Question

At the minute I'm just stuck on c) and d). I might be able to figure out d) but c) is kinda confusing.
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A space shuttle has a mass of 2.0 x 10^6 kg. At lift-off the engines generate an upward force of 3.0 x 10^7 N.
a) What is the weight of the shuttle? (2.0 x 10^7 N)
b) What is the acceleration of the shuttle when launched? (5.0 m/s/s)
c) The avg. acceleration of the shuttle during its 10 minute launch is 13 m/s/s. What velocity does it attain? (7.8 km/s)
d) As the engines burn, the mass of the fuel becomes less and less. Assuming the force exerted by the engines remains the same, would you expect the acceleration to increase, decrease, or remain the same? Why?

Answer 1

For a), I used W = Fg = mg. This becomes\[\left( 2.0 \times 10^{6} \right) 9.8\]That multiplies up to 19600000 which rounds to (with 2 sig figs): \[2.0 \times 10^{7}\]
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For b) I used \[\Sigma Fy = Fa - W = ma\]which amounts to \[\left( 3.0 x 10^{7} \right) -\left( 2.0 \times 10^{7} \right) = \left( 2.0 \times 10^{6} \right)a\]This becomes \[1.0 \times 10 ^{7} = \left( 2.0 \times 10^{6} \right)a \]which then creates the following derived equation \[a = \left(1.0 \times 10 ^{7}\right) \div \left(2.0 \times 10^{6}\right)\]This solved for the value of a makes \[a = 5.0 m/s ^{2}\]

Answer 2

Typo in the first equation of b), and the last one (derived). \[first: \left(3.0 \times 10^{7}\right) - \left(2.0 \times 10^{7}\right) = \left(2.0 \times 10^{6}\right)a\]\[derived: \frac{ 1.0 \times 10^{7} }{ 2.0 \times 10^{6} }\]

Answer 3

For c, you assume the acceleration is constant (It's not indicated anywhere it's not).
So:
\[a=\frac{ dv }{ dt }\]
\[adt=dv\]
(Separable differential equation, easy!)
\[at = v+v_{0}\]
It's initial velocity is cero, right?
\[at = v\]
Where a = m/s^2 and t = 600 s.
So, after doing lots of basic and boring stuff, v is 7.8 km/s.

Answer 4

EDIT: At the minute I'm just stuck on letters c) and d). I might be able to figure out d) but c) is kinda confusing.
In case you haven't seen the edit.

Replying to your answer, what does the d stand for? o.o

Also, for d) the acceleration should decrease because it's going up, which means deceleration, or\[-g = -9.8 m/s ^{2}\]
If I'm wrong correct me.

Answer 5

For d)
Newtons second law states that:
\[F = \frac{ dp }{ dt }\]
Where F and p are vectors. p is momentum, which is equal to mass*velocity-

Answer 6

Oh! Sorrysorrysorry. You don't know calculus. D: I thought you did. :C
Let me try again then.

Answer 7

I'm only taking elementary physics, we haven't learned an equation with a d in it... xD

Answer 8

It's the derivative. D:

Answer 9

Acceleration is defined as the change of velocity with respect to time (It's basically the same as deltaV/deltat if a is constant).
When the change is inmediate (Delta is veryveryvery small) you denote it with a d, instead of a delta.

Answer 10

Nope, haven't learned that I'm in Pre-Calculus.
Wait, what? #confused o.o

Answer 11

are you saying it's dv instead of \[\Delta v\] ?

Answer 12

It's basically the same, if the acceleration is constant. It looks prettier though.

Answer 13

I mean... dv instead of \[\Delta v = v _{f} - v _{i}\]

Answer 14

\[a = \frac{ deltaV }{ t }=\frac{ V-V_{0} }{ t }\]
Assuming V0 is cero.
\[a = \frac{ V }{ t }\]
\[V\]

Answer 15

o.o you not using a delta sign is confusing me lolol

Answer 16

but I got the gist of it, I guess.

Answer 17

Sorry, code got all ugly.
\[a = \frac{ \Delta V }{t} = \frac{ V - V_{0} }{ t }\]
Assuming V0 is cero:
\[a = \frac{ V }{ t }\]
Therefore:
\[at = V \]

Answer 18

This is d) right? o.o
Because c) has only acceleration...

Answer 19

That's c. In c they are asking for v, when you have a and t.

Answer 20

wtf NEVERMIND THAT stupid question

Answer 21

Sorry bout that.

Answer 22

Don't worry. :)
Have you seen momentum?

Answer 23

So I would get \[v = at = 13 \times 10 = 130 ...\]WHAT IS THIS

Answer 24

I messed up somewhere... the answer says 7.8 km/s

Answer 25

Oh, is it the 10 minutes? Must be 600 s then

Answer 26

\[v = at = \left(13 \frac{ meters }{ second^{2} }\right) * (600 seconds) = 7800 \frac{ meters }{ second }\]

Answer 27

My reaction: wut
I can't believe I was so stupid. Thanks so much :)
I still don't exactly get d) though...

Answer 28

Just remember to always be working in the same units everything (Minuts with minutes, seconds with seconds, etc).
What do you know about momentum and Newton's second law?

Answer 29

Momentum? D:
The second law concerns inertia right?
The stupid question has no answer...

Answer 30

Second law basically says that:
\[F = ma\]
But, you know that your force is constant and that your mass decreases with time.
If m decreases, what does v have to do so that F remains constant?

Answer 31

I do know that formula... mass decreases with time? o.o
v decreases because it's proportional ?

Answer 32

Let's say F is 10.
If m is 5. What does a have to be?
What about if m is 2?

Answer 33

a is proportional to m, so the values decrease/increase together

Answer 34

No. If F is 10 and m is 5, a is 2. (F = ma, 10 = 5*2 )
If m is 2, a is 5. (10 = 2*5)
So, when you decrease your mass, your acceleration increases.

Answer 35

.... Oh, right. #stupidme
Sorry for being an idiot.

Answer 36

It's 4 am over here = n=

Answer 37

Which makes sense, don't you think? If you're in a car, and your driving through the highway with lots of stuff like a sofa or something inside, you're going to move slower than without it.
And, yeah, I understand, It's 5 am here. ;D

Answer 38

o.o that's amazing, why are you still up XD
I'm a regular procrastinator so this is normal for me though

Answer 39

Although I did take a nap from 4 to 12 o n o

Answer 40

You can also think of it this way.
If F is constant then, let's just say F=constant = c.
So:
\[F = ma = c\]
\[a = \frac{ c }{ m }\]
So a is inveresly proportional to m.

Answer 41

So that makes it decreasing?

Answer 42

I live in a very noisy neighborhood, so I like to sleep when it's noisy outside and awake when it's quiet.
And, yes it does. Because a is inversely proportional to m, when m decreases, a increases. And when m increases, a decreases.

Answer 43

and be awake*

Answer 44

since average acceleration is given to u, just use ur acceleration equation and calculate the final velocity at the end of 10 mins!

Answer 45

average acceleration = change in velocity/ time taken

Answer 46

what .-.
I've already gotten the answer for that...?

Answer 47

i thought u had problems with c)

Answer 48

Oh, but Lessis already answered it. :)

Answer 49

oh ok..so all done in this problem?

Answer 50

I'm having problems with d) but I gotta see what Lessis said first.

Answer 51

Nah, solved.

Answer 52

I might have other questions just saying, so if you want to wait o.o

Answer 53

lolz.. close this question first ! if u are done with this!

Answer 54

You have a good point lols