Given data
*The given distance is s = 8.13 km = 8130 m
*The given time is t = 33.5 s
*The given acceleration is a = 4.6 m/s^2
(1)
The formula for the speed at the first beginning of 33.5 seconds is given as
[tex]s=ut+\frac{1}{2}at^2[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} 8130=u(33.5)+\frac{1}{2}(4.6)(33.5)^2 \\ u=165.6\text{ m/s} \end{gathered}[/tex]Hence, the speed at the first beginning of 33.5 s is u = 165.6 m/s
(2)
The formula for the speed at the end of the 33.5 s is given as
[tex]v=u+at[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} v=(165.6)+(4.6)(33.5)_{} \\ =319.7\text{ m/s} \end{gathered}[/tex]Hence, the speed at the end of the 33.5 s is v = 319.7 m/s
