Explanation:
It is given that,
Initial speed of the automobile, u = 65 km/hr =
Final speed of the automobile, v = 0
Deceleration of the automobile, [tex]a=-6\ m/s^2[/tex]
We need to find the distance covered by the car as it comes to rest. It can be calculated using third equation of motion as :
[tex]v^2-u^2=2ax[/tex]
[tex]a=\dfrac{v^2-u^2}{2x}[/tex]
[tex]a=\dfrac{0-(18.05)^2}{2\times (-6)}[/tex]
[tex]a=27.15\ m/s^2[/tex]
So, the acceleration of the car is [tex]27.15\ m/s^2[/tex]. Hence, this is the required solution.
