A car slows down from a speed of 24.0m/s to rest in 9.0s, How far did it travel in that time?

a = (vf - vi)/t

a = (0 - 24)/9
a = -(8/3) m/s²

d = vt + at²/2
d = (24)(9) + (-8/3)(9)²/2
d = 108 m

To calculate the distance traveled by the car, you can use the kinematic equation:

\[ d = v_i t + \frac{1}{2} a t^2 \]

where:
- \( d \) represents the distance traveled,
- \( v_i \) is the initial velocity of the car,
- \( t \) is the time taken,
- \( a \) represents the acceleration.

In this case, the initial velocity is 24.0 m/s, the time is 9.0 s, and the car comes to rest, so the final velocity (\( v_f \)) is 0 m/s.

Since the car slows down, the acceleration (\( a \)) is negative.

Let's calculate the distance traveled:

Step 1: Calculate the acceleration using the formula \( a = \frac{{v_f - v_i}}{t} \)
- \( v_f = 0 \, \text{m/s} \) (final velocity)
- \( v_i = 24.0 \, \text{m/s} \) (initial velocity)
- \( t = 9.0 \, \text{s} \) (time)

\( a = \frac{{0 - 24.0}}{9.0} = -2.67 \, \text{m/s}^2 \)

Step 2: Use the kinematic equation \( d = v_i t + \frac{1}{2} a t^2 \)
- \( v_i = 24.0 \, \text{m/s} \) (initial velocity)
- \( t = 9.0 \, \text{s} \) (time)
- \( a = -2.67 \, \text{m/s}^2 \) (acceleration)

\( d = (24.0 \times 9.0) + \frac{1}{2} (-2.67) \times (9.0^2) \)

Simplifying,

\( d = 216.0 - 108.0 \)

\( d = 108.0 \, \text{m} \)

Therefore, the car traveled a distance of 108.0 meters in that time.