Answer:
15.825 m
Explanation:
t = Time taken = 2.5 s
u = Initial velocity = 6.75 m/s
v = Final velocity = 5.91 m/s
s = Displacement
a = Acceleration
Equation of motion
[tex]v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{5.91-6.75}{2.5}\\\Rightarrow a=-0.336\ m/s^2[/tex]
[tex]v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{5.91^2-6.75^2}{2\times -0.336}\\\Rightarrow s=15.825\ m[/tex]
The distance Rickey slides across the ground before touching the base is 15.825 m
