Given data:
* The initial velocity given is 27.3 m/s.
* The final velocity given is 46.4 m/s.
* The value of time given is 11 second.
Solution:
The change in velocity is,
[tex]\begin{gathered} \Delta v=v_f-v_i_{} \\ \Delta v=46.4-27.3 \\ \Delta v=19.1ms^{-1} \end{gathered}[/tex]
Thus, the value of acceleration is,
[tex]a=\frac{\Delta v}{t}[/tex]
Substituting the known values,
[tex]\begin{gathered} a=\frac{19.1}{11} \\ a=1.74ms^{-2} \end{gathered}[/tex]
Thus, the value of acceleration is 1.74 meter per second squared.
In the table form, the final answers are,
