a car travels along a straight path for 9 minutes. how far is the car from the start when it is…

Explanation:

Step1: Recall the relationship between velocity - time graph and displacement

Displacement is the area under the velocity - time graph. We need to split the graph into geometric shapes and calculate their areas.

Step2: Calculate the area of the semi - circle

The semi - circle has a radius $r = 2$ and is above the time - axis. The area of a semi - circle is $A_{1}=\frac{1}{2}\pi r^{2}$. Substituting $r = 2$, we get $A_{1}=\frac{1}{2}\pi(2)^{2}=2\pi$.

Step3: Calculate the area of the triangle from $t = 4$ to $t = 5$

The base of the triangle is $b=1$ and the height is $h = 2$. The area of a triangle is $A_{2}=\frac{1}{2}bh$. So $A_{2}=\frac{1}{2}(1)(2)=1$.

Step4: Calculate the area of the triangle from $t = 5$ to $t = 9$

The base of the triangle is $b = 4$ and the height is $h=2$. The area of a triangle is $A_{3}=\frac{1}{2}bh$. So $A_{3}=\frac{1}{2}(4)(2)=4$.

Step5: Calculate the net displacement

The displacement $d$ is the sum of the positive areas (above the time - axis) minus the sum of the negative areas (below the time - axis). Here, $d=2\pi+ 1-4$. $d=2\pi - 3$. Using $\pi\approx3.14159$, we have $d=2\times3.14159 - 3=6.28318-3 = 3.28318\approx3.283$.

Answer:

$3.283$