Question

A boat travels 12.0 m while it reduces its velocity from 9.5 m/s to 5.5 m/s. What is the magnitude of the boat’s acceleration while it travels the 12.0 m?

Answer 1

Answer:

-2.5 m/s²

Explanation:

As per the provided information in the given question, we have :

• Initial velocity (u) = 9.5 m/s
• Final velocity (v) = 5.5 m/s
• Distance covered (s) = 12 m

We are asked to calculate the boat's acceleration. By using the third equation of motion,

$\longmapsto\bf {v^2 - u^2 = 2as}$

• v denotes final velocity
• u denotes initial velocity
• a denotes acceleration
• s denotes distance

$\longmapsto\rm {(5.5)^2 - (9.5)^2 = 2\times a \times 12}$

$\longmapsto\rm {30.25 - 90.25= 24 \times a }$

$\longmapsto\rm {-60= 24 \times a }$

$\longmapsto\rm {\dfrac{-60}{24}= a }$

$\longmapsto\bf {-2.5 \; ms^{-2}= a }$

∴ Acceleration of the boat is -2.5 m/s². (Negative sign shows that it is retardation.)

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More Information :

Equations of motion :

$\boxed{ \begin{array}{cc} \pmb{\sf{ \quad \: v = u + at \quad}} \\ \\ \pmb{\sf{ \quad \: s= ut + \cfrac{1}{2}a{t}^{2} \quad \: } } \\ \\ \pmb{\sf{ \quad \: {v}^{2} - {u}^{2} = 2as \quad \:}}\end{array}}$

• v denotes final velocity
• u denotes initial velocity
• a denotes acceleration
• t denotes time
• s denotes distance

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