Answer:
[tex]0.51\ \text{m/s}^2[/tex]
Explanation:
t = Time taken
g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]
r = Radius of track = 0.355 km
Displacement in [tex]y[/tex] direction is 3 m
[tex]y=ut+\dfrac{1}{2}gt^2\\\Rightarrow 3=0+\dfrac{1}{2}\times 9.81 t^2\\\Rightarrow t=\sqrt{\dfrac{3\times 2}{9.81}}\\\Rightarrow t=0.782\ \text{s}[/tex]
Displacement in [tex]x[/tex] direction
[tex]x=10.5\ \text{m}[/tex]
[tex]v=\dfrac{x}{t}\\\Rightarrow v=\dfrac{10.5}{0.782}\\\Rightarrow v=13.43\ \text{m/s}[/tex]
Centripetal acceleration is given by
[tex]a_c=\dfrac{v^2}{r}\\\Rightarrow a_c=\dfrac{13.43^2}{355}\\\Rightarrow a_c=0.51\ \text{m/s}^2[/tex]
The minimum centripetal acceleration the truck must have is [tex]0.51\ \text{m/s}^2[/tex]
