A car is traveling at 21.0 m/s. It slows to a stop at a constant rate over 5.00 s. How far does the car travel during those 5.00 seconds before it stops?

To solve this problem, we can use the formula for distance traveled under constant acceleration (or deceleration in this case), which is:

d = vt + 0.5at^2

where:

• d is the distance traveled,
• v is the initial velocity,
• a is the acceleration (which will be negative since the car is slowing down), and
• t is the time.

First, we need to find the acceleration. We know that the car slows to a stop, so its final velocity is 0 m/s. The formula for acceleration is:

a = (v_f - v_i) / t

where:

• a is the acceleration,
• v_f is the final velocity,
• v_i is the initial velocity, and
• t is the time.

Substituting the given values:

a = (0 m/s - 21.0 m/s) / 5.00 s = -4.2 m/s^2

Now we can substitute the values of v, a, and t into the distance formula:

d = (21.0 m/s * 5.00 s) + 0.5 * (-4.2 m/s^2) * (5.00 s)^2 d = 105 m - 52.5 m d = 52.5 m

So, the car travels 52.5 meters before it stops.