Question

6. A car, 1110 kg, is traveling down a horizontal road at 20.0 m/s when it locks up its brakes. The coefficient of friction between the tires and road is 0.901. How much distance will it take to bring the car to a stop?

Answer 1

Answer:

$x=22.65m$

Explanation:

We have an uniformly accelerated motion, with a negative acceleration. Thus, we use the kinematic equations to calculate the distance will it take to bring the car to a stop:

$v_f^2=v_0^2-2ax\\\frac{0^2-v_0^2}{-2a}=x\\x=\frac{v_0^2}{2a}$

The acceleration can be calculated using Newton's second law:

$\sum F_x:F_f=ma\\\sum F_y:N=mg$

Recall that the maximum force of friction is defined as $F_f=\mu N$. So, replacing this:

$\mu N=ma\\\mu mg=ma\\a=\mu g\\a=0.901(9.8\frac{m}{s^2})\\a=8.83\frac{m}{s^2}$

Now, we calculate the distance:

$x=\frac{v_0^2}{2a}\\x=\frac{(20\frac{m}{s})^2}{2(8.83\frac{m}{s^2})}\\x=22.65m$