Question

Within 4.0 s of liftoff, a spacecraft that is uniformly accelerating straight upward from rest reaches an altitude of 4.50 X 10m (up). What is the spacecraft's acceleration, and at what velocity is the spacecraft traveling when it reaches this attitude?

a. (a) 5.63 m/s², (b) 22.52 m/s
b. (a) 11.26 m/s², (b) 45.04 m/s
c. (a) 22.52 m/s², (b) 90.08 m/s
d. (a) 45.04 m/s², (b) 180.16 m/s

Answer 1

The acceleration of the spacecraft is 1.125 m/s², and it is traveling at a velocity of 4.5 m/s when it reaches an altitude of 4.50 x 10m.

Step-by-step explanation:

To find the acceleration of the spacecraft, we can use the formula:

vf = vi + at

where vf is the final velocity, vi is the initial velocity, a is the acceleration, and t is the time taken.

Since the spacecraft starts from rest, the initial velocity is 0. The final velocity can be found using the formula:

vf = vi + at

Plugging in the values, we have:

vf = 0 + at

vf = at

We know that the final velocity is 4.50 x 10 m/s and the time taken is 4.0 s. Plugging these values into the equation, we get:

4.50 x 10 = a x 4.0

Solving for a, we find that the acceleration is 1.125 m/s².

To find the velocity of the spacecraft when it reaches this altitude, we can use the formula:

vf = vi + at

Plug in the values:

vf = 0 + (1.125)(4.0)

Calculating, we find that the velocity is 4.5 m/s when it reaches this altitude.