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

  • Thread starter Thread starter eplymale3043
  • Start date Start date
  • Tags Tags
    Velocity
Click For Summary
SUMMARY

The discussion focuses on calculating the instantaneous velocity of an arrow shot upward on the Moon with an initial velocity of 59 m/s, described by the height function h(t) = 59t - 0.83t². Participants clarify that instantaneous velocity is determined by taking the derivative of the height function, resulting in h'(t) = 59 - 1.66t. After evaluating this derivative at t = 1 second, the instantaneous velocity is confirmed to be 57.34 m/s. The conversation emphasizes the importance of understanding differentiation in calculus to solve such problems.

PREREQUISITES
  • Understanding of calculus concepts, specifically differentiation
  • Familiarity with the definition of instantaneous velocity
  • Ability to apply the limit definition of a derivative
  • Knowledge of basic physics principles related to motion
NEXT STEPS
  • Study the rules of differentiation in calculus
  • Learn how to apply the limit definition of a derivative
  • Explore the concept of instantaneous velocity in physics
  • Practice solving similar problems involving motion equations
USEFUL FOR

Students studying calculus, physics enthusiasts, and anyone seeking to understand the application of derivatives in motion analysis.

eplymale3043
Messages
18
Reaction score
0

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?
 
Physics news on Phys.org
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?
 
The speed at which an object is moving at any given time. right?
 
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?
 
v =\Deltad/\Deltat
 
eplymale3043 said:
v =\Deltad/\Deltat
Correct, that would be the average velocity. For instantaneous velocity one would take the limit (substituting d for h for clarity):

v=\lim_{t\to0}\frac{\Delta h}{\Delta t} = \frac{dh}{dt}

Do you follow?
 
Yes, so far.
 
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.
 
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:

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

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!
 

Similar threads

  • · Replies 19 ·
Replies
19
Views
9K
  • · Replies 8 ·
Replies
8
Views
4K
  • · Replies 9 ·
Replies
9
Views
1K
  • · Replies 5 ·
Replies
5
Views
4K
Replies
7
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 6 ·
Replies
6
Views
5K
  • · Replies 1 ·
Replies
1
Views
4K
Replies
1
Views
3K
  • · Replies 14 ·
Replies
14
Views
6K