Explanation:
It is given that,
Initial speed of the car, u = 57.7 km/hr = 16.02 m/s
Distance between the car and the barrier, d = 22.2 m
The car hits the barrier 2.14 s later.
Final speed of the car, v = 0 (at rest)
Let a is the acceleration of the car. It can be calculated using the second equation of motion as :
[tex]d=ut+\dfrac{1}{2}at^2[/tex]
[tex]a=\dfrac{2(d-ut)}{t^2}[/tex]
[tex]a=\dfrac{2(22.2-16.02\times 2.14)}{(2.14)^2}[/tex]
[tex]a=-5.27\ m/s^2[/tex]
As the car decelerates, the acceleration of the car just before the impact is [tex]-5.27\ m/s^2[/tex]. Hence, this is the required solution.
