#1

HELP PLEASE!

I need help with this problem,

I'm stumped.

A ball is thrown into the air by a baby alien on a planet in the system of Alpha Centauri with a velocity of 35 ft/s. Its height in feet after t seconds is given by y=35t - 13(t^2).

A. Find the average velocity for the time period beginning when t=2 and lasting

.01 s, .005 s, .002 s, .001 s.

B. Estimate the instanteneous velocity when t=2

-----------------------------

Could I also get some help with these,

but the first thing is way more important.

please and thank you pit monkeys.

I need help with this problem,

I'm stumped.

A ball is thrown into the air by a baby alien on a planet in the system of Alpha Centauri with a velocity of 35 ft/s. Its height in feet after t seconds is given by y=35t - 13(t^2).

A. Find the average velocity for the time period beginning when t=2 and lasting

.01 s, .005 s, .002 s, .001 s.

B. Estimate the instanteneous velocity when t=2

-----------------------------

Could I also get some help with these,

```
1. Prove that lim(10−3x)=4.
x→2
2. Prove that lim(x+3x)=4.
x→1
```

but the first thing is way more important.

please and thank you pit monkeys.

#2

It is 6.

#3

from y=35t - 13(t^2) u can deduce that acceleration of free fall on the planet is -26ms-2.. so use v = u + at and solve..

#4

It is 6.

For some reason I don't think you're right.

RiEdit:

from y=35t - 13(t^2) u can deduce that acceleration of free fall on the planet is -26ms-2.. so use v = u + at and solve..

I don't think that is right either.

*Last edited by Mr. Rittard at Sep 8, 2009,*

#5

It's 6.

#6

I know I'm not much help.

But damn you made my brain hurt

But damn you made my brain hurt

#7

It's 6.

I'm quite sure its not 6.

You're wrong.

#8

I'm quite sure its not 6.

You're wrong.

Listen Mr. Rittard, prove to me it's not 6 and we'll go from there.

#9

Listen Mr. Rittard, prove to me it's not 6 and we'll go from there.

there needs to be more than one solution.

it can't be six.

#10

there needs to be more than one solution.

it can't be six.

Damn, touche. It's been years since I took physics, but let's see . . .

Average velocity = change in position / change in time

So for A) --> find the positions (y) at the times you have listed, then divide that by change in time.

B) Derivative of position equation = velocity equation.

Take derivative of given equation, and simply plug in.

As for the limits, you can just plug those in or draw a graph to prove it depending on what your teacher is looking for.

#11

Damn, touche. It's been years since I took physics, but let's see . . .

Average velocity = change in position / change in time

So for A) --> find the positions (y) at the times you have listed, then divide that by change in time.

B) Derivative of position equation = velocity equation.

Take derivative of given equation, and simply plug in.

Its not coming out though.

ackk.

these answers don't seem right.

#12

the question is flawed since there are no planets in the alpha centauri system.

#13

Oh I forgot that initial velocity is 35m/s. There should be an equation you can use with Vi and Vf.

**** this is basic stuff man. I don't remember it, but there should be pretty much identical sample problems in your textbook.

**** this is basic stuff man. I don't remember it, but there should be pretty much identical sample problems in your textbook.

#14

the question is flawed since there are no planets in the alpha centauri system.

I tried that approach,

but my calc teacher just laughed and told me to get to work.

#15

HELP PLEASE!

I need help with this problem,

I'm stumped.

A ball is thrown into the air by a baby alien on a planet in the system of Alpha Centauri with a velocity of 35 ft/s. Its height in feet after t seconds is given by y=35t - 13(t^2).

A. Find the average velocity for the time period beginning when t=2 and lasting

.01 s, .005 s, .002 s, .001 s.

B. Estimate the instanteneous velocity when t=2

-----------------------------

Could I also get some help with these,`1. Prove that lim(10−3x)=4.`

x→2

2. Prove that lim(x+3x)=4.

x→1

but the first thing is way more important.

please and thank you pit monkeys.

First of all, meters/seconds are cooler, embrace their magical powers TS!

Second, I don't know the acceleration of gravity in your Alpha Centauri planet.

Average velocity = distance variation/time variation.

Instantaneous velocity = v0+a*time variation = 35ft/s+g*2s (g is negative)

And your second thing, just evaluate the function in 1;1,5;1,9;1,99;1,999 and then 2,5;2,1;2,001;2,0001 and do the same thing with the second one and see how they magically approach that number.

ps. drop the ****ing Imperial System.

*Last edited by damian_91 at Sep 8, 2009,*

#16

I fail hard.

your suggestions aren't coming out correctly.

maybe I'm doing the math incorrectly.

or perhaps my teacher entered the answered incorrectly.

your suggestions aren't coming out correctly.

maybe I'm doing the math incorrectly.

or perhaps my teacher entered the answered incorrectly.

#17

Average velocity:

(y2 - y1) / (x2 - x1)

Instantaneous velocity:

Is the derivative of the displacement function, which ends up being 35 - 26t. Substitute 2 for t.

(y2 - y1) / (x2 - x1)

Instantaneous velocity:

Is the derivative of the displacement function, which ends up being 35 - 26t. Substitute 2 for t.

#18

Average velocity:

(y2 - y1) / (x2 - x1)

Instantaneous velocity:

Is the derivative of the displacement function, which ends up being 35 - 26t. Substitute 2 for t.

but how does that fit with the problem.

I'm so lost.

#19

Math isn't physics.

#20

HELP PLEASE!

I need help with this problem,

I'm stumped.

A ball is thrown into the air by a baby alien on a planet in the system of Alpha Centauri with a velocity of 35 ft/s. Its height in feet after t seconds is given by y=35t - 13(t^2).

A. Find the average velocity for the time period beginning when t=2 and lasting

.01 s, .005 s, .002 s, .001 s.

B. Estimate the instanteneous velocity when t=2

-----------------------------

Could I also get some help with these,`1. Prove that lim(10−3x)=4.`

x→2

2. Prove that lim(x+3x)=4.

x→1

but the first thing is way more important.

please and thank you pit monkeys.

you're given the equation for position (y=35t - 13(t^2)). if you derive position, you get velocity, and if you derive velocity, you get acceleration. so derive the position equation and you get y'(t)=35-26t

plug 2 into that and you get -17 (it's negative because velocity can have a direction, in this case, it's falling down at t=2 seconds).

of course, it's been a long time since i've been in calculus so i am probably completely wrong. if i am, please enlighten me as i'm actually rather interested in this problem.

#21

you're given the equation for position (y=35t - 13(t^2)). if you derive position, you get velocity, and if you derive velocity, you get acceleration. so derive the position equation and you get y'(t)=35-26t

plug 2 into that and you get -17 (it's negative because velocity can have a direction, in this case, it's falling down at t=2 seconds).

of course, it's been a long time since i've been in calculus so i am probably completely wrong. if i am, please enlighten me as i'm actually rather interested in this problem.

I think thats right!

but what about the .01 seconds and such???

#22

not really sure, that next part is worded weirdly

#23

not really sure, that next part is worded weirdly

me either.

#24

I need help with this problem,

I'm stumped.

A ball is thrown into the air by a baby alien on a planet in the system of Alpha Centauri with a velocity of 35 ft/s. Its height in feet after t seconds is given by y=35t - 13(t^2).

A. Find the average velocity for the time period beginning when t=2 and lasting

.01 s, .005 s, .002 s, .001 s.

B. Estimate the instanteneous velocity when t=2

-----------------------------

Could I also get some help with these,`1. Prove that lim(10−3x)=4.`

x→2

2. Prove that lim(x+3x)=4.

x→1

but the first thing is way more important.

please and thank you pit monkeys.

Differentiate the equation to obtain 35-26t. Put t=2 and you'll get instantaneous velocity. In this case it is negative that means it is travelling downward.

For the first part. Av vel is total displacement by total time taken. Therefore find out the position of the object in every time interval.

Eg: y at t=2 is 35(2)-13(2^2) and at an interval of 0.01 secs is 35(2.01)-13(2.01^2). Subtract these two positions to get the displacement and divide by the time interval ie 0.01. Subsequently other parts can be solve.

Other part is just simple limit problem. x->2 means that the value of x is tending towards 2. It is not EXACTLY 2. It may be 1.9999999 something. But generally limit is used to take out the slope of a graph so in order to solve limit problems such as the ones listed above. Just substitute the value of x in the expression ie 10-3x to get 10-3X2 and the answer is 10-6 ie 4. Same with the other question./

Hope it helped.

#25

Average velocity:

(y2 - y1) / (x2 - x1)

Instantaneous velocity:

Is the derivative of the displacement function, which ends up being 35 - 26t. Substitute 2 for t.

This man speaks the truth. Which reminds me I have to study my Kinematics section for the exam coming up

#26

Differentiate the equation to obtain 35-26t. Put t=2 and you'll get instantaneous velocity. In this case it is negative that means it is travelling downward.

For the first part. Av vel is total displacement by total time taken. Therefore find out the position of the object in every time interval.

Eg: y at t=2 is 35(2)-13(2^2) and at an interval of 0.01 secs is 35(2.01)-13(2.01^2). Subtract these two positions to get the displacement and divide by the time interval ie 0.01. Subsequently other parts can be solve.

Other part is just simple limit problem. x->2 means that the value of x is tending towards 2. It is not EXACTLY 2. It may be 1.9999999 something. But generally limit is used to take out the slope of a graph so in order to solve limit problems such as the ones listed above. Just substitute the value of x in the expression ie 10-3x to get 10-3X2 and the answer is 10-6 ie 4. Same with the other question./

Hope it helped.

you are a god.

#27

you are a god.

Ha Ha! I'm just a Physics God.

PS: I'm sigging you.