# Finding acceleration of an object given height

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?

• Using the same method, I get that the acceleration is about -1.69
– Hrhm
Commented May 30, 2016 at 2:46
• h''(t)=-5.9319*sin(1.3t+0.9)
– Hrhm
Commented May 30, 2016 at 2:47
• @Hrhm Thanks, this helped me see my answer was not precise enough Commented May 30, 2016 at 3:11

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.
• @SirKellington To be thorough, you should verify the validity of your solution by checking that the value of $s$ you obtained lies in the interval $0.9\pm1.3$.