# Chapter 4 - Kinematics in Two Dimensions - Exercises and Problems - Page 110: 8

(a) At t = 0: The particle's position is (0,0) The particle's speed is 2 m/s At t = 4 s: The particle's position is (0,0) The particle's speed is 8.2 m/s (b) The particle is moving at an angle of $90^{\circ}$ clockwise from the positive x-axis. The particle is moving at an angle of $14.0^{\circ}$ above the positive x-axis.

#### Work Step by Step

(a) At t = 0: $x = (\frac{1}{2}t^3-2t^2)~m$ $x = (\frac{1}{2}(0)^3-2(0)^2)~m$ $x = 0$ $y = (\frac{1}{2}t^2-2t)~m$ $y = (\frac{1}{2}(0)^2-2(0))~m$ $y = 0$ $v_x(t) = \frac{dx}{dt} = (\frac{3}{2}t^2-4t)~m/s$ $v_0 = (\frac{3}{2}(0)^2-4(0))~m/s$ $v_x = 0$ $v_y(t) = \frac{dy}{dt} = (t-2)~m/s$ $v_0 = ((0)-2)~m/s$ $v_y = -2~m/s$ The particle's position is (0,0) The particle's speed is 2 m/s At t = 4 s: $x = (\frac{1}{2}t^3-2t^2)~m$ $x = (\frac{1}{2}(4~s)^3-2(4~s)^2)~m$ $x = 0$ $y = (\frac{1}{2}t^2-2t)~m$ $y = (\frac{1}{2}(4~s)^2-2(4~s))~m$ $y = 0$ $v_x(t) = (\frac{3}{2}t^2-4t)~m/s$ $v_0 = (\frac{3}{2}(4~s)^2-4(4~s))~m/s$ $v_x = 8~m/s$ $v_y(t) = (t-2)~m/s$ $v_0 = ((4~s)-2)~m/s$ $v_y = 2~m/s$ We can find the particle's speed. $v = \sqrt{(v_x)^2+(v_y)^2}$ $v = \sqrt{(8~m/s)^2+(2~m/s)^2}$ $v = 8.2~m/s$ The particle's position is (0,0) The particle's speed is 8.2 m/s (b) At t = 0, the velocity vector points toward the negative y-axis. The particle is moving at an angle of $90^{\circ}$ clockwise from the positive x-axis. We can find the direction of motion at t = 4 seconds as an angle above the positive x-axis. $tan(\theta) = \frac{v_y}{v_x}$ $\theta = arctan(\frac{2~m/s}{8~m/s})$ $\theta = 14.0^{\circ}$

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