A race car has a speed of 98 m/s. The driver hits the breaks causing the car to slow down at a constant rate of 12 m/s^2 for a period of 8 seconds. Compute the distance traveled by the car at 2s, 4s, 6s, and 8s.

To solve this problem, we need to compute the distance traveled by the race car at 2 seconds, 4 seconds, 6 seconds, and 8 seconds, under constant deceleration.

Given:
- Initial velocity (\(v_0\)): 98 m/s
- Deceleration (\(a\)): 12 m/s² (since it's slowing down, this value is negative in the equations)
- Time intervals: 2 seconds, 4 seconds, 6 seconds, and 8 seconds

We will use the equation of motion:

\[
d(t) = v_0 t + \frac{1}{2} a t^2
\]

where:
- \(d(t)\) is the distance traveled at time \(t\),
- \(v_0\) is the initial velocity,
- \(a\) is the acceleration (which will be negative in this case because the car is decelerating),
- \(t\) is the time.

### Step-by-step calculation:

1. **At \(t = 2\) seconds**:
   \[
   d(2) = 98 \times 2 + \frac{1}{2} \times (-12) \times (2)^2
   \]
   
2. **At \(t = 4\) seconds**:
   \[
   d(4) = 98 \times 4 + \frac{1}{2} \times (-12) \times (4)^2
   \]
   
3. **At \(t = 6\) seconds**:
   \[
   d(6) = 98 \times 6 + \frac{1}{2} \times (-12) \times (6)^2
   \]
   
4. **At \(t = 8\) seconds**:
   \[
   d(8) = 98 \times 8 + \frac{1}{2} \times (-12) \times (8)^2
   \]

Now, let's compute these distances.The distances traveled by the race car at the specified times are:

1. At \( t = 2 \) seconds, the car travels **172 meters**.
2. At \( t = 4 \) seconds, the car travels **296 meters**.
3. At \( t = 6 \) seconds, the car travels **372 meters**.
4. At \( t = 8 \) seconds, the car travels **400 meters**.

Would you like further details or explanations?

Here are 5 related questions for further practice:
1. What is the final velocity of the car at \( t = 8 \) seconds?
2. How much distance does the car travel in the first second?
3. If the car continued to decelerate, when would it come to a complete stop?
4. What would the distance be if the deceleration was 10 m/s² instead of 12 m/s²?
5. How much time would it take to stop the car completely?

**Tip**: The equation of motion \( d(t) = v_0 t + \frac{1}{2} a t^2 \) is very useful for problems involving uniformly accelerated motion, even when acceleration is negative (deceleration).

A race car has a speed of 98 m/s. The driver hits the brakes causing the car to slow down at a constant rate of 12 m/s^2 for a period of 8 seconds. Compute the distance traveled by the car at 2s, 4s, 6s, and 8s.

Compute the distance traveled by a race car decelerating from 98 m/s at a rate of 12 m/s². Detailed calculations at time intervals of 2, 4, 6, and 8 seconds using kinematic equations. Perfect for high school physics students, Grades 9-12.

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