2. a particle starts at (x = 0) and (v(t)=2t + 1) for time (tgeq0). where is the particle at? 3. a particle…

2. a particle starts at (x = 0) and (v(t)=2t + 1) for time (tgeq0). where is the particle at? 3. a particle starts at (x = 0) and moves along the (x -)axis with velocity (v(t)=t^{2}+1) for time (tgeq0). where is the particle at (t = 4)? approximate the area under the curve using four rectangles of equal width and heights determined by the mid - points of the intervals, as in example 1.

2. a particle starts at (x = 0) and (v(t)=2t + 1) for time (tgeq0). where is the particle at? 3. a particle starts at (x = 0) and moves along the (x -)axis with velocity (v(t)=t^{2}+1) for time (tgeq0). where is the particle at (t = 4)? approximate the area under the curve using four rectangles of equal width and heights determined by the mid - points of the intervals, as in example 1.

Answer

Explanation:

Step1: Recall the position - velocity relationship

The position function $x(t)$ is the antiderivative of the velocity function $v(t)$. Given $v(t)=t^{2}+1$, then $x(t)=\int v(t)dt=\int(t^{2}+1)dt$. Using the power - rule for integration $\int t^{n}dt=\frac{t^{n + 1}}{n+1}+C$ ($n\neq - 1$), we have $x(t)=\frac{t^{3}}{3}+t + C$. Since the particle starts at $x = 0$ when $t = 0$, substituting these values into the position function gives $0=\frac{0^{3}}{3}+0 + C$, so $C = 0$. Thus, $x(t)=\frac{t^{3}}{3}+t$.

Step2: Find the position at $t = 4$

Substitute $t = 4$ into the position function $x(t)$. $x(4)=\frac{4^{3}}{3}+4=\frac{64}{3}+4=\frac{64 + 12}{3}=\frac{76}{3}$.

Step3: Approximate the area under the curve using mid - point rule

The interval is $[0,4]$, and we want to use $n = 4$ rectangles. The width of each rectangle $\Delta t=\frac{b - a}{n}$, where $a = 0$, $b = 4$, and $n = 4$. So $\Delta t=\frac{4-0}{4}=1$. The sub - intervals are $[0,1]$, $[1,2]$, $[2,3]$, $[3,4]$. The mid - points of these sub - intervals are $t_1=0.5$, $t_2 = 1.5$, $t_3=2.5$, $t_4=3.5$. The velocity function is $v(t)=t^{2}+1$. $v(t_1)=(0.5)^{2}+1=0.25 + 1=1.25$; $v(t_2)=(1.5)^{2}+1=2.25 + 1=3.25$; $v(t_3)=(2.5)^{2}+1=6.25 + 1=7.25$; $v(t_4)=(3.5)^{2}+1=12.25 + 1=13.25$. The area $A$ approximated by the mid - point rule is $A=\sum_{i = 1}^{4}v(t_i)\Delta t$. $A=(1.25 + 3.25+7.25 + 13.25)\times1=25$.

Answer:

The position of the particle at $t = 4$ is $\frac{76}{3}$, and the approximate area under the curve using the mid - point rule with $n = 4$ rectangles is $25$.