Question

A boat with initial position (5i -2j+ 4k) metres relative to port, acceleratesuniformly from initial velocity (4i+ j-3k) m s^-1, for 8 seconds until reaching finalvelocity (12i-7j+13k) m s^-1.a)Find the position of the object after 8 seconds.b) Find the acceleration of the object.

Answer 1

Given,

The initial position of the boat, d_i=(5i-2j+4k) m

The initial velocity of the boat, u=(4i+j-3k) m/s

The time period, t=8 s

The final velocity of the boat, v=(12i-7j+13k) m/s

To find:

a) The position of the object after 8 s

b) The acceleration of the object.

Explanation:

a)

From the equation of motion, the total distance traveled by the object is given by,

$d=\frac{1}{2}(v+u)t$

On substituting the known values,

$\begin{gathered} d=\frac{1}{2}[(12\hat{i}-7\hat{j}+13\hat{k)}+(4\hat{i}+\hat{j}-3\hat{k})]8 \\ =4\times(16\hat{i}-6\hat{j}+10\hat{k}) \\ =(64\hat{i}-24\hat{j}+40\hat{k})\text{ m} \end{gathered}$

Thus the final position of the boat is,

$\begin{gathered} d_f=d_i+d \\ =\left(5\hat{i}-2\hat{j}+4\hat{k}\right)+(64\hat{i}-24\hat{j}+40\hat{k}) \\ =(69\hat{i}-26\hat{j}+44\hat{k})\text{ m} \end{gathered}$

b)

The acceleration of an object is given by,

$a=\frac{v-u}{t}$

On substituting the known values,

$\begin{gathered} a=\frac{(12\hat{i}-7\hat{j}+13\hat{k)}-(4\hat{i}+\hat{j}-3\hat{k})}{8} \\ =\frac{8\hat{i}-8\hat{j}+16\hat{k}}{8} \\ =(\hat{i}-\hat{j}+2\hat{k})\text{ m/s}^2 \end{gathered}$

Final answer:

a) Thus the position of the boat after 8 s is

$\begin{equation*} (69\hat{i}-26\hat{j}+44\hat{k})\text{ m} \end{equation*}$

The acceleration of the boat is

$\begin{equation*} (\hat{i}-\hat{j}+2\hat{k})\text{ m/s}^2 \end{equation*}$