To solve this problem we must apply Torricelli equation:
[tex]v^2=v^2_o+2\times\alpha\times d^{}_{}[/tex]
where:
v = final velocity
vo = initial velocity
α = acceleration
d = distance
Lets use the information of the first row of the table to find a:
[tex]\begin{gathered} v^2=v^2_o+2\times\alpha\times d^{}_{} \\ 0^2=60^2+2\times\alpha\times240 \\ 0=3,600+480\times\alpha \\ -480\times\alpha=3,600 \\ \alpha=\frac{3,600}{-480} \\ \alpha=-7.5ft/s^2 \end{gathered}[/tex]
Now we can calculate distances a, b and c:
[tex]\begin{gathered} v^2=v^2_{^{}o^{}}+2\times\alpha\times d \\ 0^2=40^2+2\times(-7.5)\times d \\ 0=1,600-15\times d \\ 15\times d=1,600 \\ d=\frac{1,600}{15} \\ d\approx107ft \end{gathered}[/tex]
So distance a = 107 ft.
Now we have:
[tex]\begin{gathered} v^2=v^2_o+2\times\alpha\times d \\ 0^2=30^2+2^{}\times(-7.5)\times d \\ 0=900-15\times d \\ 15\times d=900 \\ d=\frac{900}{15} \\ d=60ft \end{gathered}[/tex]
So distance b = 60 ft.
Finally:
[tex]\begin{gathered} v^2=v^2_o+2\times\alpha\times d \\ 0^2=10^2+2\times(-7.5)\times d \\ 0=100-15\times d \\ 15\times d=100 \\ d=\frac{100}{15} \\ d\approx7ft \end{gathered}[/tex]
So distance c = 7 ft.
