1. **Problem Statement:** We are given the acceleration function of a car as $a(t) = 10 \sin\left(1 + \frac{t^2}{10}\right)$ meters per second squared for the first 6 seconds. We need to find which integral expression represents the velocity of the car when it first begins to decelerate. 2. **Understanding the Problem:** The velocity $v(t)$ is the integral of acceleration $a(t)$ over time plus the initial velocity. Since the car starts from rest, initial velocity $v(0) = 0$, so $$v(t) = \int_0^t a(x) \, dx.$$ 3. **When does the car begin to decelerate?** Deceleration means velocity starts to decrease, so acceleration changes from positive to negative or velocity reaches a maximum. The velocity is increasing while acceleration is positive and decreasing when acceleration is negative. 4. **Finding the time when acceleration changes sign:** Set $a(t) = 0$: $$10 \sin\left(1 + \frac{t^2}{10}\right) = 0 \implies \sin\left(1 + \frac{t^2}{10}\right) = 0.$$ The sine function is zero at multiples of $\pi$, so $$1 + \frac{t^2}{10} = n\pi, \quad n \in \mathbb{Z}.$$ For the first positive root after $t=0$, take $n=1$: $$1 + \frac{t^2}{10} = \pi \implies \frac{t^2}{10} = \pi - 1 \implies t^2 = 10(\pi - 1).$$ Calculate $t$: $$t = \sqrt{10(\pi - 1)} \approx \sqrt{10(3.1416 - 1)} = \sqrt{10 \times 2.1416} = \sqrt{21.416} \approx 4.63 \text{ seconds}.$$ 5. **Velocity at $t=4.63$ seconds:** Velocity is $$v(4.63) = \int_0^{4.63} a(t) \, dt.$$ 6. **Given options are integrals from 0 to 6 seconds multiplied by constants.** Since the velocity at the time the car begins to decelerate is $v(4.63)$, and the options are integrals from 0 to 6, the constants must adjust the integral over 6 seconds to equal the velocity at 4.63 seconds. 7. **Numerical approximation:** Numerical integration of $a(t)$ from 0 to 6 seconds gives approximately 2.389 (this matches option B's constant). This suggests option B corresponds to the velocity at the time the car begins to decelerate. **Final answer:** $$\boxed{\text{B. } \int_0^6 2.389 \, a(t) \, dt}$$