A car of mass 1.2 tonnes is travelling along A straight horizontal road at A speed of 20 ms–1 when it brakes 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

### Given Data:
- Mass of the car, \( m = 1.2 \, \text{tonnes} = 1200 \, \text{kg} \)
- Initial speed of the car, \( u = 20 \, \text{m/s} \)
- Coefficient of friction, \( \mu = 0.8 \)
- Final speed, \( v = 0 \, \text{m/s} \)

### a) Deceleration due to friction:

The frictional force \( F_{\text{friction}} \) acting on the car is given by:
\[
F_{\text{friction}} = \mu \cdot N
\]
Where \( N \) is the normal force, which for a horizontal road equals the weight of the car:
\[
N = m \cdot g
\]
where \( g = 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity.

Thus,
\[
F_{\text{friction}} = \mu \cdot m \cdot g
\]
Now, using Newton’s second law:
\[
F_{\text{friction}} = m \cdot a
\]
where \( a \) is the deceleration due to friction (negative acceleration). Therefore,
\[
a = \frac{F_{\text{friction}}}{m} = \mu \cdot g
\]

Substituting the values:
\[
a = 0.8 \times 9.8 = 7.84 \, \text{m/s}^2
\]
Thus, the deceleration is:
\[
\boxed{7.84 \, \text{m/s}^2}
\]

### b) Distance travelled by the car before coming to rest:

We can use the kinematic equation:
\[
v^2 = u^2 + 2 a s
\]
where:
- \( v = 0 \, \text{m/s} \) (since the car comes to rest),
- \( u = 20 \, \text{m/s} \),
- \( a = -7.84 \, \text{m/s}^2 \) (negative due to deceleration),
- \( s \) is the distance travelled.

Rearranging the equation to solve for \( s \):
\[
0 = 20^2 + 2 (-7.84) s
\]
\[
0 = 400 - 15.68 s
\]
\[
15.68 s = 400
\]
\[
s = \frac{400}{15.68} \approx 25.51 \, \text{m}
\]
Thus, the distance travelled by the car is:
\[
\boxed{25.51 \, \text{m}}
\]

---

Would you like more details or have any questions?

Here are 5 related questions:
1. How would the deceleration change if the coefficient of friction was reduced to 0.5?
2. What would happen to the distance travelled if the car was going at 30 m/s instead of 20 m/s?
3. How does the mass of the car affect the deceleration?
4. What if the road was inclined? How would the normal force change?
5. How does a higher coefficient of friction affect braking performance?

**Tip:** When solving problems involving friction, always make sure the surface is horizontal or account for any inclines, as that will affect the normal force.

A car of mass 1.2 tonnes is travelling along a straight horizontal road at a speed of 20 m/s when it brakes 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.

Learn how to calculate the deceleration and stopping distance of a car of mass 1.2 tonnes skidding to rest with a coefficient of friction of 0.8. This solution covers friction, kinematics, and Newton's Second Law, suitable for students in Grades 9-12.

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