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In 1977 off the coast of Australia, the fastest speed by a vessel on the water
was achieved. If this vessel were to undergo an average acceleration of
1.80 m/s-, it would go from rest to its top speed in 85.6 s. What was the
speed of the vessel?

Respuesta :

cryssatemp

Answer: 154.08 m/s

Explanation:

Average acceleration [tex]a_{ave}[/tex] is the variation of velocity  [tex]\Delta V[/tex] over a specified period of time  [tex]\Delta t[/tex]:

[tex]a_{ave}=\frac{\Delta V}{\Delta t}}[/tex]

Where:

[tex]a_{ave}=1.80 m/s^{2}[/tex]

[tex]\Delta V=V_{f}-V_{o}[/tex] being [tex]V_{o}=0[/tex] the initial velocity and [tex]V_{f}[/tex] the final velocity

[tex]\Delta t=85.6 s[/tex]

Then:

[tex]a_{ave}=\frac{V_{f}-V_{o}}{\Delta t}}[/tex]

Since [tex]V_{o}=0[/tex]:

[tex]a_{ave}=\frac{V_{f}}{\Delta t}}[/tex]

Finding [tex]V_{f}[/tex]:

[tex]V_{f}=a_{ave} \Delta t[/tex]

[tex]V_{f}=(1.80 m/s^{2})(85.6 s)[/tex]

Finally:

[tex]V_{f}=154.08 m/s[/tex]

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