Answer
given,
speed of the car = 18 m/s
reaches at the end of ramp = 120 m
initial velocity = 0 m/s
Using kinematic relation
a) v² = u² + 2 a s
[tex]a = \dfrac{v^2-u^2}{2 s}[/tex]
[tex]a = \dfrac{18^2-0^2}{2\times 120}[/tex]
  a = 2.7 m/s²
b) using kinematic equation
v = u + at
[tex]t = \dfrac{v}{a}[/tex]
[tex]t = \dfrac{18}{2.7}[/tex]
t = 6.67 s
c) v = 23 m/s
a = 0
[tex]d = v t +\dfrac{1}{2}at^2[/tex]
[tex]d = 23\times 6.67 +0[/tex]
d = 153.33 m