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From similarity of two right-triangles
(fig 1.13)

a n v x v x
  a  g 
n
g v v
and

m
10
m s m
a  10   5 , 3
n
s 2 m s 2
28 3 ,
s
Figure 1.13 after that

m 2 m 2 m
2
2
a  g  a  ( 10 )  5 , 3 ( )  4 , 9 .
n

s 2 s 2 s 2
So absolute value of normal acceleration is equal to
v 2
a 
n
R
We can define the radius R of curvature of trajectory
m 2
2 ( 28 3 , )
v s
R    229 m
a n 5 , 3 m

s 2

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