Question

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

Answer 1

Answer:

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

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 ($v_{o} = 0\,\frac{m}{s}$, $a = 30\,\frac{m}{s^{2}}$, $t_{o} = 0\,s$, $0\,s\le t< 30\,s$)

$v = 30\cdot t$ (2)

Free fall ($v_{o} = 900\,\frac{m}{s}$, $a = -9.807\,\frac{m}{s}$, $t_{o} = 30\,s$, $30\,s \le t \le 120\,s$)

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

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