1. **State the problem:** We have two particles A and B moving along a horizontal line. Particle A's velocity $v(t)$ is given at specific times, and Particle B's position is given by $x(t) = te^{\sin(3t)}$. We need to determine which particle is traveling faster at time $t=5$ seconds. 2. **Recall the relevant formulas:** - Velocity of Particle A at $t=5$ is given directly as $v(5) = 7.4$ cm/sec. - Velocity of Particle B is the derivative of its position function: $v_B(t) = \frac{dx}{dt}$. 3. **Find the velocity of Particle B at $t=5$:** Given $x(t) = te^{\sin(3t)}$, use the product rule: $$ v_B(t) = \frac{d}{dt} \left( t e^{\sin(3t)} \right) = e^{\sin(3t)} + t \cdot e^{\sin(3t)} \cdot \cos(3t) \cdot 3 $$ 4. **Evaluate $v_B(5)$:** Calculate each part: - $\sin(3 \times 5) = \sin(15)$ radians - $\cos(15)$ radians Using approximate values: - $\sin(15) \approx 0.6503$ - $\cos(15) \approx -0.7597$ Then: $$ v_B(5) = e^{0.6503} + 5 \times e^{0.6503} \times (-0.7597) \times 3 $$ Calculate $e^{0.6503} \approx 1.916$ So: $$ v_B(5) = 1.916 + 5 \times 1.916 \times (-0.7597) \times 3 = 1.916 - 21.77 = -19.854 $$ 5. **Interpretation:** - Particle A's velocity at $t=5$ is $7.4$ cm/sec (positive). - Particle B's velocity at $t=5$ is approximately $-19.854$ cm/sec (negative), meaning it is moving in the opposite direction. 6. **Conclusion:** At $t=5$, Particle A is traveling faster in the positive direction since $7.4 > |-19.854|$ and Particle B is moving backward. **Final answer:** Particle A is traveling faster at time $t=5$ seconds.