A drag racer, starting from rest, speeds up for 402 m with an acceleration of +24.0 m/s2. A parachute then opens, slowing the car down with an acceleration of -5.40 m/s2. How fast is the racer moving 3.75 102 m after the parachute opens
Now I did the falling rock one. This is the same idea so post your work.
To find the final velocity of the racer after the parachute opens, we can apply the laws of motion.
1. Calculate the velocity during the acceleration phase:
Using the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time:
Given: u = 0 m/s (starting from rest), a = +24.0 m/s², s = 402 m
Using the equation s = ut + 0.5at², we can solve for t:
402 = 0t + 0.5 * 24.0t²
0.5 * 24t² = 402
12t² = 402
t² = 402 / 12
t² = 33.5
t = sqrt(33.5) ≈ 5.8 s
Now, using v = u + at, we can find the velocity at this point:
v = 0 + 24.0 * 5.8
v ≈ 139.2 m/s
2. Calculate the velocity during the deceleration phase:
Given: u = 139.2 m/s, a = -5.40 m/s², s = 375 m (since the parachute opens at 402 m and we want to find the velocity at 3.75 * 10² m after the parachute opens)
Using the equation s = ut + 0.5at², we can solve for t:
375 = 139.2t + 0.5 * (-5.40)t²
0.5 * (-5.40)t² + 139.2t - 375 = 0
Solving this quadratic equation, we find the positive value of time, t ≈ 2.69 s.
Now, using v = u + at, we can find the velocity at this point:
v = 139.2 + (-5.40) * 2.69
v ≈ 125.2 m/s
Therefore, the racer is moving approximately 125.2 m/s 3.75 * 10² m after the parachute opens.