Answer:
Braking distance of vehicle is 9.97 m
Explanation:
It is given initial velocity of the vehicle u = 13.4 m/sec
Acceleration of the vehicle [tex]a=-9m/sec^2[/tex]
As the vehicle finally stops so final velocity of the vehicle v = 0 m/sec
We have to find the braking distance s
From third equation of motion
[tex]v^2=u^2+2as[/tex]
[tex]0^2=13.4^2+2\times -9\times s[/tex]
s = 9.97 m
So braking distance of the vehicle is 9.97 m
