Within 4.0 seconds of liftoff, a spacecraft that is uniformly accelerating straight upward from rest reaches an altitude of 450 m [up].

Question

a) What is the spacecraft’s acceleration?
b) At what velocity is the spacecraft travelling when it reaches this altitude?

Answer 1

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.