Question

Within 3.5 s of lift-off, a space shuttle reaches an altitude of 3.50 102 m, uniformly accelerating during the entire time. ( /7)

(a) What is its acceleration?
(b) At what speed is the shuttle traveling when it reaches this altitude?
(c) How long would it take the shuttle to travel the next 350 m if it stops accelerating after 3.5 s?

Answer 1

Hi there!

(A)

We can solve for acceleration using the following equation:

Δd = vit + 1/2at²

The shuttle lifts-off from rest, so:

Δd = 1/2at²

Rearrange to solve for 'd':

2Δd/t² = a

2(350)/(3.5²) = 57.14 m/s²

(B)

Now, we can use this equation to solve for final velocity:

vf = vi + at

vf = at

vf = 57.14 × 3.5 = 200 m/s

(C)

Now, we can use the following equation to solve for displacement:

Δd = vt

Δd/v = t

350/200 = 1.75 sec