Answer
a. $2\ m/sec$
b. negative.
c. $\frac{9}{2}=4.5\ m$
d. $t=6\ sec$
e. $t=4sec, 7sec$
f. towards the origin $6\lt t\lt 9\ sec$; away from the origin $0\lt t\lt 6\ sec$
g. right side of the origin.
Work Step by Step
Using the graph given in the Exercise and the function $s(t)=\int_0^t f(x)dx$, we have:
a. The velocity of the particle is given by $v(t)=s'(t)=f(t)$; thus $v(5)=f(5)=2\ m/sec$
b. The acceleration is given by $a(t)=v'(t)=f'(t)$. As $f'(5)\lt0$, we have $a(5)\lt0$, which means it is negative.
c. The position of the particle is given by the area under the curve. For $t=3$, the area of the triangle below the line from $0\leq t\leq 3$ is $s(3)=\frac{1}{2}\times3\times3=\frac{9}{2}=4.5\ m$
d. Since $s(t)$ is represented by the area between the curve and the axis (can be negative if the curve is below the x-axis), the largest value of $s(t)$ happens at $t=6\ sec$, where $s(6)=\int_0^6 f(x)dx$ is the integration of a positive function.
e. The acceleration $a(t)=f'(t)=0$ when $t=4sec, 7sec$, where the slope of the tangent lines on the curve is zero.
f. The particle moves towards the origin when $v=f(t)\lt 0$ which gives a time interval of $6\lt t\lt 9\ sec$. The particle moves away from the origin when $v=f(t)\gt 0$ which gives a time interval of $0\lt t\lt 6\ sec$.
g. The position $s(t)$ is represented by the area between the curve and the x-axis. At $t=9 sec$, the positive region is larger than the negative region, resulting in a positive $s(9)$, which means that the particle is on the right side of the origin.
