Question

12. A rocket, initially at rest on the ground, accelerates vertically. It accelerates uniformly until it

reaches a speed of 900 m/s after 30 s. After this period of uniform acceleration, the rocket
engine cuts out. During the next 90 s, the upward speed of the rocket decreases uniformly to
zero.
speed
m/s
times
plot a speed-time graph for the rocket for the first 120 s of it's flight ​

Answer 1

We kindly invite you to read carefully the explanation and check the image attached below.

Step-by-step explanation:

According to this problem, the rocket is accelerated uniformly due to thrust during 30 seconds and after that is decelerated due to gravity. The velocity as function of initial velocity, acceleration and time is:

`v_(f) = v_(o)+a\cdot (t-t_(o))` (1)

Where:

`v_(o)` - Initial velocity, measured in meters per second.

`v_(f)` - Final velocity, measured in meters per second.

`a` - Acceleration, measured in meters per square second.

`t_(o)` - Initial time, measured in seconds.

`t` - Final time, measured in seconds.

Now we obtain the kinematic equations for thrust and free fall stages:

Thrust (
`v_(o) = 0\,(m)/(s)`,
`a = 30\,(m)/(s^(2))`,
`t_(o) = 0\,s`,
`0\,s\le t< 30\,s`)

`v = 30\cdot t` (2)

Free fall (
`v_(o) = 900\,(m)/(s)`,
`a = -9.807\,(m)/(s)`,
`t_(o) = 30\,s`,
`30\,s \le t \le 120\,s`)

`v = 900-9.81\cdot (t-30)` (3)

Now we created the graph speed-time, which can be seen below.