Question: 14. A driver in a 1000.0 kg car traveling at 24 m/s slams on the brakes and skids to a stop. If the coefficient of friction between the tires and the road is 0.80, how long will the skid marks be?
1. **State the problem:**
A car of mass $1000.0$ kg is moving at an initial speed of $24$ m/s and comes to a stop due to friction with coefficient $\mu = 0.80$. We need to find the length of the skid marks, which is the distance the car travels while stopping.
2. **Relevant formula:**
The work-energy principle or kinematic equations can be used. Here, we use the friction force to find acceleration and then use kinematics:
Friction force $F_f = \mu mg$
Acceleration due to friction $a = \frac{F_f}{m} = \mu g$
Since friction opposes motion, acceleration is negative: $a = -\mu g$
Using kinematic equation:
$$v^2 = v_0^2 + 2 a d$$
where $v = 0$ (final velocity), $v_0 = 24$ m/s, $a = -\mu g$, and $d$ is the skid distance.
3. **Calculate acceleration:**
$$a = -\mu g = -0.80 \times 9.8 = -7.84 \text{ m/s}^2$$
4. **Apply kinematic equation:**
$$0 = (24)^2 + 2 \times (-7.84) \times d$$
$$0 = 576 - 15.68 d$$
5. **Solve for $d$:**
$$15.68 d = 576$$
$$d = \frac{576}{15.68}$$
$$d \approx 36.73 \text{ m}$$
6. **Interpretation:**
The skid marks length is approximately $36.73$ m, which rounds to $37$ m.
**Final answer:** a. 37 m