Answer:
9.22 s
Explanation:
One-quarter of a turn away is 1/4 of 2π, or π/2 which is approximately 1.57 rad
Let t (seconds) be the time it takes for the child to catch up with the horse. We would have the following equation of motion for the child and the horse:
For the child: [tex]s_c = \omega_ct = 0.233t[/tex]
For the horse: [tex]s_h = s_0 + a_ht^2/2 = 1.57 + 0.0136t^2/2 = 1.57 + 0.0068t^2[/tex]
For the child to catch up with the horse, they must cover the same angular distance within the same time t:
[tex]s_c = s_h[/tex]
[tex]0.233t = 1.57 + 0.0068t^2[/tex]
[tex]0.0068t^2 - 0.233t + 1.57 = 0[/tex]
[tex]t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]
[tex]t= \frac{0.233\pm \sqrt{(-0.233)^2 - 4*(0.0068)*(1.57)}}{2*(0.0068)}[/tex]
[tex]t= \frac{0.233\pm0.11}{0.0136}[/tex]
t = 25.05 or t = 9.22
Since we are looking for the shortest time we will pick t = 9.22 s
