Answer:
120 m/s
Step-by-step 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