Answer:
a) The acceleration of the rocket is 249 m/s².
b) The acceleration of the rocket is 25 times the acceleration of a free-falling body (25 g),
c) The distance traveled in 0.900 s was 101 m.
d) The figures are not consistent. The acceleration of the rocket was 20 g.
Explanation:
Hi there!
a) To calculate the acceleration of the rocket let's use the equation of velocity of the rocket:
v = v0 + a · t
Where:
v = velocity of the rocket.
v0 = initial velocity.
a = acceleration.
t = time.
We know that at t = 0.900 s, v = 224 m/s. The initial velocity, v0, is zero because the rocket starts from rest.
v = v0 + a · t
Solving for a:
(v - v0) / t = a
224 m/s / 0.900 s = a
a = 249 m/s²
The acceleration of the rocket is 249 m/s²
b) The acceleration of gravity is ≅ 10 m/s². The ratio of the acceleration of the rocket to the acceleration of gravity will be:
249 m/s² / 10 m/s² = 25
So, the acceleration of the rocket is 25 times the acceleration of gravity or 25 g.
c) The equation of traveled distance is the following:
x = x0 + v0 · t + 1/2 · a · t²
Where:
x = position of the rocket at time t.
x0 = initial position.
v0 = initial velocity.
t = time.
a = acceleration.
Since x0 and v0 are equal to zero, then, the equation of position gets reduced to:
x = 1/2 · a · t²
x = 1/2 · 249 m/s² · (0.900 s)²
x = 101 m
The distance traveled in 0.900 s was 101 m.
d) Now, using the equation of velocity, let's calculate the acceleration. We know that at 1.40 s the velocity of the rocket is zero and that the initial velocity is 283 m/s.
v = v0 + a · t
0 m/s = 283 m/s + a · 1.40 s
-283 m/s / 1.40 s = a
a = -202 m/s²
The figures are not consistent because 40 g is equal to an acceleration of 400 m/s² and the magnitude of the acceleration of the rocket was ≅20 g.
