To determine how many meters Maggie's car travels before coming to a stop, we can use the equations of uniformly accelerated motion. In this case, the initial velocity, final velocity, and time are given, so we can calculate the distance using the following equation:

$d = v_i t + \frac{1}{2} a t^2$

Where:

• $v_i = 15.0 \, \text{m/s}$ is the initial velocity,
• $v_f = 0.0 \, \text{m/s}$ is the final velocity,
• $t = 2.50 \, \text{s}$ is the time taken to stop,
• $a$ is the acceleration (which we can calculate).

Step 1: Calculate the acceleration

The acceleration $a$ can be found using the formula:

$a = \frac{v_f - v_i}{t}$

Substitute the known values:

$a = \frac{0.0 \, \text{m/s} - 15.0 \, \text{m/s}}{2.50 \, \text{s}} = -6.0 \, \text{m/s}^2$

Step 2: Calculate the distance

Now that we have the acceleration, we can use the equation for distance:

$d = v_i t + \frac{1}{2} a t^2$

Substitute the known values:

$d = (15.0 \, \text{m/s})(2.50 \, \text{s}) + \frac{1}{2}(-6.0 \, \text{m/s}^2)(2.50 \, \text{s})^2$

Now, let’s calculate the distance.The distance Maggie's car travels before coming to a stop is 18.75 meters. Therefore, she must apply her brakes 18.75 meters before the stop sign to stop exactly at the sign.

Do you have any further questions, or would you like more details?

Here are 5 related questions:

1. How can you determine the stopping distance if the initial velocity is doubled?
2. What happens to the stopping distance if the braking time is halved?
3. How would the result change if the car decelerated at a non-uniform rate?
4. What is the car's average velocity during the braking process?
5. How can this concept be applied to real-world driving safety?

Tip: Remember that braking distance is proportional to the square of the velocity!