A drag racer, starting from rest, speeds up for 402 m with an acceleration of +16.0 m/s2. A parachute then opens, slowing the car down with an acceleration of -5.60 m/s2. How fast is the racer moving 2.90 102 m after the parachute opens?

Vf^2 = Vo^2 + 2ad.

Vf^2 = 0 + 2*16*402 = 12,864,
Vf = 113.4m/s.

Vf^2 = (113.4)^2 + 2*(-5.6)*290 = 9616,
Vf = 98m/s.

To solve this problem, we need to break it down into two parts: the initial acceleration phase and the deceleration phase after the parachute opens.

Let's calculate the final velocity after the initial acceleration phase:
We are given:
- Initial velocity (u) = 0 m/s
- Distance traveled (s) = 402 m
- Acceleration (a) = +16.0 m/s^2 (positive because it is in the direction of motion)

We can use the equation: v^2 = u^2 + 2as, where v is the final velocity.
Rearranging the equation, we get: v = √(u^2 + 2as).

Substituting the given values, we have:
v = √(0^2 + 2 * 16.0 m/s^2 * 402 m)
v = √(0 + 12864 m^2/s^2)
v = √(12864) m/s
v ≈ 113.5 m/s

So, after the initial acceleration phase, the racer is moving at approximately 113.5 m/s.

Now, let's calculate the final velocity after the deceleration phase:
We are given:
- Distance traveled (s) = 290 m (since we are asked for the velocity 290 m after the parachute opens)
- Acceleration (a) = -5.60 m/s^2 (negative because it is opposite to the direction of motion)

Again, we can use the equation: v^2 = u^2 + 2as, where v is the final velocity.
This time our initial velocity (u) is the final velocity we calculated from the initial acceleration phase, which is 113.5 m/s.

Substituting the given values, we have:
v = √(113.5^2 + 2 * -5.60 m/s^2 * 290 m)
v = √(12862.25 + -3224 m^2/s^2 * m)
v = √(12862.25 - 3224 m^2/s^2 * m)
v = √(9638.25 m^2/s^2)
v ≈ 98.2 m/s

Therefore, the racer is moving at approximately 98.2 m/s, 290 m after the parachute opens.