Instantaneous Velocity of Arrow Shot on Moon - 59 m/s

  • Thread starter eplymale3043
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In summary: That's why I said to study differentiation in your textbook. If you can't figure out how to do that, then you might need to switch to a different course.In summary, the arrow is shot upward on the moon with a velocity of 59 m / s and its height is given by h = 59(t) - .83(t)². The instantaneous velocity is found to be 57.34 m / s.
  • #1
eplymale3043
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Homework Statement



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)². Find the instantaneous velocity after one second.


Homework Equations






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 won't take the answer 57 or 57.5.

can someone help me?
 
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  • #2
eplymale3043 said:

Homework Statement



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)². Find the instantaneous velocity after one second.


Homework Equations






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 won't 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?
 
  • #3
The speed at which an object is moving at any given time. right?
 
  • #4
eplymale3043 said:
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?
 
  • #5
v =[tex]\Delta[/tex]d/[tex]\Delta[/tex]t
 
  • #6
eplymale3043 said:
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?
 
  • #7
Yes, so far.
 
  • #8
eplymale3043 said:
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.
 
  • #9
h'(t) meaning the inverse?
 
  • #10
eplymale3043 said:
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.
 
  • #11
Hootenanny said:

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 don't understand what you mean.
 
  • #12
eplymale3043 said:
I'm sorry, I don't 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?
 
  • #13
Yes, that makes a lot more sense.

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

then solve? I am still not sure what d by itself it.
 
  • #14
eplymale3043 said:
Yes, that makes a lot more sense.

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

then solve? I am 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?
 
  • #15
Yes, but I'm really lost.

How would I find the derivative?
 
  • #16
eplymale3043 said:
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.
 
  • #17
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?
 
  • #18
eplymale3043 said:
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 :approve:
 
  • #19
yeah, it wanted me to keep finding the average velocity using. 1.5, 1.1, 1.05, 1.01, 1.005, and so on.

basically, finding the limit.
 
  • #20
You need to back up and think about the course you are taking. If they are asking you to find the instantaneous velocity, they are assuming you know how to differentiate!
 

1. What is instantaneous velocity?

Instantaneous velocity is the speed and direction of an object at a specific moment in time. It is a measure of how fast and in what direction an object is moving at a particular instant.

2. How is the instantaneous velocity of an arrow shot on the moon calculated?

The formula for calculating instantaneous velocity is: v = Δx / Δt, where v is the velocity, Δx is the change in position, and Δt is the change in time. In the case of an arrow shot on the moon at 59 m/s, the velocity would be 59 m/s.

3. Why is the instantaneous velocity of an arrow shot on the moon different from that on Earth?

The instantaneous velocity of an arrow shot on the moon is different from that on Earth because of the moon's lower gravity. The moon's gravitational pull is about 1/6th that of Earth's, so objects on the moon will experience less resistance and can travel at higher velocities.

4. Can the instantaneous velocity of an arrow shot on the moon change?

Yes, the instantaneous velocity of an arrow shot on the moon can change. It can change if the force acting on the arrow changes, such as being affected by lunar gravity or air resistance. Additionally, as the arrow travels, its velocity can also change due to changes in its direction and speed.

5. How does the instantaneous velocity of an arrow shot on the moon affect its trajectory?

The instantaneous velocity of an arrow shot on the moon affects its trajectory by determining its direction and speed. A higher velocity will result in the arrow traveling further, while a lower velocity may cause it to fall to the ground sooner. The direction of the velocity also plays a role in the arrow's trajectory as it determines the angle at which the arrow will travel.

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