Question: A car is at the bottom of a hill with an incline of $20^\circ$ and it starts to accelerate up the hill by pressing on the gas. It started at $15$ m/s, and it has a mass of $1200$ kg, and accelerates his car at $3$ m/s^2. Assume negligible friction ($k = 0$), how fast is the car going after $6$ seconds of traveling up the hill?\ncos($20^\circ$) = $0.9397$\nsin($20^\circ$) = $0.3420$\n\nA. $12.89$ m/s\nB. $17.1$ m/s\nC. $-2.1$ m/s\nD. $-22.254$ m/s

1. **State the problem:** We need to find the final velocity of a car accelerating up a hill inclined at $20^\circ$ after $6$ seconds.\n\n2. **Given data:**\n- Initial velocity, $v_0 = 15$ m/s\n- Acceleration, $a = 3$ m/s$^2$\n- Time, $t = 6$ s\n- Hill incline angle, $\theta = 20^\circ$\n- Friction is negligible ($k=0$)\n\n3. **Formula used:** The final velocity under constant acceleration is given by:\n$$v = v_0 + at$$\n\n4. **Apply the formula:**\n$$v = 15 + 3 \times 6$$\n$$v = 15 + 18$$\n$$v = 33 \text{ m/s}$$\n\n5. **Interpretation:** The car's velocity after $6$ seconds is $33$ m/s up the hill.\n\n6. **Check the options:** None of the options match $33$ m/s, so we must consider the component of velocity along the hill or if the problem expects velocity relative to horizontal or vertical. However, since acceleration and velocity are along the hill, the direct calculation applies.\n\n7. **Conclusion:** The closest option to the calculated velocity is $17.1$ m/s (option B), but this does not match the direct calculation. Since friction is negligible and acceleration is given, the final velocity is $33$ m/s.\n\n**Note:** The problem might expect velocity component along horizontal or vertical, but as stated, the velocity along the hill is $33$ m/s.