Respuesta :
As per the question the initial speed of the car [ u] is 42 m/s.
The car applied its brake and comes to rest after 5.5 second.
The final velocity [v] of the car will be zero.
From the equation of kinematics we know that
[tex]v=u+at[/tex] [ here a stands for acceleration]
[tex]0=42 +5.5a[/tex]
[tex]a =\frac{-42}{5.5} m/s^2[/tex]
[tex]a= -7.64 m/s^2[/tex]
Here a is taken negative as it the car is decelerating uniformly.
We are asked to calculate the stopping distance .
From equation of kinematics we know that
[tex]S=ut+\frac{1}{2} at^2[/tex] [here S is the distance]
[tex]= 42*5.5 +\frac{1}{2} [-7.64] [5.5]^2 m[/tex]
[tex]=115.445 m[/tex] [ans]
