Question

After 5.2 seconds of uniform acceleration from rest, a spacecraft reaches an

altitude of 5.7 x 102 m [up]. What is the final velocity of the spacecraft? What is the
acceleration of the spacecraft?

Answer 1

The final velocity of the spacecraft is
`5.7 x 10^2` m/s and the acceleration is 109.6 m/s.

Step-by-step explanation:

To find the final velocity of the spacecraft, we can use the formula:

Vf = Vi + at

where Vf is the final velocity, Vi is the initial velocity (zero in this case), a is the acceleration, and t is the time.

Given that the spacecraft reaches an altitude of
`5.7 x 10^2` m in 5.2 seconds, we know the distance
`(5.7 x 10^2 m)` and the time (5.2 seconds).

The equation can be rearranged to solve for acceleration:

a = (Vf - Vi)/t

Substituting the values:

`a = (5.7 x 10^2 m - 0 m)/(5.2 seconds)`

a = 109.6 m/s²

Therefore, the final velocity of the spacecraft is
`5.7 x 10^2` m/s and the acceleration is 109.6 m/s².

Answer 2

Final Velocity: 109.615 m/s

Acceleration: 21.08 m/s²

Step-by-step explanation:

Distance traveled is given by

Velocity v = Δx/Δt

where Δx is the distance covered and Δt the time to cover that distance

Here Δx = 5.7 x 10² m and t = 5.2 seconds

So v = 5.7 x 10²/5.2 = 109.615 m/s

Acceleration is Δv/Δt where Δv is the change in velocity from an initial velocity and Δt the time taken to reach that initial velocity

Here initial velocity is 0, so Δv= 109.615 - 0 = 109.615m/s

Time to reach this velocity = 5.2 s

So acceleration a = 109.615/5.2 = 21.08 m/s²