What’s the frequency for problem #10?

The answers penciled in on the sheet are fine ... 1500 Hz and 0.0006 sec.

If you want to get super-technical about it, you could count the number of full cycles on the WHOLE graph, from one side to the other.  At the right end, near 0.01 sec, you can notice an extra 1/4 of a cycle there.  So it's actually 15 and 1/4 cycles in 0.01 sec.

That makes the frequency (15.25 cycles) / (0.01 sec) = 1,525 Hz .

and the period ( second / 1525) = 0.000656... second.

But I think it would be OK if you didn't do that, and just call it 1500 Hz and 0.0006s.

The difference is only 1.6% , and the engineer in me says that's not enough to worry about.

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer