Calc Question

[Reaver]

Alright, I've got a Calc question because the answer to this problem isn't in the book and I'm positive a problem like this will be on the test.

A ball is launched upward at a speed of 160ft/sec from a cliff 176 feet above the ground.
a) find the velocity function v(t) and the position function s(t)
b) at what time does the ball reach its highest point? What is its maximum height above the ground?
c) when does the ball strike the ground? What is its velocity at that instant?
d) Set up the 2 definite integrals that would give the total distance traveled by the ball.

a) Well we know from the information given that V sub-zero is 160ft/sec and S sub-zero is 176ft. V'(t) which is equal to acceleration a(t) = -32feet/sec. So v(t)=-32t + 160
from that s(t) = -16t^2 + 160t +176
The 160 and 176 came from V sub-zero and S sub-zero and just solving for the constant C.

b) Ball would reach it's highest point when the velocity equals 0...so
0= -32t + 160
t=5sec
now I'd find s(5) which is 576 feet.

c) Ball strikes the ground when the position = 0, so 0=-16t^2 + 160t +176
After some factoring I got t=11 and t=-1, time can't be negative so I disregard the 2nd. To find the velocity at this point I use v(11) which is -192 , negative because it accounts for direction.

d) For this one I would use a sign chart to determine where the function is negative. So that I can figure out my intervals on my integrals. From earlier work we see that the function reaches 0 velocity at t=5 so I would use 5 on the number line like so:

<---------------->
0 5 11
v + -

from 0 to 5 velocity is positive, from 5 to 11 velocity is negative

so to set up the integral, I divide it into two integrals the first definite integral from 0 to 5, the 2nd from 5 to 11.

the first function from 0 to 5 would use the function -32t + 160 and the 2nd would use 32t - 160, since we want the distance traveled and not the displacement. Both would be added together to get the final result.

Correct me where I'm wrong here, this is how I'd do the problem.

[Zeebu]

Alright, I've got a Calc question because the answer to this problem isn't in the book and I'm positive a problem like this will be on the test.

A ball is launched upward at a speed of 160ft/sec from a cliff 176 feet above the ground.
a) find the velocity function v(t) and the position function s(t)
b) at what time does the ball reach its highest point? What is its maximum height above the ground?
c) when does the ball strike the ground? What is its velocity at that instant?
d) Set up the 2 definite integrals that would give the total distance traveled by the ball.

a) Well we know from the information given that V sub-zero is 160ft/sec and S sub-zero is 176ft. V'(t) which is equal to acceleration a(t) = -32feet/sec. So v(t)=-32t + 160
from that s(t) = -16t^2 + 160t +176
The 160 and 176 came from V sub-zero and S sub-zero and just solving for the constant C.
so far so good

b) Ball would reach it's highest point when the velocity equals 0...so
0= -32t + 160
t=5sec good again

now I'd find s(5) which is 576 feet. looks good

c) Ball strikes the ground when the position = 0, so 0=-16t^2 + 160t +176
After some factoring I got t=11 and t=-1, time can't be negative so I disregard the 2nd. To find the velocity at this point I use v(11) which is -192 , negative because it accounts for direction. looks good

d) For this one I would use a sign chart to determine where the function is negative. So that I can figure out my intervals on my integrals. From earlier work we see that the function reaches 0 velocity at t=5 so I would use 5 on the number line like so:

<---------------->
0 5 11
v + -

from 0 to 5 velocity is positive, from 5 to 11 velocity is negative

so to set up the integral, I divide it into two integrals the first definite integral from 0 to 5, the 2nd from 5 to 11.

the first function from 0 to 5 would use the function -32t + 160 and the 2nd would use 32t - 160, since we want the distance traveled and not the displacement. Both would be added together to get the final result.
looks good again

Correct me where I'm wrong here, this is how I'd do the problem.

looks good to me, but if you wanted to make it a little less complicated, on the ball's way down, instead of using 32t-160| t=5..11 you could just use 32t|t=0..6

if you were allowed... both will give you 576

but all in all, youre golden unless i suck at life

[Reaver]

Thanks for taking the time to check it out =) I was fairly confident in doing all parts though the last one I was a little sketchy about

[PjOtTeR]

Alright, I've got a Calc question because the answer to this problem isn't in the book and I'm positive a problem like this will be on the test.

A ball is launched upward at a speed of 160ft/sec from a cliff 176 feet above the ground.
a) find the velocity function v(t) and the position function s(t)
b) at what time does the ball reach its highest point? What is its maximum height above the ground?
c) when does the ball strike the ground? What is its velocity at that instant?
d) Set up the 2 definite integrals that would give the total distance traveled by the ball.

a) Well we know from the information given that V sub-zero is 160ft/sec and S sub-zero is 176ft. V'(t) which is equal to acceleration a(t) = -32feet/sec. So v(t)=-32t + 160
from that s(t) = -16t^2 + 160t +176
The 160 and 176 came from V sub-zero and S sub-zero and just solving for the constant C.

b) Ball would reach it's highest point when the velocity equals 0...so
0= -32t + 160
t=5sec
now I'd find s(5) which is 576 feet.

c) Ball strikes the ground when the position = 0, so 0=-16t^2 + 160t +176
After some factoring I got t=11 and t=-1, time can't be negative so I disregard the 2nd. To find the velocity at this point I use v(11) which is -192 , negative because it accounts for direction.

d) For this one I would use a sign chart to determine where the function is negative. So that I can figure out my intervals on my integrals. From earlier work we see that the function reaches 0 velocity at t=5 so I would use 5 on the number line like so:

<---------------->
0 5 11
v + -

from 0 to 5 velocity is positive, from 5 to 11 velocity is negative

so to set up the integral, I divide it into two integrals the first definite integral from 0 to 5, the 2nd from 5 to 11.

the first function from 0 to 5 would use the function -32t + 160 and the 2nd would use 32t - 160, since we want the distance traveled and not the displacement. Both would be added together to get the final result.

Correct me where I'm wrong here, this is how I'd do the problem.

Isn't D just what you travel up and what you travel down added together?
Which just makes: 2* distance startingpoint to top + startingpoint to endpoint...
which is 2 * 400 + 176 = 1328
But i dont really understand... you need the integrals for both of parts, that are the formula's?

So you basicly get for 0 to 5: -16t^2 + 160t
And from 5 to 11: 16t^2 (if you start this graph at 0 instead of 5, else it's 16(t-5)^2...)

If you ask me, calculating velocity/speed when they ask for distance traveled is rather dumb... while the formula for distance is also given...

[Reaver]

Isn't D just what you travel up and what you travel down added together?
Which just makes: 2* distance startingpoint to top + startingpoint to endpoint...
which is 2 * 400 + 176 = 1328
But i dont really understand... you need the integrals for both of parts, that are the formula's?

So you basicly get for 0 to 5: -16t^2 + 160t
And from 5 to 11: 16t^2 (if you start this graph at 0 instead of 5, else it's 16(t-5)^2...)

If you ask me, calculating velocity/speed when they ask for distance traveled is rather dumb... while the formula for distance is also given...

The area under the curve of a velocity graph with respect to the x-axis is the distance traveled, if it's positive, if it's negative then it gives you the net area or displacement. I know there's multiple ways to do the problem but that's how they want it solved.

[PjOtTeR]

The area under the curve of a velocity graph with respect to the x-axis is the distance traveled, if it's positive, if it's negative then it gives you the net area or displacement. I know there's multiple ways to do the problem but that's how they want it solved.

You are still wrong. I think I made clear I didn't really understand the question. If you do it your way, it's still just 32t on the second part...

So from 0 to 5: -32t + 160
and from 5 to 11: 32t or 32(t-5) if you start it on 5.. which basicly is your -160.. hmm

And if you take the integrals of those formula's which are the area's under the curves of those velocity graphs with respect to their x-axis'....
Using the formula: 1/(1+a)x^(a+1)
You ofcourse get my aformentioned formula's then... which are:
From 0 to 5: -16t^2 + 160t
And from 5 to 11: 16t^2

[Reaver]

You are still wrong. I think I made clear I didn't really understand the question. If you do it your way, it's still just 32t on the second part...

So from 0 to 5: -32t + 160
and from 5 to 11: 32t

And if you take the integrals of those formula's which are the area's under the curves of those velocity graphs with respect to their x-axis'....
Using the formula: 1/(1+a)x^(a+1)
You ofcourse get my aformentioned formula's then... which are: 1/(1+a
From 0 to 5: -16t^2 + 160t
And from 5 to 11: 16t^2

Yes I understand how to get the antiderivative, but as the problem states I don't need the antiderivative I just need to set it up. I'm not actually solving anything in part D but just setting up the problem, so there's no need to calculate the antiderivative let alone figure out it's values on the interval 0 to 5 and 5 to 11.

Edit: You didn't comment on a through c, I'm taking it that you agree with those?

[Izor]

Physics Question*

[Reaver]

Physics Question*

Sure it's a physics question, from my calculus book from my calculus class =)

[Zeebu]

Isn't D just what you travel up and what you travel down added together?
Which just makes: 2* distance startingpoint to top + startingpoint to endpoint...
which is 2 * 400 + 176 = 1328
But i dont really understand... you need the integrals for both of parts, that are the formula's?

So you basicly get for 0 to 5: -16t^2 + 160t
And from 5 to 11: 16t^2 (if you start this graph at 0 instead of 5, else it's 16(t-5)^2...)

If you ask me, calculating velocity/speed when they ask for distance traveled is rather dumb... while the formula for distance is also given...

you broke math =(

You are still wrong. I think I made clear I didn't really understand the question. If you do it your way, it's still just 32t on the second part...

So from 0 to 5: -32t + 160
and from 5 to 11: 32t or 32(t-5) if you start it on 5.. which basicly is your -160.. hmm

And if you take the integrals of those formula's which are the area's under the curves of those velocity graphs with respect to their x-axis'....
Using the formula: 1/(1+a)x^(a+1)
You ofcourse get my aformentioned formula's then... which are:
From 0 to 5: -16t^2 + 160t
And from 5 to 11: 16t^2

no, without changing the time from 5-11 to 0-6 you cant integrate just the 32t and expect to get the correct answer

the total distance will be 976... 400 up and then 576 down

we'll skip straight to part D... i think we can all agree the rest is pretty solid

the only equation we really need is the s(t) one... we can differentiate the rest
s(t) = -16t^2+160t+174
v(t) = -32t+160
a(t) = -32

found out that it reaches highes point at 5 seconds (576 in air 400 above cliff)
found out it stops moving at 11 seconds (0 in air 176 below cliff)

the way you want to you it pjot would give is an incorrect answer to part D

-16(5)^2 + 160(5) = 400 up
the ball is currently not moving 576 feet above the ground and 400 feet above the cliff ledge
at this point you have 2 options
you can continue this same problem with the same time scheme (5-11 seconds) or creat a new problem (0-6 seconds) we'll look at both

0-6: to get the right answer, you only need to plug 6 into 16t^2 and you'll get 576 till it hits the ground (v(0)=0 and s(0)=0 relatively)

5-11: to do this you will need both an accel and a velocity
so the equation is 16t^2-160t where t goes form 5 to 11
so [16(11)^2-160(11)]-[16(5)^2-160(5)]
do the math and you'll get [176]-[-400] which is 576

add either of those together and theres your total distance traveled... 976

had you done it your pay pjot, the ball would have gone too far and burrowed into the ground by like 960 feet
16(11)^2-16(5)^2 = 1536
1536-576 = 960

[PjOtTeR]

you broke math =(



no, without changing the time from 5-11 to 0-6 you cant integrate just the 32t and expect to get the correct answer

the total distance will be 976... 400 up and then 576 down

we'll skip straight to part D... i think we can all agree the rest is pretty solid

the only equation we really need is the s(t) one... we can differentiate the rest
s(t) = -16t^2+160t+174
v(t) = -32t+160
a(t) = -32

found out that it reaches highes point at 5 seconds (576 in air 400 above cliff)
found out it stops moving at 11 seconds (0 in air 176 below cliff)

the way you want to you it pjot would give is an incorrect answer to part D

-16(5)^2 + 160(5) = 400 up
the ball is currently not moving 576 feet above the ground and 400 feet above the cliff ledge
at this point you have 2 options
you can continue this same problem with the same time scheme (5-11 seconds) or creat a new problem (0-6 seconds) we'll look at both

0-6: to get the right answer, you only need to plug 6 into 16t^2 and you'll get 576 till it hits the ground (v(0)=0 and s(0)=0 relatively)

5-11: to do this you will need both an accel and a velocity
so the equation is 16t^2-160t where t goes form 5 to 11
so [16(11)^2-160(11)]-[16(5)^2-160(5)]
do the math and you'll get [176]-[-400] which is 576

add either of those together and theres your total distance traveled... 976

had you done it your pay pjot, the ball would have gone too far and burrowed into the ground by like 960 feet
16(11)^2-16(5)^2 = 1536
1536-576 = 960

My way... I already said if you restart it at 0 you can use 16t^2
If you want to keep it at 5 to 11... you have to do 16(t-5)^2
This gives:
[16(11-5)^2-160(11-5)]-[16(5-5)^2-160(5-5)] =
[16(6)^2 - 160(6) ]-[16(0)^2 -160(0)]
This gives: 576 - 0 = 576

And that 1328 = 576 * 2 + 176
I at first just simply took the height as between startingpoint and endpoint... then realised it was wrong but forgot to chance the solution to it. nicely noticed.
So I indeed was correct with the 2nd fault you pointed out... yet you were right on my first fault. If I would've looked it over, I would've probably seen it.

[Zeebu]

ooo i missed the t-5 shizzle =)

what do you do pjot? still in school for something or out in real world

same reaver

what are you studying?

[PjOtTeR]

ooo i missed the t-5 shizzle =)

what do you do pjot? still in school for something or out in real world

same reaver

what are you studying?

Im currently studying Technology Management. I will probably make sure I get my propedeuce this year and then see what I will do. Im thinking of going more to the economical side... Im also thinking of going to USA to play soccer for some college or to try to become a pro here in the Netherlands... Cause I can go to some pro-club, but it's one of the worst in it's kind.

[Zeebu]

nifty

well... i am nearly a year out of a very good school with my mechanical engineering degree (as well as some other schtuff)

but if youre pro level soccer and lookin into that... dont go to my school... but if you want to go into engineering of any kind or get a good economics degree, talk to me

[Reaver]

Civil Engineering, I didn't start out as a Civil Engineer, I've actually changed major so I should be a junior but I'm actually a freshman, though many of my credits should transfer. I'm going to a local college most likely for the first 2-3 years, and then after that I'll transfer to a reputable 4 year college and graduate there, to get the name of the college on my degree =)