiating z'(t), we find z''(t) = 2E.
(f) To find when the particle's velocity is zero, we set z'(t) = 0 and solve for t. -2D + 2Et = 0 => Et = D => t = D/E.