A boat takes off from a dock at 2.5 m/s and speeds up at 4.2 m/s squared for six seconds how far has the most traveled

Respuesta :

The boat's position [tex]x[/tex] relative to its starting point [tex]x_0=0[/tex] is determined by

[tex]x=x_0+v_0t+\dfrac12at^2[/tex]

where [tex]v_0[/tex] is its initial velocity, [tex]a[/tex] is its acceleration, and [tex]t[/tex] is time. After [tex]t=6\,\mathrm s[/tex], the boat has traveled

[tex]x=\left(2.5\,\dfrac{\mathrm m}{\mathrm s}\right)(6\,\mathrm s)+\dfrac12\left(4.2\,\dfrac{\mathrm m}{\mathrm s^2}\right)(6\,\mathrm s)^2[/tex]

[tex]\implies x=91\,\mathrm m[/tex]

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