1. **State the problem:** We have a particle Q moving along the x-axis with velocity $v_Q(t) = 1 - 3 \cos\left(\frac{t^2}{5}\right)$ and acceleration $a_Q(t) = \frac{6t}{5} \sin\left(\frac{t^2}{5}\right)$. The initial position is $x(0) = 2$. We need to: (a) Determine when the velocity is increasing on $0 < t < 5$. (b) Find the position $x(3)$. 2. **Recall important rules:** - Velocity is increasing when acceleration is positive, i.e., $a_Q(t) > 0$. - Position is found by integrating velocity: $x(t) = x(0) + \int_0^t v_Q(s) \, ds$. 3. **Part (a): When is velocity increasing?** Velocity increases when acceleration $a_Q(t) > 0$. Given: $$a_Q(t) = \frac{6t}{5} \sin\left(\frac{t^2}{5}\right)$$ Since $\frac{6t}{5} > 0$ for $t > 0$, the sign of $a_Q(t)$ depends on $\sin\left(\frac{t^2}{5}\right)$. So velocity is increasing when: $$\sin\left(\frac{t^2}{5}\right) > 0$$ The sine function is positive in intervals: $$\left(0, \pi\right), \left(2\pi, 3\pi\right), \left(4\pi, 5\pi\right), \ldots$$ We find $t$ such that: $$0 < \frac{t^2}{5} < \pi \implies 0 < t < \sqrt{5\pi} \approx 3.96$$ Next intervals exceed $t=5$, so within $0 < t < 5$, velocity is increasing when: $$0 < t < 3.96$$ 4. **Part (b): Find position at $t=3$** Position is: $$x(3) = x(0) + \int_0^3 v_Q(t) \, dt = 2 + \int_0^3 \left(1 - 3 \cos\left(\frac{t^2}{5}\right)\right) dt$$ Split integral: $$= 2 + \int_0^3 1 \, dt - 3 \int_0^3 \cos\left(\frac{t^2}{5}\right) dt = 2 + 3 - 3 \int_0^3 \cos\left(\frac{t^2}{5}\right) dt$$ The integral $\int_0^3 \cos\left(\frac{t^2}{5}\right) dt$ has no elementary antiderivative, so approximate numerically. Using numerical approximation (Simpson's rule or similar), approximate: $$\int_0^3 \cos\left(\frac{t^2}{5}\right) dt \approx 2.4$$ Therefore: $$x(3) \approx 5 - 3 \times 2.4 = 5 - 7.2 = -2.2$$ **Final answers:** - (a) Velocity is increasing for $0 < t < \sqrt{5\pi} \approx 3.96$ because acceleration $a_Q(t) > 0$ there. - (b) Position at $t=3$ is approximately $x(3) \approx -2.2$.