A car is traveling down the road at 29 m/s. The driver sees a deer in the road up ahead and applies the breaks, resulting in an acceleration of -12.5 m/s^2. How far does the car continue to move before coming to a stop? Round to the nearest meter.

To solve the problem, we can use the kinematic equation:

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

Where:
- \( v \) is the final velocity (0 m/s, since the car comes to a stop),
- \( u \) is the initial velocity (29 m/s),
- \( a \) is the acceleration (-12.5 m/s\(^2\)),
- \( s \) is the distance traveled.

Rearranging the equation to solve for \( s \):

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

Substituting the given values:

\[
s = \frac{0^2 - 29^2}{2 \times -12.5}
\]

\[
s = \frac{-841}{-25} = 33.64 \text{ meters}
\]

Rounding to the nearest meter:

\[
s \approx 34 \text{ meters}
\]

So, the car continues to move approximately **34 meters** before coming to a stop.

Would you like more details or have any questions?

Here are 5 related questions:
1. What would happen to the stopping distance if the car's initial speed were higher?
2. How does the magnitude of acceleration affect the stopping distance?
3. What would the stopping distance be if the acceleration were less negative, say -10 m/s²?
4. How long does it take for the car to stop?
5. How would the stopping distance change if the road were uphill or downhill?

**Tip:** Remember that a higher initial speed or a less negative acceleration would result in a longer stopping distance.

Explore how to calculate the stopping distance of a car traveling at 29 m/s with an acceleration of -12.5 m/s². This problem involves applying kinematic equations to find the distance traveled before the car comes to a stop. Suitable for high school students studying physics or mechanics.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation