Question

Consider two particles A and B. The angular position of particle A, with constant angular acceleration, depends on time according to \theta_{\rm A}(t)=\theta_0+\omega_0t+\frac{_1}{^2}\alpha t^2. At time t=t_1, particle B, which also undergoes constant angular acceleration, has twice the angular acceleration, half the angular velocity, and the same angular position that particle A had at time t=0.Which of the following equations describes the angular position of particle B? A) theta_{\rm B}(t)=\theta_0+2\omega_0t+\frac{1}{4}\alpha t^2B) theta_{\rm B}(t)=\theta_0+\frac{1}{2}\omega_0t+\alpha t^2C) theta_{\rm B}(t)=\theta_0+2\omega_0(t-t_1) +\frac{1}{4}\alpha (t-t_1)^2D) theta_{\rm B}(t)=\theta_0+\frac{1}{2}\omega_0(t-t_1)+\alpha (t-t_1)^2E) theta_{\rm B}(t)=\theta_0+2\omega_0(t+t_1) +\frac{1}{4}\alpha (t+t_1)^2F) theta_{\rm B}(t)=\theta_0+\frac{1}{2}\omega_0(t+t_1)+\alpha (t+t_1)^2

Answer 1

Answer:

θ = θ₀ + ½ w₀ (t -t_1) + α (t -t_1)²

Explanation:

This is an angular kinematic exercise the equation for the angular position

the particle A

       θ = θ₀ + ω₀ t + ½ α t²

They say for the particle B

     w₀B = ½ w₀

     αB = 2 α

In addition, the particle begins at a time t_1 after particle A, in order to use the same timer, we must subtract this time from the initial

      t´ = t - t_1

l

et's write the equation of particle B

      θ = θ₀ + w₀B t´ + ½ αB t´2

replace

     θ = θ₀ + ½ w₀ (t -t_1) + ½ 2α (t -t_1)²

     θ = θ₀ + ½ w₀ (t -t_1) + α (t -t_1)²