Question

Review. While learning to drive, you are in a $\mathbf{1200}\mathbf{-}\mathit{k}\mathit{g}$car moving at$\mathbf{20}\mathbf{.}\mathbf{0}\mathbf{}\mathit{m}\mathit{l}\mathit{s}$across a large, vacant, level parking lot. Suddenly you realize you are heading straight toward the brick sidewall of a large supermarket and are in danger of running into it. The pavement can exert a maximum horizontal force of$\mathbf{7000}\mathbf{}\mathit{N}$on the car. (a) Explain why you should expect the force to have a well-defined maximum value. (b) Suppose you apply the brakes and do not turn the steering wheel. Find the minimum distance you must be from the wall to avoid a collision. (c) If you do not brake but instead maintain constant speed and turn the steering wheel, what is the minimum distance you must be from the wall to avoid a collision? (d) Of the two methods in parts (b) and (c), which is better for avoiding a collision? Or should you use both the brakes and the steering wheel, or neither? Explain. (e) Does the conclusion in part (d) depend on the numerical values given in this problem, or is it true in general? Explain.

Short answer

a) The force should be well defined because the maximum horizontal force exerted by the pavement on the car is frictional force which is only due to the rough surface of the pavement.

b) The minimum distance from the wall to avoid the collision in case we apply the brake is $34.3m$.

c) The minimum distance from the wall to avoid the collision in case we do not apply brake is $68.6m$.

d) The best method to stop the car for avoiding collision is to apply a brake.

e) Yes the conclusion in part (d) is correct.

Step-by-step solution

Collision

Whenever two particles tries to attend the same position at a time then collision will occur. Collision can be elastic or inelastic.

The expression for the force from the Newton’s second law is given by.

$F=ma$

Here$F$is the force,$m$is the mass and$a$is the acceleration.

The expression for the centripetal force is given by

$F=m\frac{v^2}{r}$

Here$F$ is the centripetal force,$m$ is the mass,$v$ is the velocity and$r$ is the radius of the turn.

Explanation for part (a)

(a)

The force should be well defined because the maximum horizontal force exerted by the pavement on the car is frictional force which is only due to the rough surface of the pavement.

Calculation of minimum distance to avoid collision in case we apply brake

(b)

The expression for the force is given by

$F=ma$

Here$F$is the force,$m$is the mass and$a$is the acceleration.

Substitute $-7000N$(frictional force acting opposite to the motion of the car) for$F$and$1200kg$for$m$in the above equation.

$$\begin{gathered} -7000=\left(1200\right)a \\ a=-5.83m/s^2 \end{gathered}$$

Here negative sign of acceleration shows that the car is getting de-accelerated.

The expression for the third law of motion is given by

$v_f^2-v_i^2=2as$

Here$v_f$ is the final velocity,$v_i$is the initial velocity, a is the acceleration and$s$ is the distance travelled by the car.

As the car is getting stopped, substitute$0m/s$ for $v_f$,$20m/s$ for$v_i$ and $-5.83m/s^2$for a in the above equation.

$$\begin{gathered} 0^2-{20}^2=2\left(-5.83\right)s \\ s=34.3 \text{m} \end{gathered}$$

Therefore, the minimum distance from the wall to avoid the collision in case we apply the brake is $34.3m$.

Calculation of minimum distance to avoid collision in case we do not apply brake

(c)

The expression for the centripetal force is given by

$F=m\frac{v^2}{r}$

Here $F$is the centripetal force,$m$ is the mass,$v$ is the velocity and$r$ is the radius of the turn.

Substitute $7000N$for $F$, $20m/s$for $v$and $1200kg$for $m$in the above equation.

$$\begin{gathered} 7000=1200\frac{{20}^2}{r} \\ r=68.6 \text{m} \end{gathered}$$

Therefore, the minimum distance from the wall to avoid the collision in case we do not apply brake is $68.6m$.

Explanation for part (d)

(d)

The best method to stop the car for avoiding collision is to apply a brake because by applying brake car will stop in shortest distance.

Explanation for part (e)

(e)

Yes the conclusion in part (d) is correct because radius of turn is twice of the shortest distance of stopping the car.