# Trying to find instantaneous velocity at a point

1. Sep 3, 2013

### mileena

1. The problem statement, all variables and given/known data

Given the function s(t) = -16t2 + 100t which represents the velocity of an object in meters, what is the instantaneous velocity at t = 3 seconds?

2. Relevant equations

I think these are correct:

Average velocity equals the slope of the secant line connecting any two points on the graph if there is acceleration is not 0 (i.e., the graph is curved or exponential). (If acceleration is 0, so that the graph of the position vs. time graph is linear, just take the slope of the line.)

Instantaneous velocity equals the slope of the line tangent to a point on the graph

Instantaneous velocity also equals:

lim f(a)
x→a

3. The attempt at a solution

I did the slope for secant lines connecting the following pairs of time points:

2, 3
2.9, 3
2.99, 3
2.999, 3
2.9999, 3

The respective slopes went from:

20
5.6
4.16
4.016
4.0016

So I am postulating the limit at 3 is 4. So instantaneous velocity at 3 seconds is 4 m/s.

However, if I do:

lim (-16t2 + 100t)
x→3

I get 156. So the instantaneous velocity at 3 seconds is 156 m/s.

What am I doing wrong? Is the instantaneous velocity at 3 seconds 4 m/s or 156 m/s?

2. Sep 3, 2013

### Staff: Mentor

No, this isn't right for a couple of reasons.
1. $\lim_{x \to a}f(a) = f(a)$
f(a) is a constant, so its limit is just itself.
2. Instantaneous velocity at t = a is this:
$$\lim_{h \to 0}\frac{s(a + h) - s(a)}{h}$$
In other words, this is the limit of the slopes of the secant lines as the points that define the ends of the secant lines grow closer together. This is the same as what you're doing below.
-16t2 + 100t is NOT the velocity, so evaluating it at t = 3 will not give you the instantaneous velocity at that time. You need to use the difference quotient that I wrote above.

3. Sep 3, 2013

### mileena

So, if I just have one time value, I can't calculate the instantaneous velocity? I have to have two time values (a and h)??

4. Sep 3, 2013

### Staff: Mentor

You have the formula for the position, s(t). To get the instantaneous velocity at t = 3, you need to take the limit.
$$v(3) = s'(3) = \lim_{h \to 0} \frac{s(3 + h) - s(3)}{h}$$

Expand the s(3 + h) term and the s(3) term using the function formula, and combine them.
Divide by h.
Take the limit as h → 0.

That will give you the instantaneous velocity at time t = 3.

5. Sep 3, 2013

### mileena

Ok, thanks again.

I am really sorry for asking this, but am I supposed to replace h above in the formulas with the function formula in order to expand s(3 + h) and s(3):

-16t2 + 100t ?

And what is h?

I know the above sounds like I don't know what I am doing, which is true.

Maybe I should stick to just getting repeated slopes of secant lines to find the limit.

6. Sep 3, 2013

### Staff: Mentor

You have s(t) = -16t2 + 100t. Do you know how to work with function notation?

s(3) = -16(3)2 + 100(3)

So s(3 + h) would be what?

Don't worry about h - it represents a number reasonably close to zero.

7. Sep 3, 2013

### mileena

Oh duh, now that you put it that way, I can do it. I just got confused with the terminology!

So:

s(3 + h) =

-16(3 + h)2 + 100(3 + h) =

-16(9 + 6h + h2) + 300 + 100h =

144 -96h - 16h2 + 300 + 100h =

-16h2 + 4h + 444

and

s(3) =

-16(3)2 + 100(3) =

144 + 300 =

444

so

s[(3 + h) - s(3)]/h =

(-16h2 + 4h + 444 - 444)/h =

(-16h2 + 4h)/h

so

lim s[(3 + h) - s(3)]/h = +∞ m/s
h→0

But that answer does not agree with my earlier answer of 4 m/s

Last edited: Sep 3, 2013
8. Sep 3, 2013

### mileena

Oh, I forgot to factor above:

(-16h2 + 4h)/h =

[-4h(4h - 1)]/h =

-4(4h - 1) =

4 m/s

Thank you!

So I know now instantaneous velocity at a point p is the lim as h→0 of the difference quotient!

9. Sep 4, 2013

### verty

The other way is to use the nax^(n-1) formula to find the derivative of s. But I take it you were not allowed to use that in this question. Usually when the question wants you to use the difference quotient, it'll say something like "find the instantaneous velocity from first principles".