You testify as an expert witness in a case involving an accident in which car A slid into the rear of car B, which was stopped at a red light along a road headed down a hill (Fig. 6-25). You find that the slope of the hill is ${}^{\mathbf{\theta }\mathbf{=}\mathbf{12}\mathbf{.}{\mathbf{0}}^{\mathbf{0}}}$, that the cars were separated by distance ${}^{\mathbf{d}\mathbf{=}\mathbf{24}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}}$${}^{{\mathbf{v}}_{\mathbf{0}}\mathbf{}\mathbf{=}\mathbf{18}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}}$when the driver of car A put the car into a slide (it lacked any automatic anti-brake-lock system), and that the speed of car A at the onset of braking was ${}^{{\mathbf{v}}_{\mathbf{0}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{18}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}}$.With what speed did car A hit car B if the coefficient of kinetic friction was (a) ${}^{\mathbf{0}\mathbf{.}\mathbf{60}}$(dry road surface) and (b) ${}^{\mathbf{0}\mathbf{.}\mathbf{10}}$(road surface covered with wet leaves)?

Distance between the cars,$d=24.0m$

Initial velocity of the car,$v_o=18.0m/s$

Coefficient of kinetic friction, ${\mu }_k-0.60$(On dry road surface)

${\mu }_k=0.10$(Road surface covered with wet leaves)

Slope of the hill, $\theta ={12}^0$

The problem is based on Newton’s second law of motion which states that the rate of change of momentum of a body is equal in both magnitude and direction of the force acting on it. Also, it deals with the kinematic equation of motion that describe the motion at constant acceleration. Use Newton's 2nd law of motion along horizontal and vertical direction to find the acceleration of the car on both surfaces and then by using the Kinematic equation, find the final speed of car A to hit car B.

Formula:

$F_{net}=\sum ma$

where, F is the net force, m is mass and a is an acceleration.

for dry road surface

The Kinetic frictional force,$f_k={\mu }_{k}N$

By using Newton’s 2nd law of motion along vertical direction,$N=W\cos \left({12}^0\right)$

$$\begin{gathered} f_k={\mu }_{k}N \\ ={\mu }_k\left(W\cos \left({12}^0\right)\right) \end{gathered}$$

Thus,

Then by using the Newton’s 2nd law along horizontal direction,

$$\begin{gathered} \sum F_x=m{a}_x \\ W\sin \left({12}^0\right)-f_k=m{a}_x \\ mg\sin \left({12}^0\right)-{\mu }_{k}mg\cos \left({12}^0\right)=m{a}_x \\ {a}_x=g\sin \left({12}^0\right)-{\mu }_{k}g\cos \left({12}^0\right) \\ =-3.7m/s^2 \end{gathered}$$

By using the kinematic equations on motion, the final velocity of the car is,

$$\begin{gathered} v^2={v_o}^2+2ad \\ ={(18.0)}^2+2(-3.7)(24.0) \\ =146.4 \end{gathered}$$

$v=12.09m/s$

Thus, the final velocity of the car A to hit the car B

If road surface covered with wet leaves:

Acceleration of the car is,

In this case, motion of the car and acceleration is along the same direction.So, acceleration is positive.

Then, final velocity of the car is A to hit the car B is,

$$\begin{gathered} v^2={v_o}^2+2ad \\ ={(18.0)}^2+2(1.08)(24.0) \\ v=19.38m/s \end{gathered}$$

Thus, the final velocity of the car A to hit the car B is, ${}^{v=12.09m/s}$