Answer:
D. 4 m
Explanation:
According to the work-energy theorem, the work done by the force acting on a body modifies its kinetic energy.
[tex]W=\Delta K\\W=K_f-K_i\\W=\frac{mv_f^2}{2}-\frac{mv_i^2}{2}[/tex]
The car comes to rest after a given distance, so [tex]v_f=0[/tex]. Recall the definition of work [tex]W=Fdcos\theta[/tex], here F is the constant force acting on the car, d is its traveled distance and [tex]\theta[/tex] is the angle between the force and the displacement, since friction force acts opposite to the direction of motion [tex]\theta=180^\circ[/tex] :
[tex]-Fd=-\frac{mv_i^2}{2}\\d=\frac{mv_i^2}{2F}[/tex]
We have:
[tex]v_i'=1\frac{m}{s}\\v_i=0.5\frac{m}{s}\\v_i'=2vi[/tex]
Thus:
[tex]d'=\frac{2m(v_i')^2}{2F}\\d'=\frac{2m(2v_i)^2}{2F}\\d'=4\frac{2mv_i^2}{2F}\\d'=4d\\d'=4(1m)\\d'=4m[/tex]
