 ###### Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

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# Instantaneous Velocity - Problem 2

Norm Prokup ###### Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

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By finding the slope between two points that are very close together on a position curve, you can find an approximate value for the instantaneous velocity of an object. Another way of finding the information needed to estimate instantaneous velocity is to look at the position curve to find the object's location at certain times. By finding two points on the curve that are very close and calculating the average velocity, you can estimate the instantaneous velocity at one of those points.

Instantaneous velocity, like average velocity, can be calculated to be negative or positive.

In a previous example, we talked about how instantaneous velocity can be interpreted as the slope of the position curve. Now in this problem, I’ve been given a position curve and I’m going to use that idea of slope as instantaneous velocity to solve the problem.

The height h of t of a bungee jumper in feet, over time t in seconds is given by this graph. So all I know I know is that the bungee jumper starts at t equals 0 at a height of 300 and then jumps off. And his height goes down, down, down to 50 at t equals 5 and then back up to 250 at t equals 9. Let’s see if we can answer the questions based on that information.

Part a says what is the instantaneous velocity at t equals 5 and at t equals 9. Well let’s go back. Here is t equals 9 and notice that both of these points have something in common. If you’re to draw a tangent line to the curve at these points, you’d get a horizontal tangent, which means that a slope of 0. So the slope of the curve at both of these points is 0 and that’s the instantaneous velocity, 0. So both of those points, for an instant the bungee jumper is at rest.

Now part b, between 0 and 9, when is the velocity positive and when is negative? Well you just have to think in terms of slopes. Here, everywhere on this interval from 0 to 5, if you drew a tangent line, the tangent line would have a negative slope everywhere, except maybe at the endpoints. Between 0 and 5, we have a negative slope. That’s when the velocity is negative. I’ve answered the second part first, between 0 and 5.

When is it positive? Well right in here is where the tangent would have a positive slope. So that’s where the velocity’s positive, between 5 and 9.

For t between 0 and 9, estimate when the velocity’s greatest and when it’s least. Here we already know that between 0 and 9 the velocity’s negative in this interval, and positive in this interval. So the greatest interval is going to be somewhere in here. You just have to sort of estimate when the slope is greatest.

Imagine tangent lines. The tangent line with the steepest slope represents the point with the greatest velocity. You just have to kind of eyeball it in a graph like this where we don’t have any particular numbers. I’m going to say somewhere in here, right around t equals 7. The velocity is greatest at t approximately 7 seconds.

When is it least? This is tricky because remember, the velocity is negative here. Don’t confuse velocity with speed. The guy could be falling very fast, but his velocity is negative, he’s going in the downward direction, so his velocity is negative. The velocity is actually least when it’s most negative. That’s going to be somewhere in this interval. Let’s look for the steepest downward slope. Right about there, right about at t equals 3. So the velocity is least at t is about 3 seconds.

Remember when you’re looking at a graph that shows you position over time, the slope of that graph gives you instantaneous velocity.