Given:
he initial velcicty of the runner is
[tex]v_{ir}=\text{ 0 m/s}[/tex]The final velocity of the runner is
[tex]v_{fr}=\text{ 2.6 m/s}[/tex]The time taken is t = 2.1 s
The initial velocity of the motorcycle is
[tex]v_{im}=\text{ 37 m/s}[/tex]The final velocity of the motorcycle is
[tex]v_{fm}=\text{ 44 m/s}[/tex]Required:
Acceleration of the motorcycle.
The difference in distance between runner and motorcycle.
Explanation:
cceleration of the runner is
[tex]\begin{gathered} a_r=\frac{v_{fr}-v_{ir}}{t} \\ =\frac{2.6-0}{2.1} \\ =1.238\text{ m/s}^2 \end{gathered}[/tex]Acceleration of the motorcycle is
[tex]\begin{gathered} a_m=\frac{v_{fm}-v_{im}}{t} \\ =\frac{44-37}{2.1} \\ =3.33\text{ m/s}^2 \end{gathered}[/tex]The distance travelled by the runner is
[tex]\begin{gathered} S_r=v_{ir}t+\frac{1}{2}a_rt^2 \\ =0\times2.1+\frac{1}{2}\times1.238\times(2.1)^2 \\ =2.73\text{ m} \end{gathered}[/tex]The distance travelled by motorcycle is
[tex]\begin{gathered} S_m=v_{im}t+\frac{1}{2}a_mt^2 \\ =37\times2.1+\frac{1}{2}\times3.33\times(2.1)^2 \\ =77.7+7.34 \\ =85.04\text{ m} \end{gathered}[/tex]Thus, the motorcycle travels more distance than the runner.
The
difference between the distance travelled is
[tex]\begin{gathered} \Delta S=S_m-S_r \\ =85.04-2.73 \\ =82.31\text{ m} \end{gathered}[/tex]Final Answer:
The acceleration of the runner is 1.238 m/s^2
The acceleration of the motorcycle is 3.33 m/s^2
The difference in distance between runner and motorcycle is 82.31 m.
