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Sagot :
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,
[tex]d=\frac{1}{2}(v+u)t[/tex]On substituting the known values,
[tex]\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}[/tex]Thus the final position of the boat is,
[tex]\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}[/tex]b)
The acceleration of an object is given by,
[tex]a=\frac{v-u}{t}[/tex]On substituting the known values,
[tex]\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}[/tex]Final answer:
a) Thus the position of the boat after 8 s is
[tex]\begin{equation*} (69\hat{i}-26\hat{j}+44\hat{k})\text{ m} \end{equation*}[/tex]The acceleration of the boat is
[tex]\begin{equation*} (\hat{i}-\hat{j}+2\hat{k})\text{ m/s}^2 \end{equation*}[/tex]
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