Science:Math Exam Resources/Courses/MATH102/December 2017/Question 15 (b)/Solution 1

According to the Hint, let us compute the acceleration, which is the second derviative: applying the chain rule, we get

$$h(t)=3+3\sin \left({\dfrac {5}{3}}t\right)$$

$$h'(t)=3\cos \left({\dfrac {5}{3}}t\right)\cdot {\dfrac {5}{3}}=5\cos \left({\dfrac {5}{3}}t\right)$$

$$h''(t)=-5\sin \left({\dfrac {5}{3}}t\right)\cdot {\dfrac {5}{3}}=-{\dfrac {25}{3}}\sin \left({\dfrac {5}{3}}t\right).$$

As ${\textstyle t}$ varies, we see that ${\textstyle h''(t)}$ oscillates between ${\textstyle -{\dfrac {25}{3}}}$ and ${\textstyle {\dfrac {25}{3}}}$. Since ${\textstyle {\dfrac {25}{3}}>5}$, the grasshopper can repeatedly (recall that the function ${\textstyle h}$ is periodic of period ${\textstyle {\dfrac {6\pi }{5}}}$ from part (a)) reach a vertical acceleration of ${\textstyle 5{\text{cm}}/{\text{m}}^{2}}$. So the grasshopper can free itself.

Answer: ${\textstyle {\color {blue}{\mbox{Yes, since its vertical acceleration can reach up to }}{\dfrac {25}{3}}.}}$