When you brake your car with your brand new tyres, your acceleration is \(5 \mathrm{~m} / \mathrm{s}^{2}\). (a) Find an expression for the distance you need to stop the car as a function of the starting velocity. With your old tires, the acceleration is only two thirds of the acceleration with the new tyres. (b) How does this affect the braking distance? (c) Your reaction time is \(0.5 \mathrm{~s}\). If a child jumps into the street \(30 \mathrm{~m}\) ahead of you when you are driving \(50 \mathrm{~km} / \mathrm{h}\), are you able to stop with your new tires? What would happen if you did not change tyres?

Braking distance refers to the distance a vehicle travels from the point where the brakes are applied to the point where it fully stops. This distance is crucial for understanding vehicle safety and driving dynamics. There are several factors that affect braking distance, including:

• Initial speed of the vehicle
• Coefficient of friction between tires and road surfaces
• Condition of the brakes

In the problem presented, the braking distance changes based on the condition of the tires, which affects the car's acceleration during braking. The faster you are driving, the longer the braking distance will be. This is why it's essential to always maintain a safe speed and ensure your tires and brakes are in good condition.

Kinematic equations describe the motion of objects without considering the forces that cause the motion. They are particularly useful in solving problems involving linear motion, where acceleration is constant. In this exercise, the key kinematic equation used is:

\[v^2 = u^2 + 2as\]

Where:

• \(v\) is the final velocity
• \(u\) is the initial velocity
• \(a\) is the acceleration
• \(s\) is the distance

By setting the final velocity \( v \) to zero (since the car stops), we can rearrange the equation to solve for the stopping distance \( s \):

\[ s = \frac{u^2}{2a} \]

This equation shows that the stopping distance is inversely proportional to the acceleration and directly proportional to the square of the initial velocity.

Acceleration is the rate at which the velocity of an object changes over time. It is a vector quantity, meaning it has both magnitude and direction. In the context of this exercise, acceleration occurs in the negative direction (deceleration) because the car is slowing down. With new tires, the car's acceleration during braking is given as \( 5 \text{ m/s}^2 \). For old tires, this acceleration is only two-thirds of the new tires, which equals:

\[ \frac{2}{3} \times 5 \text{ m/s}^2 = \frac{10}{3} \text{ m/s}^2 \]

This reduced acceleration means the car will take a longer distance to stop when using old tires compared to new ones. The relationship between stopping distance and acceleration highlights the importance of maintaining good tire conditions to ensure shorter and safer stopping distances.

Stopping distance is the total distance a vehicle travels during the driver's reaction time plus the braking distance. It is affected by several factors, such as driver's reaction time, vehicle speed, and braking efficiency. We can break down stopping distance into two main components:

• **Reaction Distance:** The distance the car travels during the driver's reaction time. For example, with a reaction time of 0.5 seconds and a speed of 13.89 m/s (equivalent to 50 km/h), the reaction distance is:
  
  \[ d_{\text{reaction}} = v \times t_{\text{reaction}} = 13.89 \text{ m/s} \times 0.5 \text{ s} = 6.945 \text{ m} \]
• **Braking Distance:** The distance calculated using the kinematic equations when the brakes are applied:
  
  For new tires:
  \[ s_{\text{new}} = \frac{13.89^2}{10} = 19.29 \text{ m} \]
  
  For old tires:
  \[ s_{\text{old}} = \frac{3 \times 13.89^2}{20} = 28.935 \text{ m} \]

Combining these distances gives the total stopping distance, which includes both the reaction distance and the braking distance. For new tires, the total stopping distance is 26.235 m, whereas for old tires, it is 35.88 m. This difference emphasizes the importance of tire conditions in ensuring a vehicle can stop safely in time to avoid hazards.