**Answer:**

It takes the car 2.5 s to slow down to a final speed of 10.0 m/s.

The car has moved 31.25 m during the braking period.

**Explanation:**

Assuming that motion is a straight line and knowing that the accelaration is a constant (-2.0 m/s²), these equations apply:

v(t) = v0 +a*t

X(t) = X0 + v0*t + 1/2*a*t²

where:

v(t) = velocity at time "t".

v0 = initial velocity.

a = acceleration.

t = time.

X(t) = position at time "t"

X0 = initial position

Given data:

v0 = 15.0 m/s

a = 2.0 m/s²

v(t) = 10.0 m/s

We need to find "t" at which the speed is 10 m/s.

From the equation of velocity above:

v(t) = v0 + a*t

solving the equation for "t"

v(t) - v0 = a*t

(v(t) - v0)/a = t

(10 m/s - 15 m/s) / (-2m/s²) = t

2.5 s = t

Now that we know how long it takes to slow down to a speed of 10 m/s, we can calculate how far has the car moved in that period using the equation for position at a given time:

X(t) = X0 + v0*t + 1/2*a*t²

Since we only want to know how many meters has the car moved during 2.5 s with an acceleration of -2m/s², we can make X0(initial position) = 0.

Then:

X(2.5 s) = 0 m + 15 m/s * 2.5 s + 1/2 * (-2m/s²)* (2.5 s)²

X(2.5 s) = 37.5 m + (-6.25 m) = 31.25 m