Answer
The detailed explanations are below.
Work Step by Step
$s=200t-16t^2$, $0\le t\le12.5$
1) Find $v$ and $a$:
$$v=\frac{ds}{dt}=\frac{d}{dt}(200t-16t^2)=200-32t$$
$$a=\frac{dv}{dt}=\frac{d}{dt}(200-32t)=-32$$
The graphs of $s, v$ and $a$ are shown below.
2) We examine each element of the graph:
The curve $s$:
- The curve $s$ starts from the origin at $t=0$, increases constantly before reaching its peak, whose value is over $600ft$, at $t=6.25$. According to the given information about an object fired from the Earth's surface, we know that at the first $6.25$ seconds, the object continues to climb up higher and higher, and ultimately reaches a peak height of $600ft$.
- From $t=6.25$ to $t=12.5$, the curve decreases constantly before reaching back $0$ at $t=12.5$. So after reaching its peak at $t=6.25$, the object changes direction and falls down constantly and returns to the ground after $12.5$ seconds from the moment it was fired.
Overall, the object is furthest from the origin at $t=6.25$ seconds after it was fired, which is when it reached its peak of height.
The line $v$:
- The line $v$ decreases constantly on $[0,12.5]$. From $t=0$ to $t=6.25$, $v$ is positive and decreases constantly before reaching $0$ at $t=6.25$.
The object has its maximum speed at the time it was fired. Then as it climbed up $(v\gt0)$, it went more and more slowly before having the velocity $0$, or we can say it is at rest, at $t=6.25$ seconds after it was fired.
- From $t=6.25$ to $t=12.5$, $v$ is negative and decreases constantly. After the object reached its peak and stopped at $t=6.25$, it started to fall down ($v\lt0$) and its velocity rose constantly and reached its maximum as it hit the ground.
The line $a$:
The line $a$ is a horizontal line $a=-32$ throughout the whole interval. Since $a$ is negative, we know that the movement trend of velocity is decrease on the whole interval, and since $a$ does not change its value at all, this decrease trend is constant.
