**Problem 16** 1. The problem gives the particle position as functions of time: $$x=2t, \quad y=-\frac{t^2}{2}+8$$ To find the trajectory equation (relation between $x$ and $y$ eliminating $t$), solve for $t$: $$t=\frac{x}{2}$$ Substitute into $y$: $$y=-\frac{1}{2}\left(\frac{x}{2}\right)^2+8 = -\frac{x^2}{8} + 8$$ This is a parabola opening downward. 2. Velocity vector at $t_1=2.0$ s is the first derivative of position with respect to $t$: $$v_x=\frac{dx}{dt}=2, \quad v_y= \frac{dy}{dt} = -t$$ At $t=2$ s: $$v = (2, -2) \quad \text{units: m/s}$$ Vector magnitude: $$|v|=\sqrt{2^2 + (-2)^2} = \sqrt{8} = 2\sqrt{2} \approx 2.83$$ 3. Acceleration vector is the derivative of velocity: $$a_x = \frac{d^2x}{dt^2} = 0, \quad a_y = \frac{d^2y}{dt^2} = -1$$ At $t=2$ s, acceleration is: $$a = (0, -1)$$ Intrinsic components: - Tangential acceleration $\tilde{a}_t$ is the projection of $\mathbf{a}$ on velocity direction: $$\tilde{a}_t = \frac{\mathbf{a} \cdot \mathbf{v}}{|\mathbf{v}|} = \frac{0 \cdot 2 + (-1)(-2)}{2.83} = \frac{2}{2.83} \approx 0.71$$ - Normal acceleration $\tilde{a}_n$ using Pythagoras: $$|a| = \sqrt{0^2 + (-1)^2} = 1$$ $$\tilde{a}_n = \sqrt{|a|^2 - \tilde{a}_t^2} = \sqrt{1 - 0.71^2} = \sqrt{1-0.5} \approx 0.71$$ 4. Curvature radius $\rho$ at $t=2$ s is given by: $$\rho = \frac{|\mathbf{v}|^3}{|\mathbf{v} \times \mathbf{a}|}$$ Cross product magnitude in 2D: $$|\mathbf{v} \times \mathbf{a}| = |v_x a_y - v_y a_x| = |2(-1) - (-2)(0)| = 2$$ Thus, $$\rho = \frac{(2.83)^3}{2} = \frac{22.63}{2} = 11.31$$ --- **Problem 17** 1. Given: $$x=3t, \quad y=-t^2+4t$$ Solve for $t$: $$t=\frac{x}{3}$$ Substitute in $y$: $$y = -\left(\frac{x}{3}\right)^2 + 4 \left( \frac{x}{3} \right) = -\frac{x^2}{9} + \frac{4x}{3}$$ This parabola opens downward qualitatively for $x>0, y>0$. 2. Velocity components: $$v_x = \frac{dx}{dt} = 3, \quad v_y = \frac{dy}{dt} = -2t +4$$ Velocity vector: $$\mathbf{v} = (3, -2t + 4)$$ Velocity magnitude: $$|\mathbf{v}|= \sqrt{3^2 + (-2t+4)^2} = \sqrt{9 + (4 - 2t)^2}$$ 3. Average velocity magnitude from $t=0$ to $t=3$: Position at $t=0$: $$x_0=0, \quad y_0=0$$ Position at $t=3$: $$x_3=9, \quad y_3 = -9 + 12 = 3$$ Displacement: $$\Delta r = \sqrt{(9-0)^2 + (3-0)^2} = \sqrt{81 +9} = \sqrt{90} = 3 \sqrt{10} \approx 9.49$$ Average velocity magnitude: $$v_{avg} = \frac{\Delta r}{3} = \frac{9.49}{3} \approx 3.16$$ 4. Acceleration components: $$a_x = \frac{d^2x}{dt^2} = 0, \quad a_y = \frac{d^2y}{dt^2} = -2$$ Acceleration vector: $$\mathbf{a} = (0, -2)$$ Magnitude: $$|\mathbf{a}| = 2$$ 5. Angle $\alpha$ between $\mathbf{v}$ and $\mathbf{a}$ at $t=0$: $$\mathbf{v}(0) = (3,4), \quad \mathbf{a} = (0,-2)$$ Dot product: $$\mathbf{v} \cdot \mathbf{a} = 3*0 + 4*(-2) = -8$$ Magnitudes: $$|\mathbf{v}(0)| = 5, \quad |\mathbf{a}|= 2$$ $$\cos \alpha = \frac{-8}{5*2} = -0.8$$ $$\alpha = \cos^{-1}(-0.8) \approx 143.13^\circ$$ 6. Tangential acceleration component at $t=0$: $$\tilde{a}_t = \frac{\mathbf{a} \cdot \mathbf{v}}{|\mathbf{v}|} = \frac{-8}{5} = -1.6$$ --- **Final answers:** - Trajectory 16: $$y = -\frac{x^2}{8} + 8$$ - Velocity at $t=2$ for 16: $(2,-2)$ - Acceleration components at $t=2$ for 16: $\tilde{a}_t \approx 0.71$, $\tilde{a}_n \approx 0.71$ - Curvature radius at $t=2$ for 16: $11.31$ - Trajectory 17: $$y = -\frac{x^2}{9} + \frac{4x}{3}$$ - Velocity vector 17: $(3, -2t+4)$, magnitude as above - Average velocity magnitude 17 (0 to 3 s): $3.16$ - Acceleration vector 17: $(0,-2)$, magnitude 2 - Angle between velocity and acceleration at $t=0$: $143.13^\circ$ - Tangential acceleration at $t=0$: $-1.6$