Question

A space shuttle sits on the launch pad for 2.0 minutes, and then goes from rest to 4600 m/s in 8.0 minutes. Treat its motion as straight-line motion. What is the average acceleration of the shuttle (a) during the first 2.0 minutes, (b) during the 8.0 minutes the shuttle moves, and (c) during the entire 10 minute period?

Answer 1

a.) a = 0 ms⁻²

b.) a = 9.58 ms⁻²

c.) a = 7.67 ms⁻²

Step-by-step explanation:

a.)

Acceleration (a) is defined as the time rate of change of velocity

`a = \frac{v_(2) - v_(1) } {t}`

Given data

Final velocity = v₂ = 0 m/s

Initial velocity = v ₁ = 0 m/s

As the space shuttle remain at rest for the first 2 minutes i.e there is no change in velocity so,

a = 0 ms⁻²

b.)

Given data

As the space shuttle start from rest, So initial velocity is zero

Initial velocity = v₁ = 0 ms⁻¹

Final velocity = v₂ = 4600 ms⁻¹

Time = t = 8 min = 480 s

By the definition of Acceleration (a)

`a = \frac{v_(2) - v_(1) } {t}`

`a = \frac{4600 - 0 } {480}`

a = 9.58 ms⁻²

c.)

Given data

As the space shuttle is at rest for first 2 min then start moving, So initial velocity is zero

Initial velocity = v₁ = 0 ms⁻¹

Final velocity = v₂ = 4600 ms⁻¹

Time = t = 10 min = 600 s

By the definition of Acceleration (a)

`a = \frac{v_(2) - v_(1) } {t}`

`a = \frac{4600 - 0 } {600}`

a = 7.67 ms⁻²