Static Friction

Question

27. The coefficient of static friction between car’s tires and a level road is 0.80. If the car is to be stopped in a maximum time of 3.0 s, its maximum speed is (a) 2.4 m/s (b) 23.5 m/s (c) 7.8 m/s (d) 2.6 m/s

Perhaps an easy question for some, but I simply do not understand how to apply the formula for static equilibrium in this case without being given any mass or force.

Please help out!

Answer 1

\begin{align*} ma &= -\mu mg \\ a &= -\mu g \\ \frac{v-u}{t} &= -\mu g \\ \frac{0-u}{3} &= -0.8 \times 9.8 \\ u &= 23.5 \end{align*}

Answer 2

If $M$ is the mass of the car, the frictional force is $\mu M g$, where $\mu$ is the coefficient of friction. This is the only force in horizontal direction. Therefore, if $a$ is the acceleration then $Ma = \mu M g$ or $a = \mu g$. If $u$ is the initial velocity and $v$ is the velocity after time $t$ then $v = u + at$. Since the car stops after $t = 3.0$ seconds, $v = 0$ at that time. Therefore the maximum $u$ is $|at| = |\mu g t|$. Plugging in the values gives $u = 23.52$ meters per second.

Answer 3

The acceleration will be given by $$ \frac{F_f}{m} = a = \frac{v_f -v_i}{\Delta t} $$ or since $\mu =\frac{F_f}{F_n}$ $$ \frac{\mu F_n}{m} = \frac{v_f-v_i}{\Delta t} $$ then $$ \mu g = \frac{v_f-v_i}{\Delta t} $$ or $$ v_i = - \mu g \Delta t $$