Answer:
[tex]t =4.8sec[/tex]
Explanation:
From the question we are told that
Velocity of speed boat [tex]V=29.0m/s[/tex]
Distance to Marker [tex]d=100[/tex]
Acceleration of [tex]a=-3.4m/s^2[/tex]
Generally the Newtons 3rd motion equation is given as
[tex]v^2 = u^2 + 2 * a* s[/tex]
[tex]v^2 = 29^2 + 2 * -3.4* 100[/tex]
[tex]v = \sqrt{161}[/tex]
[tex]v=12.68m/s[/tex]
Generally the Newton's first equation of motion is given as
[tex]v = u + a*t[/tex]
[tex]12.68 = 29 -3.4*t[/tex]
[tex]12.68-29 = -3.4t[/tex]
[tex]-16.32 = -3.4t[/tex]
[tex]t =\frac{-16.32}{-3.4}[/tex]
[tex]t =4.8sec[/tex] .
