Sorry - I looked at this earlier, but had to go. Here's how to fix things, if you haven't already done so.
Your function 10-8cos(0.4t) has the correct period, but is a minimum at t=0. Part (c) of this problem indicates you want to be at axle height when t=0.
That would be 10+8sin(0.4t), which starts at the center, and rises.
But, since cos(x) = sin(π/2 -x) you want
h(t) = 10 - 8cos(π/2 - 0.4t) = 10 - 8cos(0.4 (π/5 - t))
or, since cos(-x) = cos(x), it might look nicer as
h(t) = 10 - 8cos(0.4 (t - π/5))
That also helps to show the shift of 1/4 period.
To fix the function to have the desired behavior, you need to make some adjustments:
1. Start with the original function: h(t) = 10 - 8cos(0.4t)
2. To ensure that the function starts at axle height when t = 0, you need to shift the function up by 8 units. This can be achieved by adding 8 to the original function:
h(t) = 10 - 8cos(0.4t) + 8
3. However, this modification will also shift the entire graph up by 8 units. To counteract this effect, you need to subtract 8sin(0.4t) from the function, as sin(x) is equal to cos(π/2 - x):
h(t) = 10 - 8cos(0.4t) + 8 - 8sin(0.4t)
4. Simplifying further, you can rewrite this equation as:
h(t) = 18 - 8cos(0.4t) - 8sin(0.4t)
5. Alternatively, using the identity cos(-x) = cos(x), you can rewrite the equation as:
h(t) = 18 - 8cos(0.4(t - π/5))
This final equation will result in a function that starts at axle height when t = 0, and has the desired behavior.