1. **Problem statement:** Given velocity function $v(t) = - (t + 1) \sin\left(\frac{t^2}{2}\right)$ and initial position $x(0) = 1$, find: (a) acceleration at $t=2$ and whether speed is increasing at $t=2$. (b) all times $t$ in $(0,3)$ when the particle changes direction. 2. **Formulas and rules:** - Acceleration: $a(t) = v'(t)$. - Speed increases if velocity and acceleration have the same sign. - Particle changes direction when velocity changes sign, i.e., $v(t)=0$ and sign changes. 3. **Find acceleration $a(t)$:** Use product rule on $v(t) = - (t+1) \sin\left(\frac{t^2}{2}\right)$: $$a(t) = - \frac{d}{dt} \left[(t+1) \sin\left(\frac{t^2}{2}\right)\right] = - \left(1 \cdot \sin\left(\frac{t^2}{2}\right) + (t+1) \cdot t \cdot \cos\left(\frac{t^2}{2}\right)\right) = - \sin\left(\frac{t^2}{2}\right) - t(t+1) \cos\left(\frac{t^2}{2}\right).$$ 4. **Evaluate $a(2)$:** $$a(2) = - \sin(2) - 2 \times 3 \times \cos(2) = - \sin(2) - 6 \cos(2).$$ Numerically, $\sin(2) \approx 0.9093$, $\cos(2) \approx -0.4161$, so $$a(2) \approx -0.9093 + 2.4966 = 1.5873.$$ 5. **Evaluate $v(2)$:** $$v(2) = -3 \sin(2) = -3 \times 0.9093 = -2.7279.$$ Velocity is negative, acceleration positive. 6. **Is speed increasing at $t=2$?** Since velocity and acceleration have opposite signs, speed is decreasing at $t=2$. 7. **Find times when particle changes direction in $(0,3)$:** Solve $v(t) = 0$: $$- (t+1) \sin\left(\frac{t^2}{2}\right) = 0 \implies \sin\left(\frac{t^2}{2}\right) = 0,$$ since $t+1 > 0$ for $t>0$. Zeros of sine at $\frac{t^2}{2} = n \pi$, $n=0,1,2,\ldots$ $$t = \sqrt{2 n \pi}.$$ For $0 < t < 3$, find $n$: $$0 < \sqrt{2 n \pi} < 3 \implies 0 < 2 n \pi < 9 \implies 0 < n < \frac{9}{2\pi} \approx 1.432.$$ So $n=1$ valid. 8. **Check sign change at $t=\sqrt{2\pi}$:** Velocity changes sign from negative to positive, so particle changes direction at $$t = \sqrt{2 \pi} \approx 2.5066.$$