Answer:
a) Minimum acceleration is [tex]a=2.31\frac{m}{s^{2} }[/tex].
b) It will take [tex]t_{f}=14.41s[/tex].
Explanation:
Let's order the information.
Initial velocity: [tex]v_{i}=0m/s[/tex]
Final velocity: [tex]v_{f}=33.3m/s[/tex]
Initial position: [tex]x_{i}=0m[/tex]
Final position: [tex]x_{f}=240m[/tex]
a) We can use velocity's equation:
[tex]v_{f}^{2} = v_{i}^{2} +2a(x_{f}-x_{i})[/tex]
⇒ [tex]a=\frac{v_{f}^{2}-v_{i}^{2}}{2(x_{f}-x_{i})}[/tex]
⇒ [tex]a=2.31\frac{m}{s^{2} }[/tex].
b) For this, equation for average acceleration will be helpful. Taking [tex]t_{i}=0[/tex] and having [tex]t_{f}[/tex] as the unknown time it becomes airborne:
[tex]a=\frac{v_{f}-v_{i}}{t_{f}-t_{i}} =\frac{v_{f} }{t_{f}}[/tex]
⇒ [tex]t_{f}=\frac{v_{f}}{a}=\frac{33.3\frac{m}{s}}{2.31\frac{m}{s^{2}}}[/tex]
⇒ [tex]t_{f}=14.41s[/tex].
