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.