Let the initial velocity of the car be u.
Final velocity of the car (v) = 5.43 m/s
deceleration (a) = - 2.78 m/s^2
Time taken (t) = 2.26 s
Using the first equation of motion:
v = u + at
u = v - at
![u = 5.43 - (-2.78 * 2.26)](https://img.qammunity.org/2019/formulas/physics/high-school/1qw622x17vvj3hs86ev1szxlufe8giig83.png)
u = 5.43 + 6.28
u = 11.71 m/s
Let the car's displacement be x.
Using second equation of motion:
![x = ut + (1)/(2)at^2](https://img.qammunity.org/2019/formulas/physics/high-school/yrxv9oesdwtrwi2d3wvgxtmvhc6c7br9j0.png)
![x = 11.71 * 2.26 - (1)/(2) * 2.78 * 2.26^2](https://img.qammunity.org/2019/formulas/physics/high-school/ralftw0yyq5f828r2rfj7q8ilu1xav6zme.png)
x = 26.4646 - 7.0995
x = 19.3651 meters
Hence, the displacement of the car is 19.36 meters