at time t, 0 < t < 2, the velocity of a particle moving along the x - axis is given by v(t)=t sin(t²). let t…

Explanation:

Step1: Find the time $b$

The particle changes direction when $v(t)$ changes sign. Set $v(t)=t\sin(t^{2}) = 0$. Since $t\in(0,2)$, then $t^{2}=k\pi$ ($k\in\mathbb{Z}$). For $t\in(0,2)$, when $k = 1$, $t^{2}=\pi$, so $t = b=\sqrt{\pi}\approx1.77$. The particle changes direction when $v(t)$ goes from negative to positive.

Step2: Recall the distance - velocity relationship

The distance $d$ traveled by a particle over the interval $[a,b]$ is given by $d=\int_{a}^{b}|v(t)|dt$. Since $v(t)=t\sin(t^{2})$ and for $t\in(0,\sqrt{\pi})$, we know that $v(t)$ is non - positive on some sub - intervals and non - negative on others. Let $u = t^{2}$, then $du=2tdt$ and $t\sin(t^{2})dt=\frac{1}{2}\sin(u)du$.

Step3: Evaluate the integral for distance

We need to evaluate $\int_{0}^{\sqrt{\pi}}|t\sin(t^{2})|dt$. Since $v(t)=t\sin(t^{2})$, and using the substitution $u = t^{2}$, when $t = 0$, $u = 0$; when $t=\sqrt{\pi}$, $u=\pi$. [ \begin{align*} \int_{0}^{\sqrt{\pi}}|t\sin(t^{2})|dt&=\int_{0}^{\sqrt{\pi}}| \frac{1}{2}\sin(u)|du\ &=\frac{1}{2}\int_{0}^{\pi}|\sin(u)|du \end{align*} ] We know that $\sin(u)\geq0$ for $u\in[0,\pi]$, so $\frac{1}{2}\int_{0}^{\pi}\sin(u)du$. [ \begin{align*} \frac{1}{2}[-\cos(u)]_{0}^{\pi}&=\frac{1}{2}[-\cos(\pi)+\cos(0)]\ &=\frac{1}{2}(1 + 1)\ &= 1 \end{align*} ] The closest value to $1$ among the options is $1.011$.

Answer:

C. 1.011