Question

Barney was driving at 25 mph. A car pulls out 30 ft ahead of him. The coefficient of friction, f, is 0.7. Which statement is true? A Barney hits the car. B Barney stops less than a foot from the car. C Barney misses the car by 3 ft. D Barney's skid marks measure 23 ft.

Answer 1

Answer:

B Barney stops less than a foot from the car.

Step-by-step explanation:

From the information given:

The initial speed of barney $v_o$ = 25 mph

Coefficient of friction $\mu$ = 0.7

Recall that

1 mph = 0.447 m/s

Then

$v_o$ = 25 × 0.447 m/s

$v_o$ = 11.175 m/s

The force of friction acting on the barney is expressed as:

$F = \mu N \\ \\ F = \mu mg$

$ma = \mu mg$

$a = \mu g$

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;

$v^2 - v_o^2 = 2as$

$0 - (11.17)^2 = 2 \times (-6.86) s$

$-124.76 = -13.72 s$

$s = \dfrac{124.76}{13.72}$

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.