Calculating Car Stopping Distance with Retardation

This problem asks us to find the distance a car travels while braking, given the time it takes to stop and the rate of retardation (negative acceleration).

We are given the following information:

• Time taken to stop, $\(t = 20 \text{ S}\)$.
• Final velocity, since the car stops, $\(v = 0 \text{ m/s}\)$.
• Retardation (deceleration), $\(a = -0.6 \text{ m/s}^2\)$. Note that retardation is negative acceleration, so we use a negative sign.

We need to find the distance traveled, $\(s\)$.

Using Kinematic Equations for Uniformly Accelerated Motion

We can use the standard kinematic equations to solve this problem. The relevant equations are:

1. $\(v = u + at\)$
2. $\(s = ut + \frac{1}{2}at^2\)$
3. $\(v^2 = u^2 + 2as\)$

Here, $\(u\)$ is the initial velocity of the car just before the brakes were applied.

Step 1: Find the Initial Velocity (\(u\))

We know $\(v\)$, $\(t\)$, and $\(a\)$. We can use the first equation, $\(v = u + at\)$, to find the initial velocity $\(u\)$.

Substitute the given values into the equation:

$\(0 \text{ m/s} = u + (-0.6 \text{ m/s}^2)(20 \text{ S})\)$

$\(0 \text{ m/s} = u - 12 \text{ m/s}\)$

Solving for $\(u\)$:

$\(u = 12 \text{ m/s}\)$

So, the initial velocity of the car was 12 m/s.

Step 2: Find the Distance Traveled (\(s\))

Now that we know the initial velocity $\(u\)$, we can find the distance traveled $\(s\)$ using either the second or third kinematic equation. Let's use the second equation: $\(s = ut + \frac{1}{2}at^2\)$.

Substitute the values of $\(u\)$, $\(t\)$, and $\(a\)$ into the equation:

$\(s = (12 \text{ m/s})(20 \text{ S}) + \frac{1}{2}(-0.6 \text{ m/s}^2)(20 \text{ S})^2\)$

$\(s = 240 \text{ m} + \frac{1}{2}(-0.6 \text{ m/s}^2)(400 \text{ S}^2)\)$

$\(s = 240 \text{ m} + (-0.3 \text{ m/s}^2)(400 \text{ S}^2)\)$

$\(s = 240 \text{ m} - 120 \text{ m}\)$

$\(s = 120 \text{ m}\)$

Alternatively, using the third equation $\(v^2 = u^2 + 2as\)$:

$\((0 \text{ m/s})^2 = (12 \text{ m/s})^2 + 2(-0.6 \text{ m/s}^2)s\)$

$\(0 = 144 \text{ m}^2/\text{S}^2 - 1.2 \text{ m/s}^2 \cdot s\)$

$\(1.2 \text{ m/s}^2 \cdot s = 144 \text{ m}^2/\text{S}^2\)$

$\(s = \frac{144 \text{ m}^2/\text{S}^2}{1.2 \text{ m/s}^2}\)$

$\(s = 120 \text{ m}\)$

Both methods give the same result.

The distance the car travels during the 20 seconds while the brakes are applied is 120 m.

Revision Table: Kinematic Equations for Uniform Acceleration

Additional Information: Understanding Retardation and Braking Distance

Retardation is simply acceleration in the direction opposite to the velocity. It causes an object to slow down. In this problem, the retardation of 0.6 m/s² means the car's velocity decreases by 0.6 m/s every second.

Braking distance is the distance a vehicle travels from the point where the brakes are fully applied to the point where it comes to a complete stop. It is a critical factor in road safety. The braking distance depends on several factors, including:

• Initial speed of the vehicle ($\(u\)$): Higher speed means longer braking distance (it increases with the square of the initial velocity).
• Deceleration rate ($\(a\)$): Higher retardation (more effective braking) means shorter braking distance.
• Condition of the road surface (friction).
• Condition of the tires.
• Mass of the vehicle (though mass cancels out in the kinematic equations if the braking force is proportional to mass, like friction often is).

Understanding how these factors affect braking distance is important for safe driving and in physics problems related to motion and forces.