1. **Problem statement:** A particle moves along the x-axis with velocity given by $$v(t) = - (t + 1) \sin\left(\frac{t^2}{2}\right).$$ At time $t=0$, the position is $x=1$. (a) Find the acceleration at $t=2$ and determine if the speed is increasing at $t=2$. (b) Find all times $t$ in the interval $0 < t < 3$ when the particle changes direction. --- 2. **Recall formulas and rules:** - Acceleration is the derivative of velocity: $$a(t) = v'(t).$$ - Speed is the absolute value of velocity: $$\text{speed} = |v(t)|.$$ - Speed increases when velocity and acceleration have the same sign. - Particle changes direction when velocity changes sign, i.e., when $v(t) = 0$ and velocity crosses zero. --- 3. **Find acceleration $a(t)$:** Given $$v(t) = - (t + 1) \sin\left(\frac{t^2}{2}\right),$$ use product rule: $$a(t) = \frac{d}{dt} v(t) = - \frac{d}{dt} \left[(t+1) \sin\left(\frac{t^2}{2}\right)\right].$$ Let $$u = t+1, \quad w = \sin\left(\frac{t^2}{2}\right).$$ Then $$a(t) = - \left(u' w + u w'\right).$$ Calculate derivatives: $$u' = 1,$$ $$w' = \cos\left(\frac{t^2}{2}\right) \cdot \frac{d}{dt} \left(\frac{t^2}{2}\right) = \cos\left(\frac{t^2}{2}\right) \cdot t.$$ So, $$a(t) = - \left(1 \cdot \sin\left(\frac{t^2}{2}\right) + (t+1) \cdot t \cdot \cos\left(\frac{t^2}{2}\right)\right) = - \sin\left(\frac{t^2}{2}\right) - t(t+1) \cos\left(\frac{t^2}{2}\right).$$ --- 4. **Evaluate acceleration at $t=2$:** Calculate each term: $$\sin\left(\frac{2^2}{2}\right) = \sin(2),$$ $$\cos\left(\frac{2^2}{2}\right) = \cos(2),$$ $$t(t+1) = 2 \times 3 = 6.$$ Therefore, $$a(2) = - \sin(2) - 6 \cos(2).$$ Numerical approximations: $$\sin(2) \approx 0.9093,$$ $$\cos(2) \approx -0.4161,$$ so $$a(2) \approx -0.9093 - 6 \times (-0.4161) = -0.9093 + 2.4966 = 1.5873.$$ --- 5. **Check velocity at $t=2$:** $$v(2) = - (2+1) \sin(2) = -3 \times 0.9093 = -2.7279.$$ Velocity is negative, acceleration is positive. --- 6. **Is speed increasing at $t=2$?** Speed increases if velocity and acceleration have the same sign. Here, $v(2) < 0$ and $a(2) > 0$, so they have opposite signs. Therefore, speed is **decreasing** at $t=2$. --- 7. **Find times when particle changes direction in $0 < t < 3$:** Particle changes direction when velocity changes sign, i.e., when $$v(t) = - (t+1) \sin\left(\frac{t^2}{2}\right) = 0.$$ Since $t+1 > 0$ for $t > 0$, the zeros come from $$\sin\left(\frac{t^2}{2}\right) = 0.$$ The sine function is zero at integer multiples of $\pi$: $$\frac{t^2}{2} = n \pi, \quad n = 0,1,2,\ldots$$ Solve for $t$: $$t = \sqrt{2 n \pi}.$$ We want $0 < t < 3$, so find $n$ such that $$0 < \sqrt{2 n \pi} < 3 \implies 0 < 2 n \pi < 9 \implies 0 < n < \frac{9}{2\pi} \approx 1.432.$$ So $n=1$ is valid. Thus, $$t = \sqrt{2 \pi} \approx 2.5066.$$ --- 8. **Check if velocity changes sign at $t = \sqrt{2\pi}$:** - For $t$ slightly less than $2.5066$, $\sin\left(\frac{t^2}{2}\right)$ is positive (since sine crosses zero from positive to negative at $\pi$ multiples). - For $t$ slightly greater than $2.5066$, $\sin\left(\frac{t^2}{2}\right)$ is negative. Since $v(t) = - (t+1) \sin\left(\frac{t^2}{2}\right)$, the sign of $v(t)$ changes from negative to positive at $t=\sqrt{2\pi}$. Therefore, the particle changes direction at $$t = \sqrt{2 \pi}.$$ --- **Final answers:** (a) $$a(2) = - \sin(2) - 6 \cos(2) \approx 1.5873.$$ Speed is decreasing at $t=2$ because velocity and acceleration have opposite signs. (b) Particle changes direction at $$t = \sqrt{2 \pi} \approx 2.5066$$ within the interval $(0,3)$.