Answer:
Vmax = 12.21m/s
[tex]a = 4.07m/s^{2}[/tex]
Explanation:
For the first 3 seconds:
[tex]V_{3}=V_{o}+a*t=0+a*3=3a[/tex] This is the maximum speed. But we need acceleration.
[tex]X_{3}=V_{o}*t+\frac{a*t^{2}}{2} =\frac{9*a}{2}[/tex]
For the other 6.69s with constant speed:
[tex]X_{t}=100=X_{3}+V_{3}*t[/tex]
[tex]100=\frac{9*a}{2} +3*a*6.69[/tex] Solving for a:
[tex]a = 4.07m/s^{2}[/tex] Now we replace this value on [tex]V_{3}=3a[/tex]:
[tex]V_{3}=V_{max}=12.21m/s[/tex]
