1. A Simple Model of a Rocket

2. How it gets into the air
The engine produces thrust from burning of powder
The trust from the burning of the powder results in kinetic energy.

3. First … Thrust Thrust is in Newtons for a given time
To clarify use a simplified curve
8 Newtons for 1.8 seconds
This is like a finger pushing the rocket into the air

4. Kinetic Energy and Work Done
Work by a constant applied force: 𝑊= 𝐹 𝑥 ∆𝑥
So then the change in kinetic energy is: ∆𝐾=𝑊𝑛𝑒𝑡
The result is that the work done by the rocket engine, we are looking at it as a constant applied force is then turned into kinetic energy

5. First Part of the Model Launch, and powered ascent
Launch occurs at time t=0
End of powered ascent is at 𝑇 𝑓𝑜
T=0, K=0
Work done by engine results in 𝑊 𝑒𝑛𝑔𝑖𝑛𝑒 =𝐾 𝑚𝑎𝑥 at 𝑡=𝑇 𝑓𝑜

6. Need to Find ∆𝑥 𝐹 𝑡 𝐹 𝑔 The engine burns for 1.8 seconds
Free body diagram: 𝐹= 𝐹 𝑡 − 𝐹 𝑔
𝐹 𝑡 =8𝑁 𝐹 𝑔 =𝑚𝑔
𝐹=𝑚𝑎 or 𝑎= 𝐹 𝑚
𝑎= ( 𝐹 𝑡 − 𝐹 𝑔 ) 𝑚
With the mass of the rocket at ~80g
But now we have a value for constant
acceleration …… we want distance ∆𝑥
𝐹 𝑔

7. The result of this exercise …
The distance an object travels is ∆𝑥=𝑣∗𝑡 𝑡ℎ𝑒 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑡𝑖𝑚𝑒𝑠 𝑡ℎ𝑒 𝑡𝑖𝑚𝑒
The acceleration is related to the velocity, and we know the time (1.8 seconds)
From calculus and physics
𝑣= 𝑎 𝑑𝑡
What does this mean for us?

8. We will do this graphically
At time zero, the rocket is standing still
At time 𝑡=0.1𝑠𝑒𝑐 the rocket has been accelerating for 0.1 sec and, in this case, 10 𝑚 𝑠 2 .
Every 0.1 second the speed increases by 1 𝑚 𝑠

9. So how far did we go?
Remember, this is only a model up to 𝑇 𝑓𝑜 , or 1.8 seconds
For example (not the answer) the average velocity is 𝑣 𝑎𝑣𝑔 =9 𝑚 𝑠 for 1.8 seconds
The distance is
∆𝑥= 𝑣 𝑎𝑣𝑔 ∗ 𝑇 𝑓𝑜
Why did we want this?

10. The work done is converted to kinetic energy ….
𝑊=𝐹∗∆𝑥 with:
𝐹= 𝐹 𝑡 − 𝐹 𝑔 and:
∆𝑥= 𝑣 𝑎𝑣𝑔 ∗ 𝑇 𝑓𝑜
And
𝑎= ( 𝐹 𝑡 − 𝐹 𝑔 ) 𝑚
Now you do it!!!