# Chapter 4 - Kinematics in Two Dimensions - Exercises and Problems: 9

(a) The puck is moving at an angle of $61.9^{\circ}$ above the positive x-axis. (b) The puck is a distance of 180 cm from the origin.

#### Work Step by Step

(a) The acceleration in the x-direction is the slope of the $v_x$ versus time graph; $a_x = \frac{40~cm/s}{5~s} = 8~cm/s^2$ At t = 2 s: $v_x = a_x~t = (8~cm/s^2)(2~s)$ $v_x = 16~cm/s$ $v_y = 30~cm/s$ We can find the angle $\theta$ above the positive x-axis; $tan(\theta) = \frac{30~cm/s}{16~cm/s}$ $\theta = arctan(\frac{30~cm/s}{16~cm/s})$ $\theta = 61.9^{\circ}$ The puck is moving at an angle of $61.9^{\circ}$ above the positive x-axis. (b) The displacement in the x-direction is the area under the $v_x$ versus time graph. $x = \frac{1}{2}(40~cm/s)(5~s)$ $x = 100~cm$ The displacement in the y-direction is the area under the $v_y$ versus time graph. $y = v_y~t$ $y = (30~cm/s)(5~s)$ $y = 150~cm$ We can find the distance $d$ from the origin. $d = \sqrt{(x)^2+(y)^2}$ $d = \sqrt{(100~cm)^2+(150~cm)^2}$ $d = 180~cm$ The puck is a distance of 180 cm from the origin.

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