D1 = distance 1 = 11.10 m/s t
D2 = 10m + 11.10 m/s t
D2 = u2 t + 1/2 a2 t^2
u2= speed 2 = 9.50m/s
10m + 11.10 m/s t = 9.50 m/s t + 1/2 (1.20 m/s^2) t^2
Solve for t
10m + 11.10 m/s t = 9.50 m/s t + 0.6 m/s^2 t^2
10 + 11.10 t = 9.5 t + 0.6 t^2
0 = 0.6t^2 - 1.6t - 10
Apply quadratic equation:
[tex]t=\frac{-b\pm\sqrt[\placeholder{⬚}]{b^2-4\times a\times c}}{2a}[/tex][tex]t=\frac{1.6\pm\sqrt[\placeholder{⬚}]{(-1.6)^2-4(0.6)(-10)}}{2(0.6)}[/tex][tex]t=\frac{1.6\pm\sqrt[\placeholder{⬚}]{26.56}}{1.2}[/tex][tex]t=\frac{1.6+5.15}{1.2}=5.63\text{ s}[/tex]Answers: 5.63 sec
