Problem: Your school science club has devised a special event for homecoming. You've attached a rocket to the rear of a small car that has been decorated in the blue-and-gold school colors. The rocket provides a constant acceleration for 9.0s. As the rocket shuts off, a parachute opens and slows the car at a rate of 5.0m/s2. The car passes the judges' box in the center of the grandstand, 990mfrom the starting line, exactly 12s after you fire the rocket. What is the car's speed as it passes the judges?

120 m/s

Explanation:

- Let "a" denote the acceleration due to the rocket's motion.

- Let x1 denote the distance traveled while the rocket after launching which is the first 9 seconds

- Let x2 denote the distance traveled between the period when the rocket shuts off and when the judges are reached which is the last 3 seconds

- Let v1 denote the velocity of the car just when the rocket shuts off.

- Let v2 denote the final velocity

Total distance from the starting line at which the car passes the judges' box in the center of the grandstand would be x1 + x2

Thus;

x1 + x2 = 990

We are told that the rocket provides a constant acceleration for 9 seconds.

Thus, distance travelled after 9 seconds will be gotten from Newton's equation of motion;

x = ut + ½at²

Initial velocity = 0 m/s and t = 9. Thus;

x1 = ½at² = (1/2)(a)(9²)

x1 = (1/2)(a)(81)

x1 = 40.5a

Likewise, final velocity after 9 seconds will be gotten from;

v = u + at

Thus;

v1 = 0 + a(9)

v1 = (a)(9)

v1 = 9a

We are told that the rocket now slows down at a rate of 5 m/s² which is a Deceleration

Thus, distance during this negative acceleration is;

x2 = v1(t) + (-½at²)

x2 = v1(t) - ½at²

Putting 9a for v1, 5 m/s² for a, 3 for t, we have;

x2 = 9a(3) - (1/2)(5)(3²)

x2 = (9a)(3) - (1/2)(5)(9)

x2 = 27a - 22.5

From earlier, we said x1 + x2 = 990

Thus;

40.5a + 27a - 22.5 = 990

67.5a = 990 + 22.5

67.5a = 1012.5

a = 1012.5/67.5

a = 15 m/s²

From earlier, since v1 = 9a

Then, v1 = 9(15)

v1 = 135 m/s

The cars speed when it passes the judges would be;

v = u + at

Putting 135 for u, -5 m/s²(negative because of Deceleration), and 3 s for t, we have;

v = 135 + (-5 × 3)

v = 135 - 15

v = 120 m/s

b.

hope this : )

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