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).