Trying to find instantaneous velocity at a point

[mileena]

Homework Statement

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?

Homework 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

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?

 

[Mark44]

Homework Statement

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?

Homework 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

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.

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?

-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.

 

[mileena]

Thanks for replying Mark44!

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)??

 

[Mark44]

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.

 

[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.

 

[Mark44]

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.

 

[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 =

444so

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

 

[mileena]

Oh, I forgot to factor above:

(-16h² + 4h)/h =

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

-4(4h - 1) =

4 m/s

Ok, now the answers agree!

Thank you!

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

 

[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".