Brake or turn? Figure 6- 44 depicts an overhead view of a car’s path as the car travels toward a wall. Assume that the driver begins to brake the car when the distance to the wall is d = 107 m, and take the car’s mass as m = 1400kg, its initial speed as v₀ = 35m/s, and the coefficient of static friction as ${\mathit{\mu }}_{\mathbf{s}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{50}$. Assume that the car’s weight is distributed evenly on the four wheels, even during braking.

(a) What magnitude of static friction is needed (between tires and road) to stop the car just as it reaches the wall?

(b) What is the maximum possible static friction f_s,max?

(c) If the coefficient of kinetic friction between the (sliding) tires and the road is${\mathit{\mu }}_{\mathbf{k}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{40}$, at what speed will the car hit the wall? To avoid the crash, a driver could Select to turn the car so that it just barely misses the wall, as shown in the figure.

(d) What magnitude of frictional force would be required to keep the car in a circular path of radius dand at the given speed v₀, so that the car moves in a quarter circle and then parallel to the wall?

(e) Is the required force less than f_s.maxso that a circular path is possible?

• Distance to wall, d = 107m.
• Mass of car, m = 1400kg.
• Initial speed, v₀ = 35m/s.
• Coefficient of static friction, ${\mu }_s=0.50$.

The problem is based on the kinematic equations of motion in which the motion is described at constant acceleration. Also, it deals with the centripetal force. It is a force that makes a body follow a curved path.

Formula:

The velocity in kinematic equations is given by,

$v^2=v_0^2+2ad$

The centripetal force is given by,

$F_r=\frac{m{v}_0^2}{r}$

Using the kinematic equation, the deceleration of the car is calculated as,

$v^2=v_0^2+2ad$

Substitute the values, and we get,

$$\begin{gathered} 0={\left(35m/s\right)}^2+2a\left(107m\right) \\ a=-5.72m/s^2 \end{gathered}$$

Thus, the force of friction required to stop by car is given by,

$f=m\left|a\right|$

Substitute the values, and we get,

$$\begin{gathered} f=\left(1400kg\right)\left(5.72m/s^2\right) \\ f\approx 8.0\times {10}^{3}N \end{gathered}$$

Thus, the magnitude of static friction is $8.0\times {10}^{3}N$.

The maximum possible static friction is given by,

$f_{s,m}={\mu }_{s}mg$

Substitute the values, and we get,

$$\begin{gathered} f_{s,m}=\left(0.50\right)\left(1400kg\right)\left(9.80m/s^2\right) \\ {f}_{s,m}\approx 6.9\times {10}^{3}N. \end{gathered}$$

Thus, the maximum possible static friction is $6.9\times {10}^{3}N$.

If ${\mu }_k=0.40,then{f}_k={\mu }_{k}mg$, and the deceleration is $a=-{\mu }_{k}g$.

Therefore, the speed of the car when it hits the wall is,

$v=\sqrt{v_0^2+2ad}$

Substitute the values, and we get,

$$\begin{gathered} v=\sqrt{{\left(35m/s\right)}^2-2\left(0.40\right)\left(9.8m/s^2\right)\left(107m\right)} \\ v\approx 20m/s \end{gathered}$$

Thus, the speed at which the car will hit the wall 20m/s.

The force required to keep the motion circular is,

$F_r=\frac{m{v}_0^2}{r}$

Substitute the values, and we get,

$$\begin{gathered} F_r=\frac{\left(1400kg\right){\left(35.0m/s\right)}^2}{107m} \\ {F}_r=1.6\times {10}^{4}N \end{gathered}$$

Thus, the magnitude of frictional force is $1.6\times {10}^{4}N$.

From the calculations, we can say that since F_r > f_s,max that means no circular path is possible.