# Instanteous velocity

1. ### eplymale3043

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)². 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?

2. ### Hootenanny

9,676
Staff Emeritus
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. ### eplymale3043

20
The speed at which an object is moving at any given time. right?

4. ### Hootenanny

9,676
Staff Emeritus
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. ### eplymale3043

20
v =$$\Delta$$d/$$\Delta$$t

6. ### Hootenanny

9,676
Staff Emeritus
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?

20
Yes, so far.

8. ### Hootenanny

9,676
Staff Emeritus
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. ### eplymale3043

20
h'(t) meaning the inverse?

10. ### Hootenanny

9,676
Staff Emeritus

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. ### eplymale3043

20
I'm sorry, I dont understand what you mean.

12. ### Hootenanny

9,676
Staff Emeritus
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. ### eplymale3043

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.

14. ### Hootenanny

9,676
Staff Emeritus
No, you need to take the derivative of h(t) and then plug in the numbers. You are studying calculus aren't you?

15. ### eplymale3043

20
Yes, but I'm really lost.

How would I find the derivative?

16. ### Hootenanny

9,676
Staff Emeritus
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. ### eplymale3043

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?

18. ### Hootenanny

9,676
Staff Emeritus
Sounds spot on to me

19. ### eplymale3043

20
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. ### HallsofIvy

41,270
Staff Emeritus
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!