Question

A 1600 kg car moves along a horizontal road
at speed v0 = 17.3 m/s. The road is wet,
so the static friction coefficient between the
tires and the road is only μs = 0.311 and
the kinetic friction coefficient is even lower,
μk = 0.2177.
The acceleration of gravity is 9.8 m/s2 . Assume: No aerodynamic forces; g =
9.8 m/s2, forward is the positive direction.
What is the highest possible deceleration of
the car under such conditions?
Answer in units of m/s2

Answer 1

A 1600 kg car moves along a horizontal road at speed $v_0 = 17.3 \, \text{m/s}$. The road is wet, so the static friction coefficient between the tires and the road is only $\mu_s = 0.311$ and the kinetic friction coefficient is even lower, $\mu_k = 0.2177$.

The acceleration of gravity is $9.8 \, \text{m/s}^2$. Assume: No aerodynamic forces; $g = 9.8 \, \text{m/s}^2$, forward is the positive direction.

What is the highest possible deceleration of the car under such conditions?

Answer in units of m/s²

**Solution:**

$$m = 1600 \, \text{kg} - \text{mass of the car};$$ $$V_0 = 17.7 \, \frac{\text{m}}{\text{s}} - \text{speed of the car};$$ $$g = 9.8 \, \frac{\text{m}}{\text{s}^2} - \text{acceleration due to gravity};$$ $$a_{\max} - \text{highest possible deceleration of the car}$$ $$\mu_s = 0.311 - \text{static friction coefficient between the tires and the road};$$ $$\mu_k = 0.2177 - \text{kinetic friction coefficient};$$

Second Newton's law for the car (N - reaction force):

$$y: N = mg$$ $$x: F_{fr} = m a_{\max}$$

Formula for the friction force:

$$F_{fr} = N \mu_k = m g \mu_k \quad (\text{we use } \mu_k \text{ because car moves})$$

(2)in(1):

$$m g \mu_k = m a_{\max}$$ $$a_{\max} = g \mu_k = 9.8 \, \frac{\text{m}}{\text{s}^2} \cdot 0.2177 = 2.13 \, \frac{\text{m}}{\text{s}^2}$$

**Answer:** the highest possible deceleration of the car under such conditions is $2.13 \, \frac{\text{m}}{\text{s}^2}$.