A drag racer, starting from rest, speeds up for 402 m with an acceleration of +20.0 m/s2. A parachute then opens, slowing the car down with an acceleration of -5.20 m/s2. How fast is the racer moving 3.75 102 m after the parachute opens?

First compute the maximum speed using

V(max) = sqrt(2aX)
a is the acceleration and x = 402 m.

After the chute opens (which I will call t=0), the distance moved is
X = Vmax*t - 2.6 t^2

Solve for t when X = 375 m.
Then use that t to get the velocity at that time.

V(t) = Vmax - 5.20 t

To solve this problem, we need to break it down into different parts and find the velocity at each part. Let's follow these steps:

Step 1: Find the final velocity during the acceleration phase.
We know that the initial velocity (u) is 0 m/s, and the acceleration (a) is +20.0 m/s^2. The distance covered (s) during this phase is 402 m. We can use the following equation to find the final velocity (v) during acceleration:

v^2 = u^2 + 2a*s

Substituting the given values:

v^2 = 0^2 + 2 * 20.0 * 402

v^2 = 2 * 20.0 * 402

v^2 = 16080

Taking the square root of both sides:

v = sqrt(16080)

v ≈ 126.89 m/s

So, the final velocity during the acceleration phase is approximately 126.89 m/s.

Step 2: Find the time taken during the acceleration phase.
We can use the following equation to find the time (t) taken during acceleration:

v = u + a*t

From step 1, we know that v = 126.89 m/s, u = 0 m/s, and a = +20.0 m/s^2. Substituting these values into the equation:

126.89 = 0 + 20.0 * t

20.0t = 126.89

t = 126.89 / 20.0

t ≈ 6.3445 s

So, the time taken during the acceleration phase is approximately 6.3445 seconds.

Step 3: Find the final velocity during the deceleration phase.
We know that the acceleration during the deceleration phase is -5.20 m/s^2. The time taken (t) during this phase can be calculated by subtracting the time taken during the acceleration phase from the total time:

t = total time - time during acceleration
t = 3.75 * 10^2 - 6.3445

t ≈ 369.6555 s

Now we can find the final velocity (v) during deceleration using the following equation:

v = u + a*t

where u is the velocity at the end of the acceleration phase.

From step 2, u = 126.89 m/s, a = -5.20 m/s^2, and t ≈ 369.6555 s. Substituting these values into the equation:

v = 126.89 + (-5.20) * 369.6555

v ≈ 126.89 - 1923.2446

v ≈ -1796.3546 m/s

So, the final velocity during the deceleration phase is approximately -1796.3546 m/s.

Step 4: Find the total final velocity.
The total final velocity is the final velocity during the deceleration phase because the car slows down after the parachute is opened. Therefore, the total final velocity is approximately -1796.3546 m/s.