The height of a moving object is given as a function of time.
$$h(t) = 3.0 + 2.7 \sin(1.3t + 0.9)$$
$t$ is measured in seconds and $h$ is measured in feet.
Given this, I've found the velocity and acceleration as follows:
- $h'(t)=3.51 \cos(1.3t+0.9)$
- $h''(t)=-4.563 \sin(1.3t+0.9)$
I am trying to find the acceleration at the first instant when $t > 0$ and the height is $4$ feet.
I have tried finding time by:
- Setting $h(t)=4$ and solving algebraically
- Graphing $h(t)$ and finding an intersection at $y=4$
This yields ~$ t = 1.432 s$
Graph provided here: https://www.desmos.com/calculator/gxytwo3kan
I then plugged $t = 1.432 s$ into my acceleration function giving:
- $h''(1.432) = -3.47142 ft/s^2$
This is the answer I am confident in, but is being marked incorrect on my assignment. Can someone help me understand what I might be missing here?
You don’t have to find $t$ to solve this. Since the term $s=\sin(1.3t+0.9)$ appears in both the position and acceleration, rewrite them as $h(s)=3.0+2.7s$ and $a(s)=-4.563s$ and solve from there. This will give you an exact solution instead of the approximation that you got from the graph.