A driver in a car, originally moving at 13.1 m/s, applies the brakes until the car comes to a stop. The car moves a distance of 34.1 m while braking. How much time did it take for the car to stop? Assume constant acceleration during braking.

To find the time it took for the car to stop, we can use the equation of motion that relates distance, initial velocity, acceleration, and time:

distance = initial velocity * time + (1/2) * acceleration * time^2

In this case, the initial velocity is 13.1 m/s, the distance is 34.1 m, and the acceleration is unknown. However, we know that the car eventually comes to a stop, so its final velocity is 0 m/s.

The equation can be rearranged to solve for time:

distance = initial velocity * time + (1/2) * acceleration * time^2

34.1 = 13.1 * time + (1/2) * acceleration * time^2

Since the car comes to a stop, the final velocity is 0 m/s:

0 = 13.1 + acceleration * time

Now we have two equations:

34.1 = 13.1 * time + (1/2) * acceleration * time^2
0 = 13.1 + acceleration * time

We can solve these two equations simultaneously to find the values of time and acceleration.

However, since we are assuming constant acceleration during braking, we can simplify the problem by using the formula that relates initial velocity, final velocity, acceleration, and distance:

final velocity^2 = initial velocity^2 + 2 * acceleration * distance

Since the final velocity is 0 m/s and the initial velocity is 13.1 m/s, we can rearrange the formula to solve for acceleration:

acceleration = - (initial velocity^2) / (2 * distance)

Plugging in the values:

acceleration = - (13.1^2) / (2 * 34.1)

acceleration = - 214.81 / 68.2

acceleration = - 3.149 m/s^2

Now, we can use the formula for time:

distance = initial velocity * time + (1/2) * acceleration * time^2

34.1 = 13.1 * time + (1/2) * (-3.149) * time^2

This equation is a quadratic equation. Rearranging it, we get:

-3.149 * time^2 + 13.1 * time - 34.1 = 0

We can solve this quadratic equation to find the values of time. Using the quadratic formula,

time = (-b ± √(b^2 - 4ac)) / 2a

where a = -3.149, b = 13.1, and c = -34.1.

Plugging in the values, we get:

time = (-13.1 ± √(13.1^2 - 4 * (-3.149) * (-34.1))) / 2 * (-3.149)

Simplifying this equation will give us the two possible values of time.

To solve this problem, we can use the equation:

\(v^2 = u^2 + 2as\)

where:
- \(v\) is the final velocity (which is 0 m/s since the car comes to a stop)
- \(u\) is the initial velocity (13.1 m/s)
- \(a\) is the acceleration (which is negative since the car is decelerating)
- \(s\) is the distance traveled (34.1 m)

Rearranging the equation, we have:

\(a = \frac{{v^2 - u^2}}{{2s}}\)

Substituting the given values, we have:

\(a = \frac{{0^2 - 13.1^2}}{{2 \cdot 34.1}}\)

Calculating this expression, we can find the acceleration.

the average velocity during braking

... (13.1 + 0) / 2

divide the stopping distance by the average velocity to find the time