Answer:
B Barney stops less than a foot from the car.
Step-by-step explanation:
From the information given:
The initial speed of barney [tex]v_o[/tex] = 25 mph
Coefficient of friction [tex]\mu[/tex] = 0.7
Recall that
1 mph = 0.447 m/s
Then
[tex]v_o[/tex] = 25 × 0.447 m/s
[tex]v_o[/tex] = 11.175 m/s
The force of friction acting on the barney is expressed as:
[tex]F = \mu N \\ \\ F = \mu mg[/tex]
[tex]ma = \mu mg[/tex]
[tex]a = \mu g[/tex]
a = 0.7 × 9.8 m/s
a = 6.86 m/s²
Since the force acts in an opposite direction to the motion of Barney, Thus the acceleration must also be negative.
a = -6.86 m/s²
Now;
Suppose Barney stops after traveling a distance (s), so that the final speed is zero.
Then;
Using the kinematic equation;
[tex]v^2 - v_o^2 = 2as[/tex]
[tex]0 - (11.17)^2 = 2 \times (-6.86) s[/tex]
[tex]-124.76 = -13.72 s[/tex]
[tex]s = \dfrac{124.76}{13.72}[/tex]
s = 9.09 m
Recall that;
1 m = 3.28 ft
So;
s = 9.09 m = 9.09 × 3.28 ft
s = 29.81 ft
∴
The distance traveled by the Barney is 29.81 ft which implies that Barney stops less than a foot from the car.
