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Instanteous velocity

by eplymale3043
Tags: instanteous, velocity
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eplymale3043
#1
Sep4-08, 07:47 PM
P: 20
1. The problem statement, all variables and given/known data

If an arrow is shot upward on the moon with a velocity of 59 m / s, its height in meters after t seconds is given by h = 59(t) - .83(t). Find the instantaneous velocity after one second.


2. Relevant equations




3. The attempt at a solution

h = 59(1) - .83(1)2
h = 58.17

h = 59(1.0000001)- .83(1.0000001)2
h = 59.0000059 - 0.8300001660000083
h = 58.1700057339999917

v = 58.1700057339999917 - 58.17 / .0000001
v = 57.339999917

thats as far as i get. i think the instant vel would be 57.3 m / s because it wont take the answer 57 or 57.5.

can someone help me?
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Hootenanny
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Sep4-08, 07:53 PM
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Quote Quote by eplymale3043 View Post
1. The problem statement, all variables and given/known data

If an arrow is shot upward on the moon with a velocity of 59 m / s, its height in meters after t seconds is given by h = 59(t) - .83(t). Find the instantaneous velocity after one second.


2. Relevant equations




3. The attempt at a solution

h = 59(1) - .83(1)2
h = 58.17

h = 59(1.0000001)- .83(1.0000001)2
h = 59.0000059 - 0.8300001660000083
h = 58.1700057339999917

v = 58.1700057339999917 - 58.17 / .0000001
v = 57.339999917

thats as far as i get. i think the instant vel would be 57.3 m / s because it wont take the answer 57 or 57.5.

can someone help me?
Welcome to PF eplymale3043,

Since you've posted this in the calculus forums, I'm guessing that we should use calculus. So, what is the definition of instantaneous velocity, or velocity in general?
eplymale3043
#3
Sep4-08, 07:56 PM
P: 20
The speed at which an object is moving at any given time. right?

Hootenanny
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Sep4-08, 08:02 PM
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Instanteous velocity

Quote Quote by eplymale3043 View Post
The speed at which an object is moving at any given time. right?
Speed and velocity are two different quantities, but that's not what I was getting at. What is the definition of velocity in terms of distance and time?
eplymale3043
#5
Sep4-08, 08:05 PM
P: 20
v =[tex]\Delta[/tex]d/[tex]\Delta[/tex]t
Hootenanny
#6
Sep4-08, 08:09 PM
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Quote Quote by eplymale3043 View Post
v =[tex]\Delta[/tex]d/[tex]\Delta[/tex]t
Correct, that would be the average velocity. For instantaneous velocity one would take the limit (substituting d for h for clarity):

[tex]v=\lim_{t\to0}\frac{\Delta h}{\Delta t} = \frac{dh}{dt}[/tex]

Do you follow?
eplymale3043
#7
Sep4-08, 08:11 PM
P: 20
Yes, so far.
Hootenanny
#8
Sep4-08, 08:13 PM
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Quote Quote by eplymale3043 View Post
Yes, so far.
Now it's your turn to do some work. You are given h(t) in the question, all you need to do is find h'(t) and then plug in the numbers.
eplymale3043
#9
Sep4-08, 08:21 PM
P: 20
h'(t) meaning the inverse?
Hootenanny
#10
Sep4-08, 08:30 PM
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Quote Quote by eplymale3043 View Post
h'(t) meaning the inverse?

h'(t)
meaning the first derivative of h with respect to t. As I said in my previous post, velocity is defined as the rate of change of displacement with time.
eplymale3043
#11
Sep4-08, 08:35 PM
P: 20
Quote Quote by Hootenanny View Post

h'(t)
meaning the first derivative of h with respect to t. As I said in my previous post, velocity is defined as the rate of change of displacement with time.
I'm sorry, I dont understand what you mean.
Hootenanny
#12
Sep4-08, 08:43 PM
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Quote Quote by eplymale3043 View Post
I'm sorry, I dont understand what you mean.
You are given h(t), which is a function describing how the height [displacement] of the arrow varies with time. Velocity is defined as the time derivative of displacement (height in this case), hence you need to take the derivative of h(t) with respect to t. In other words evaluate:

[tex]\frac{dh}{dt} = \frac{d}{dt}\left(59t - 0.83t^2\right)[/tex]

Does that make sense?
eplymale3043
#13
Sep4-08, 08:50 PM
P: 20
Yes, that makes alot more sense.

so, basically. I need to plug in the change in height for dh and change in time for dt.

then solve? im still not sure what d by itself it.
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#14
Sep4-08, 08:52 PM
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Quote Quote by eplymale3043 View Post
Yes, that makes alot more sense.

so, basically. I need to plug in the change in height for dh and change in time for dt.

then solve? im still not sure what d by itself it.
No, you need to take the derivative of h(t) and then plug in the numbers. You are studying calculus aren't you?
eplymale3043
#15
Sep4-08, 08:56 PM
P: 20
Yes, but I'm really lost.

How would I find the derivative?
Hootenanny
#16
Sep4-08, 08:59 PM
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Quote Quote by eplymale3043 View Post
Yes, but I'm really lost.

How would I find the derivative?
There are plenty of resources available on the internet regarding calculus, but I would suggest that you sit down a read the first few chapters on differentiation from your textbook. I'm more than happy to help you though this problem, but there is little more than can be done without knowledge of differentiation.
eplymale3043
#17
Sep4-08, 10:03 PM
P: 20
After studying the chapter and doing some research in my calculus book, I have calculated that the velocity at 1 second is 57.34 m / s.

Does this sound right to you?
Hootenanny
#18
Sep5-08, 02:13 AM
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Quote Quote by eplymale3043 View Post
After studying the chapter and doing some research in my calculus book, I have calculated that the velocity at 1 second is 57.34 m / s.

Does this sound right to you?
Sounds spot on to me


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