To solve this problem, we need to determine how far Maggie's car travels while decelerating uniformly from 15.0 m/s to 0 m/s in 2.50 seconds.

We can use the following kinematic equation to find the distance:

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

Where:

• $d$ is the distance (the unknown we want to find),
• $v_i = 15.0 \, \text{m/s}$ is the initial velocity,
• $t = 2.50 \, \text{seconds}$ is the time,
• $a$ is the acceleration (which we need to find).

Step 1: Find acceleration ($a$)

We use the equation:

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

Where:

• $v_f = 0 \, \text{m/s}$ (final velocity),
• $v_i = 15.0 \, \text{m/s}$ (initial velocity),
• $t = 2.50 \, \text{seconds}$.

$a = \frac{0 - 15.0}{2.50} = -6.0 \, \text{m/s}^2$

The acceleration is $-6.0 \, \text{m/s}^2$, which means the car is decelerating.

Step 2: Find the distance ($d$)

Now that we have the acceleration, we can use the kinematic equation to find the distance traveled during deceleration:

$d = (15.0)(2.50) + \frac{1}{2} (-6.0)(2.50)^2$

Let's calculate that.

$d = 37.5 + \frac{1}{2}(-6.0)(6.25)$ $d = 37.5 - 18.75$ $d = 18.75 \, \text{meters}$

Conclusion:

Maggie must apply her brakes 18.75 meters before the stop sign to stop at the sign.

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Would you like more details on any part of the solution, or have any questions? Here are some related questions to consider:

1. How is uniform deceleration different from non-uniform deceleration?
2. What other kinematic equations are useful for uniformly accelerated motion?
3. How would this problem change if the final velocity wasn't zero?
4. How can we determine the stopping distance if the car was decelerating at a different rate?
5. What is the importance of knowing both acceleration and time in these problems?

Tip: When solving kinematics problems, it's often useful to first identify the variables and decide which kinematic equation fits the known data.