Your diagram probably shows a picture of a circle.
a. The velocity vector points in a direction that is always tangent to the circle.
b. The acceleration vector points toward the center of the circle.
c. In uniform circular motion, we have
[tex]a=\dfrac{v^2}r=\dfrac{\left(13.5\,\frac{\rm m}{\rm s}\right)^2}{7.1\,\rm m}=25.7\,\dfrac{\rm m}{\mathrm s^2}[/tex]
d. With [tex]g=9.80\,\dfrac{\rm m}{\mathrm s^2}[/tex], we have
[tex]14.5g=142\,\dfrac{\rm m}{\mathrm s^2}[/tex]
e. The required velocity [tex]v[/tex] satisfies
[tex]14.5g=\dfrac{v^2}{7.1\,\rm m}\implies v=\sqrt{102.95g}=31.8\,\dfrac{\rm m}{\rm s}[/tex]
a) The velocity vector is always perpendicular to the radius of the circle
b) Since centripetal acceleration makes the velocity vector rotate, a.cent is directed to the center of the circle along its radius.
c) a=v^2/R=13.5^2/7.1=25.67 m/s/s
d) a.max=14.5*9.8=142.1 m/s/s
e) [tex]v=\sqrt{a.max*R}=\sqrt{142.1*7.1}=31.76[/tex] m/s
