To solve this problem, we need to find the acceleration and the distance traveled by the race car before it comes to a complete stop.

1. Finding the Acceleration

Acceleration $a$ can be found using the formula:

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

Where:

• $v_f$ is the final velocity (0 m/s, since the car stops),
• $v_i$ is the initial velocity (50 m/s),
• $t$ is the time taken to stop (7 seconds).

Substituting the values:

$a = \frac{{0 \text{ m/s} - 50 \text{ m/s}}}{{7 \text{ s}}} = \frac{{-50 \text{ m/s}}}{{7 \text{ s}}} \approx -7.14 \text{ m/s}^2$

So, the acceleration is approximately $-7.14 \text{ m/s}^2$.

2. Finding the Distance Traveled

The distance traveled before coming to a stop can be found using the formula:

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

Substituting the known values:

$d = 50 \text{ m/s} \cdot 7 \text{ s} + \frac{1}{2} \cdot (-7.14 \text{ m/s}^2) \cdot (7 \text{ s})^2$

First, calculate $50 \text{ m/s} \cdot 7 \text{ s} = 350 \text{ m}$.

Then, calculate $\frac{1}{2} \cdot (-7.14 \text{ m/s}^2) \cdot 49 \text{ s}^2 = -175.86 \text{ m}$.

Now, summing these up:

$d = 350 \text{ m} - 175.86 \text{ m} = 174.14 \text{ m}$

So, the distance traveled before stopping is approximately 174.14 meters.

Summary:

• Acceleration: $-7.14 \text{ m/s}^2$
• Distance Traveled: $174.14 \text{ m}$

Would you like to dive deeper into any of these concepts or have any questions?

Here are 5 related questions you might find interesting:

1. How would the stopping distance change if the initial speed were doubled?
2. What is the significance of negative acceleration in this context?
3. How would the stopping time change if the brakes could apply twice the force?
4. Can you derive the formula for distance traveled from the kinematic equations?
5. How would the stopping distance differ if the road were inclined?

Tip: Always keep track of the signs when dealing with acceleration, as it indicates the direction of the force applied.