A car of mass 1200kg is travelling along a straight horizontal road at a speed of 20 ms–1 when it brakes

Question

sharply then skids. Friction brings the car to rest. If the coefficient of friction between the tyres and road is 0.8,
calculate a) the deceleration b) the distance travelled by the car before it comes to rest.
c) the time taken for the car to come to rest

Answer 1

To solve the problem, we'll follow these steps:

(a) Calculate the deceleration

The deceleration can be calculated using the formula that relates the force of friction to the weight of the car. The force of friction ($F_f$) acting on the car is given by:

$$F_f=\mu \cdot N$$

Where:

• $\mu$ is the coefficient of friction (0.8),
• $N$ is the normal force, which equals the weight of the car when on a horizontal surface, given by $N=mg$ (where $m$ is the mass and $g$ is the acceleration due to gravity, approximately $9.81,{\text{m/s}}^2$).

Now we calculate the normal force:

$$N=m\cdot g=1200,\text{kg}\cdot 9.81,{\text{m/s}}^2=11772,\text{N}$$

Now calculate the frictional force:

$$F_f=\mu \cdot N=0.8\cdot 11772,\text{N}=9417.6,\text{N}$$

Using Newton's second law, $F=ma$, we can find the deceleration ($a$):

$$F_f=m\cdot a⟹a=\frac{F_f}{m}=\frac{9417.6,\text{N}}{1200,\text{kg}}\approx 7.848,{\text{m/s}}^2$$

The deceleration is approximately:

$$a\approx 7.85,{\text{m/s}}^2$$

(b) Calculate the distance travelled by the car before it comes to rest

We can use the kinematic equation relating initial velocity, final velocity, acceleration, and distance:

$$v^2=u^2+2as$$

Where:

• $v$ is the final velocity (0 m/s, since the car comes to rest),
• $u$ is the initial velocity (20 m/s),
• $a$ is the deceleration (-7.85 m/s², we take it as negative since it's a deceleration),
• $s$ is the distance travelled.

Rearranging the equation to find $s$:

$$0=(20,\text{m/s}{)}^2+2(-7.85,{\text{m/s}}^2)s$$ $$0=400-15.7s$$ $$15.7s=400$$ $$s=\frac{400}{15.7}\approx 25.5,\text{m}$$

The distance travelled before coming to rest is approximately:

$$s\approx 25.5,\text{m}$$

(c) Calculate the time taken for the car to come to rest

We can use another kinematic equation that relates initial velocity, final velocity, acceleration, and time:

$$v=u+at$$

Rearranging for $t$:

$$t=\frac{v-u}{a}$$

Substituting in the known values:

$$t=\frac{0-20,\text{m/s}}{-7.85,{\text{m/s}}^2}$$ $$t\approx \frac{-20}{-7.85}\approx 2.55,\text{s}$$

The time taken for the car to come to rest is approximately:

$$t\approx 2.55,\text{s}$$

Summary of Results:

• Deceleration: $\approx 7.85,{\text{m/s}}^2$
• Distance travelled: $\approx 25.5,\text{m}$
• Time taken: $\approx 2.55,\text{s}$