The stopping distance is 45.0 m
Explanation:
First of all, we find the acceleration of the truck, by using Newton's second law:
[tex]F=ma[/tex]
where
[tex]F=-1.26\cdot 10^4 N[/tex] is the net force on the truck
[tex]m=2.3\cdot 10^3 kg[/tex] is the mass of the truck
a is its acceleration
Solving for a,
[tex]a=\frac{F}{m}=\frac{-1.26\cdot 10^4}{2.3\cdot 10^3}=-5.48 m/s^2[/tex]
where the negative sign means the acceleration is opposite to the direction of motion.
Now, since the motion of the truck is at constant acceleration, we can apply the following suvat equation:
[tex]v^2-u^2=2as[/tex]
where
v = 0 is the final velocity of the truck
u = 22.2 m/s is the initial velocity
[tex]a=-5.48 m/s^2[/tex] is the acceleration
s is the stopping distance
And solving for s,
[tex]s=\frac{v^2-u^2}{2a}=\frac{0-(22.2)^2}{2(-5.48)}=45.0 m[/tex]
Learn more about accelerated motion:
brainly.com/question/9527152
brainly.com/question/11181826
brainly.com/question/2506873
brainly.com/question/2562700
#LearnwithBrainly
