Answer
The graphs are as follow:
Work Step by Step
We have given acceleration $a=-5.2m/s^2$ and initial velocity $u=137km/h=\frac{685}{18}m/s$.
Using equation of motion $x=ut+\frac{1}{2}at^2$.
We get the equation for this slowing $x=\frac{685}{18}t-\frac{1}{2}\times5.2t^2$ or $x=\frac{685}{18}t-2.6t^2$.
By ploting this equation we get the x versus t graph given below.
We have also given final velocity $v=90km/h=\frac{450}{18}m/s$.
Using the equation of motion $v=u+at$.
We get the equation for this slowing $v=\frac{685}{18}-5.2t$.
This is an equation of line with points $(0,\frac{685}{18})$ and $(2.51,\frac{450}{18})$.
By joining these end points we get the v versus t graph given below.
