Respuesta :
Given Data:
*The initial speed of the motorcycle is:
[tex]u=28.1\text{ ft/s}[/tex]*The final speed of the motorcycle is:
[tex]v=53.5\text{ ft/s}[/tex]*The distance is:
[tex]s=181\text{ ft}[/tex]Explanation:
Using the third kinematics equation of motion, we get:
[tex]2as=v^2-u^2[/tex]Substituting the known values, we get:
[tex]\begin{gathered} (53.5\text{ ft/s})^2-(28.1\text{ ft/s})^2=2a(181\text{ ft}) \\ 2862.25\text{ ft}^2\text{/s}^2-789.61\text{ ft}^2\text{/s}^2=a(362\text{ ft}) \\ 2072.64\text{ ft}^2\text{/s}^2=a(362\text{ ft}) \\ a=\frac{2072.64\text{ ft}^2\text{/s}^2}{362\text{ ft}} \\ a=5.726\text{ ft/s}^2 \end{gathered}[/tex]Using the first kinematics equation, we get:
[tex]v=u+at[/tex]Substituting the known values, we get:
[tex]\begin{gathered} 53.5\text{ ft/s = 28.1 ft/s +\lparen5.726 ft/s}^2\text{\rparen t} \\ 25.4\text{ ft/s = \lparen5.726 ft/s}^2)t \\ t=\frac{25.4\text{ ft/s}}{\text{5.726 ft/s}^2} \\ t=4.436\text{ s} \end{gathered}[/tex]Final Answer:
The time elapsed while the motorcycle moves this distance is:
[tex]4.436\text{ s}[/tex]