Physics Help!

[allies]

Heyo... for all you physics buffs (even if you're not a buff, you probably still know how to do these) I have two questions that I don't understand how to do:

A 1000 kg weather rocket is launched straight up. The rocket motor provides a constant acceleration for 18 s, then the motor stops. The rocket altitude 22 s after launch is 5300 m. You can ignore any effects of air resistance.

(a) What was the rocket's acceleration during the first 18 s?
(b) What is the rocket's speed as it passes through a cloud 5300 m above the ground?

So far I have:
mass is extraneous
ti = 18s
tf = 22s
delta t = 4s
xi = ?
xf = 5300m

Vavg = (5300- xi)/4


I don't know how to get either a or b


2nd question:
Careful measurements have been made of Olympic sprinters in the 100 meter dash. A simple but reasonably accurate model is that a sprinter accelerates at 3.7 m/s2 for 3.36 s, then runs at constant velocity to the finish line.
(a)What is the race time for a sprinter who follows this model?
9.53s
(b) A sprinter could run a faster race by accelerating faster at the beginning, thus reaching top speed sooner. If a sprinter's top speed is the same as in part a, what acceleration would he need to run the 100 meter dash in 9.63 s?
(c) By what percent did the sprinter need to increase his acceleration in order to decrease his time by 1%?
%

I have:
top speed = 12.432 m/s

and that's about it

All I'm asking is for at least a little direction towards the answer, I don't learn things if I'm just given it... so help would be appreciated!

 

[silverpig]

Hint for 1(a): The acceleration of the rocket is -9.81 m/s^2 from t=18 to t=22. Break the problem up into two segments, one from t=0 to t=18 and one from t=18 to t=22

 

[Howard]

ti generally refers to initial time, but why do you set initial time to 18s when you are not finding a difference between ti and tf, or of functions of ti and tf?

You should always start with a free-body diagram for physics problems, unless you're sure they're not necessary. Even then, they help to order the problem-solving process.

 

[Sam334]

Originally posted by: silverpig
Hint for 1(a): The acceleration of the rocket is -9.81 m/s^2 from t=18 to t=22. Break the problem up into two segments, one from t=0 to t=18 and one from t=18 to t=22

What he said. And remember accel isn't the same as velocity. It means the change in velocity is constant for the first 18 seconds.

 

[Howard]

1.a) = 29.9m/s^2?

2.a) Your answer is wrong.

 

[allies]

Originally posted by: silverpig
Hint for 1(a): The acceleration of the rocket is -9.81 m/s^2 from t=18 to t=22. Break the problem up into two segments, one from t=0 to t=18 and one from t=18 to t=22

i understand that, but i'm lacking enough variables to the point where I'm still confused on how to solve either a or b

 

[TuxDave]

Do you know your kinematic equations?

d = 0.5*a*t^2+v*t+di

 

[Howard]

Originally posted by: allies

Originally posted by: silverpig
Hint for 1(a): The acceleration of the rocket is -9.81 m/s^2 from t=18 to t=22. Break the problem up into two segments, one from t=0 to t=18 and one from t=18 to t=22

i understand that, but i'm lacking enough variables to the point where I'm still confused on how to solve either a or b

Solve for a from t=0 to t=18
Solve for v at t=18
Solve for height at t=18
Solve for a from t=18 to t=22
Solve for v at t=22
Solve for height at t=22 (height is given so fill out the rest of the equation)
Solve for a from t=0 to t=18

a should remain as a variable from the first step to the second last step.

 

[allies]

Originally posted by: Howard
1.a) = 29.9m/s^2?

2.a) Your answer is wrong.

29.9 = no

how is my answer wrong?

 

[TuxDave]

Originally posted by: allies

Originally posted by: Howard
1.a) = 29.9m/s^2?

2.a) Your answer is wrong.

29.9 = no

how is my answer wrong?

You're in the right ballpark so you probably did something right. Double check your math.

 

[allies]

... shoot i'm getting tired / panicky... time for a shower then back to work on this

 

[DrPizza]

lets see if I can help... I'm leaving units out. Sorry, but it's late here. That would be an unforgiveable error in my own physics class.
a is the acceleration from the motor of the rocket.
fvck the formulas (yes, this statement is coming from a physics teacher)... especially s = s0 + vit + 1/2 at^2. They don't help you to understand anything.

Let's use a as the acceleration of the rocket during the first 18 seconds.
The rocket's velocity at 18 seconds is 18a.
For the first 18 seconds, the average velocity of the rocket is 9a.
For the first 18 seconds, the rocket will have travelled 9a*18
from 18 to 22 seconds, the rocket will have slowed by 4*9.8
The average velocity from 18 to 22 seconds will be (18a + (18a - 39.2) )/2 or 18a - 19.6
the distance the rocket will have travelled during those 4 seconds is 4(9a - 19.6)

Soooo.... 9a*18 + 4(18a -19.6) = whatever the hell the altitude of the rocket was at 22 seconds.

incidentally, working the problem out this way, only using the most basic formulas (or I'd like to call them concepts) leads to the velocity of the rocket at 22 seconds as well.

 

[DrPizza]

note: if anyone has a way of getting students to understand problems that require use of the formula vf^2 = vi^2 + 2as, without actually using that formula, I'd love to know. When I introduce those problems, I derive the formula for the students from the most basic formulas, then show them a couple of problems that I simply term "plug and chug" problems. They are about the only type of problems that I hate.

 

[TuxDave]

Originally posted by: DrPizza
note: if anyone has a way of getting students to understand problems that require use of the formula vf^2 = vi^2 + 2as, without actually using that formula, I'd love to know. When I introduce those problems, I derive the formula for the students from the most basic formulas, then show them a couple of problems that I simply term "plug and chug" problems. They are about the only type of problems that I hate.

How about doing things graphically using a velocity vs time graph?

 

[DrPizza]

oh, one more thing...
Thanks for the problem. I'm going to "borrow" it for class tomorrow.
It'll work perfectly - I have 3 new concepts to teach tomorrow; that'll make a great warm-up problem and will lead to the other concepts.

 

[DrPizza]

Originally posted by: TuxDave

Originally posted by: DrPizza
note: if anyone has a way of getting students to understand problems that require use of the formula vf^2 = vi^2 + 2as, without actually using that formula, I'd love to know. When I introduce those problems, I derive the formula for the students from the most basic formulas, then show them a couple of problems that I simply term "plug and chug" problems. They are about the only type of problems that I hate.

How about doing things graphically using a velocity vs time graph?

Hey, not a bad idea! I've been using a lot of classroom experiments and data collection so far. The students are pretty used to the graphs. I could always just save those problems until slightly later when I introduce them.

 

[allies]

i'm still completely lost!

 

[DrPizza]

okay, I'll put notes at the end of the other post.

 

[DrPizza]

Originally posted by: DrPizza
lets see if I can help... I'm leaving units out. Sorry, but it's late here. That would be an unforgiveable error in my own physics class.
a is the acceleration from the motor of the rocket.
fvck the formulas (yes, this statement is coming from a physics teacher)... especially s = s0 + vit + 1/2 at^2. They don't help you to understand anything.

Let's use a as the acceleration of the rocket during the first 18 seconds.
The rocket's velocity at 18 seconds is 18a.
<acceleration is the time rate of change of velocity. if the acceleration is a, then every second, the velocity increases by a units. So, after 18 seconds, the velocity will have increased by "a", 18 times.>
For the first 18 seconds, the average velocity of the rocket is 9a.
<I assume that the rocket started on a launch pad, i.e. from rest. The initial velocity was zero and the velocity at 18 seconds is equal to 18a. So, to find the average velocity, add them together, divide by two.>
For the first 18 seconds, the rocket will have travelled 9a*18
<this is simply displacement equals average velocity * time>
from 18 to 22 seconds, the rocket will have slowed by 4*9.8
<the velocity decreases by 9.8 meters per second each second due to gravity. So, over 4 seconds, it will have decreased by 9.8 meters per second 4 times. >
The average velocity from 18 to 22 seconds will be (18a + (18a - 39.2) )/2 or 18a - 19.6
<at 18 seconds, the velocity was 9a. At 22 seconds, the velocity was 9a - 4*9.8. Again, the average velocity - add them together, divide by two.>
the distance the rocket will have travelled during those 4 seconds is 4(9a - 19.6)
<average velocity * time again.>
Soooo.... 9a*18 + 4(18a -19.6) = whatever the hell the altitude of the rocket was at 22 seconds.
<add the two displacements together. Solve the resulting simple algebraic expression for a>
<now remember, to find the velocity at 22 seconds, it's 18a - 4*9.8>


incidentally, working the problem out this way, only using the most basic formulas (or I'd like to call them concepts) leads to the velocity of the rocket at 22 seconds as well.

ooops, bolded section is where I had an error.
velocity from 18 to 22 seconds decreases from 18a to 18a - 4*9.8

 

[DrPizza]

2nd question:
Careful measurements have been made of Olympic sprinters in the 100 meter dash. A simple but reasonably accurate model is that a sprinter accelerates at 3.7 m/s2 for 3.36 s, then runs at constant velocity to the finish line.
(a)What is the race time for a sprinter who follows this model?
9.53s

Okay, here are some starters again.

Assuming the sprinter doesn't get a running start - they generally start from rest, right? The initial velocity is zero.

After 3.36 seconds, the velocity will be 3.36 * 3.7 m/s.

The average velocity for the first 3.36 seconds is (again, add them together, divide by 2)
So, 3.36 * 3.7 m/s (whatever that is) divided by 2.
Take the average velocity, multiply by the time, and you have the displacement for the first 3.36 seconds.
Hey, how much farther does your sprinter have to go to reach 100 meters?
Remember, he's running the rest of that distance at 3.36 * 3.7 m/s.




Did that help?

 

[allies]

adding the two displacements:: 198a-78.4=5300
a = 26.3717 which is apparently the wrong answer

wtf am i doing wrong?!

 

[DrPizza]

bedtime for me... good luck.

 

[DrPizza]

Originally posted by: allies
adding the two displacements:: 198a-78.4=5300
a = 26.3717 which is apparently the wrong answer

wtf am i doing wrong?!

beats me... I'm tired; your answer looks right to me. Maybe I blundered somewhere from being tired. I generally don't make such mistakes though.

edit: no, I found your mistake.

 

[DrPizza]

Originally posted by: allies
adding the two displacements:: 198a-78.4=5300
a = 26.3717 which is apparently the wrong answer

wtf am i doing wrong?!

How about you solved that wrong (or else my calculator is wrong)??
5300 + 78.4 = 5378.4
divide by 198, and my calculator is showing 27.1636 repeating.

What you did wrong is you subtracted the 78.4 from 5300 instead of added it.
Hopefully this was just one of those mistakes that you point at and laugh, saying "I'm such an idiot for making that mistake," Otherwise, I'd start reading the policy on dropping classes.

 

[allies]

hahah yea that's one of those mistakes, i'm running on like 5 hours of sleep as we speak... sorry