Question

The boat is traveling along the circular path with a speed of v=(0.0625t2) m/s, where t is in seconds. (Figure 1)

Part A

If ? = 42 m , determine the magnitude of its acceleration when t = 12 s .

Express your answer to three significant figures and include the appropriate units.

Answer 1

Concepts and Reason

When an object is moving in a circle, it has two components of acceleration. One component is along the tangent of the circle and the other is towards the center of the circle. They are called tangential and radial/normal acceleration respectively.

The acceleration of the object is the resultant of the radial/normal and tangential accelerations.

Fundamentals

The tangential acceleration follows the laws of motion and is given by,

${a_t} = \frac{{dv}}{{dt}}$

Here, the tangential velocity is v and the time is t.

The normal acceleration is directed towards the center of rotation and is given by,

${a_n} = \frac{{{v^2}}}{R}$

Here, the radius of the circle is R.

Calculate the tangential velocity when $t = 12{\rm{ s}}$.

$\begin{array}{c}\\{\left. v \right|_{t = 12}} = 0.0625{\left( {12} \right)^2}\\\\ = 9{\rm{ m/s}}\\\end{array}$

Calculate the normal acceleration at this instant.

${a_n} = \frac{{{{\left( {{{\left. v \right|}_{t = 12}}} \right)}^2}}}{\rho }$

Here, the radius of the circular path is $\rho$.

Substitute 9 m/s for ${\left. v \right|_{t = 12}}$ and 42 m for $\rho$.

$\begin{array}{c}\\{a_n} = \frac{{{{\left( 9 \right)}^2}}}{{42}}\\\\ = 1.9286{\rm{ m/}}{{\rm{s}}^2}\\\end{array}$

Calculate the tangential acceleration of the boat when $t = 12{\rm{ s}}$.

${a_t} = \frac{{dv}}{{dt}}$

Substitute $0.0625{t^2}$ for v.

$\begin{array}{c}\\{a_t} = \frac{{d\left( {0.0625{t^2}} \right)}}{{dt}}\\\\ = 0.125t\\\end{array}$

$\begin{array}{c}\\{\left. {{a_t}} \right|_{t = 12}} = 0.125 \times 12\\\\ = 1.5{\rm{ m/}}{{\rm{s}}^2}\\\end{array}$

Calculate the total acceleration of the boat.

$a = \sqrt {a_n^2 + a_t^2}$

Substitute $1.9286{\rm{ m/}}{{\rm{s}}^2}$ for ${a_n}$ and $1.5{\rm{ m/}}{{\rm{s}}^2}$ for ${a_t}$.

$\begin{array}{c}\\a = \sqrt {{{1.9286}^2} + {{1.5}^2}} \\\\ = 2.44{\rm{ m/}}{{\rm{s}}^2}\\\end{array}$

Ans:

The magnitude of acceleration of the boat when $t = 12{\rm{ s}}$ is $2.44{\rm{ m/}}{{\rm{s}}^2}$.