Question

Within 4.0 seconds of liftoff, a spacecraft that is uniformly accelerating straight upward from rest reaches an altitude of 450 m [up].
a) What is the spacecraft’s acceleration?
b) At what velocity is the spacecraft travelling when it reaches this altitude?

Answers

GPT 3.5
To solve this problem, we will use the following kinematic equation:

h = ut + (1/2)at^2

Where:
h = altitude
u = initial velocity (0 m/s, as the spacecraft is at rest)
a = acceleration
t = time

Given:
h = 450 m
t = 4.0 s
u = 0 m/s

a) To find the acceleration:
Using the given values in the equation, we get:

h = ut + (1/2)at^2
450 = 0 + (1/2)a(4.0)^2
450 = 8a

Divide both sides by 8 to isolate "a":
450/8 = a
a ≈ 56.25 m/s^2

Therefore, the spacecraft's acceleration is approximately 56.25 m/s^2.

b) To find the velocity:
We can use the formula for velocity:

v = u + at

Given:
u = 0 m/s
t = 4.0 s
a = 56.25 m/s^2

Substituting the values into the equation:

v = 0 + 56.25(4.0)
v ≈ 225 m/s

Therefore, the spacecraft is traveling at a velocity of approximately 225 m/s when it reaches an altitude of 450 m.