1. **State the problem:** Given the position function $$x(t) = 0.050 \cos(10t + 6\pi)$$ in SI units, find the velocity $$v$$ and acceleration $$a$$ at time $$t = 0.10$$ seconds. 2. **Formulas used:** - Velocity is the first derivative of position with respect to time: $$v(t) = \frac{dx}{dt}$$. - Acceleration is the second derivative of position with respect to time: $$a(t) = \frac{d^2x}{dt^2}$$. 3. **Calculate velocity:** $$v(t) = \frac{d}{dt} \left(0.050 \cos(10t + 6\pi)\right) = 0.050 \times (-\sin(10t + 6\pi)) \times 10 = -0.50 \sin(10t + 6\pi)$$ 4. **Calculate acceleration:** $$a(t) = \frac{d}{dt} v(t) = \frac{d}{dt} \left(-0.50 \sin(10t + 6\pi)\right) = -0.50 \times \cos(10t + 6\pi) \times 10 = -5.0 \cos(10t + 6\pi)$$ 5. **Evaluate at $$t=0.10$$ s:** Calculate the argument: $$10 \times 0.10 + 6\pi = 1 + 6\pi$$ Since $$\sin(\theta)$$ and $$\cos(\theta)$$ are periodic with period $$2\pi$$, we can simplify: $$6\pi = 3 \times 2\pi$$ so $$\sin(1 + 6\pi) = \sin(1) \approx 0.8415$$ $$\cos(1 + 6\pi) = \cos(1) \approx 0.5403$$ 6. **Calculate numeric values:** $$v(0.10) = -0.50 \times 0.8415 = -0.42075 \approx -0.42$$ $$a(0.10) = -5.0 \times 0.5403 = -2.7015 \approx -2.70$$ 7. **Compare with options:** None of the options exactly match these values, but the closest velocity is approximately $$-0.50$$ and acceleration approximately $$-0.25$$ in option (a). However, our acceleration is about $$-2.70$$, which is much larger in magnitude. **Re-examining step 4:** The acceleration formula should be: $$a(t) = \frac{d}{dt} v(t) = \frac{d}{dt} \left(-0.50 \sin(10t + 6\pi)\right) = -0.50 \times 10 \cos(10t + 6\pi) = -5.0 \cos(10t + 6\pi)$$ This is correct. **Re-examining step 5:** The original position amplitude is 0.050, so velocity amplitude is 0.50, acceleration amplitude is 5.0. Given the options, the only one with velocity near -0.50 and acceleration near -0.25 is (a), but acceleration magnitude is off. **Possibility:** The problem might expect acceleration as $$a = -\omega^2 x$$, so: $$a(0.10) = -10^2 \times x(0.10) = -100 \times 0.050 \cos(1 + 6\pi) = -5.0 \times 0.5403 = -2.70$$ This matches our previous calculation. Therefore, the correct values are approximately: $$v \approx -0.42$$ $$a \approx -2.70$$ None of the options exactly match, but closest velocity is negative near -0.5 and acceleration is negative and larger in magnitude. **Final answer:** Velocity $$v \approx -0.42$$, acceleration $$a \approx -2.70$$ at $$t=0.10$$ s.