a particle with velocity v(t) = t² + 3, where t is time in seconds, moves in a straight line. how far does…

Explanation:

Step1: Recall the displacement - velocity relationship

The displacement $s$ of a particle with velocity $v(t)$ over the interval $[a,b]$ is given by $s=\int_{a}^{b}v(t)dt$. Here, $v(t)=t^{2}+3$, $a = 2$, and $b = 3$.

Step2: Calculate the definite integral

We know that $\int(t^{2}+3)dt=\frac{t^{3}}{3}+3t + C$. Then, $\int_{2}^{3}(t^{2}+3)dt=\left[\frac{t^{3}}{3}+3t\right]_{2}^{3}$.

Step3: Evaluate the definite - integral

First, substitute $t = 3$ into $\frac{t^{3}}{3}+3t$: $\frac{3^{3}}{3}+3\times3=\frac{27}{3}+9 = 9 + 9=18$. Then substitute $t = 2$ into $\frac{t^{3}}{3}+3t$: $\frac{2^{3}}{3}+3\times2=\frac{8}{3}+6=\frac{8 + 18}{3}=\frac{26}{3}$. Now, subtract: $18-\frac{26}{3}=\frac{54 - 26}{3}=\frac{28}{3}$.

Answer:

$\frac{28}{3}$