We use the kinematic equations,
[tex]v=u+at[/tex] (A)
[tex]S= ut + \frac{1}{2} at^2[/tex] (B)
Here, u is initial velocity, v is final velocity, a is acceleration and t is time.
Given, [tex]u=0[/tex], [tex]a=0.21 \ m/s^2[/tex] and [tex]s= 280 m[/tex].
Substituting these values in equation (B), we get
[tex]280 \ m = 0 +\frac{1}{2} (0.21 m/s^2) t^2 \\\\ t^2 = \frac{280 \times 2}{0.21 } \\\\ t= 51.63 \ s[/tex].
Therefore from equation (A),
[tex]v = 0 + (0.21) \times (51.63 s)= 10.84 \ m/s[/tex]
Thus, the magnitude of the boat's final velocity is 10.84 m/s and the time taken by boat to travel the distance 280 m is 51.63 s
The boat's final velocity is 10.84 m/s and it takes 51.6 seconds for the boat to travel this distance.
The sail boat starts from rest, hence the initial velocity (u) = 0, the acceleration (a) = 0.21 m/s², the distance (s) = 280 m.
To find the boat final velocity, use the equation:
v² = u² + 2as
v² = 0² + 2(0.21)(280)
v² = 117.6
v = 10.84 m/s
The time taken is given as:
v = u + at
10.84 = 0 + 0.21t
t = 51.6 seconds
Hence the boat's final velocity is 10.84 m/s and it takes 51.6 seconds for the boat to travel this distance.
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