# Chapter 3 - Kinematics in Two Dimensions; Vectors - General Problems: 53

Horizontal = 3.9m/s$^2$ to the left Vertical = 1.9 m/s$^2$ down.

#### Work Step by Step

The problem takes place along a straight line. The starting velocity is 30.6 m/s and the ending velocity is 0 m/s. Use 2.11a to find the acceleration of $-4.37 \frac{m}{s^{2}}$. This is directed downhill, at an angle of 206 degrees from the positive x-axis (180 degrees away from uphill, which is at 26 degrees). Horizontal acceleration: $a_{x} = 4.37 \frac{m}{s^{2}}cos 206 ^{\circ} = -3.9 \frac{m}{s^{2}}$ Vertical acceleration: $a_{y} = 4.37 \frac{m}{s^{2}}sin 206 ^{\circ} = -1.9 \frac{m}{s^{2}}$

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