# Tangent Lines/Instaneous Velocity

## Homework Statement

I need to use the following data table to:
1. make a position-time graph
2. draw tangents (it isn't specified how many, but a previous practice question only drew 3 out of a possible 5)
3. create a time-velocity table.

0 --- 0
0.25 --- 0.29
0.50 --- 1.15
0.75 --- 2.59
1.00 --- 4.60
1.25 --- 7.19
1.50 --- 10.35

vinst = d/t

## The Attempt at a Solution

The position-time graph is attached. I think that's right, but what I'm really having trouble with is drawing the tangent lines and calculating the instantaneous velocity. I feel like I didn't draw the tangents right but I'm not sure how to do it differently.
Here's my calculations:
0.50: 7 / 1.75-0.25 = 3.7
1.00: 10 / 1.65-0.50 = 5.5
1.50: 10 / 1.50-0.75 = 6.2

#### Attachments

• 37.2 KB Views: 288

Related Introductory Physics Homework Help News on Phys.org
Andrew Mason
Homework Helper

## Homework Statement

The position-time graph is attached. I think that's right, but what I'm really having trouble with is drawing the tangent lines and calculating the instantaneous velocity. I feel like I didn't draw the tangents right but I'm not sure how to do it differently.
Here's my calculations:
0.50: 7 / 1.75-0.25 = 3.7
1.00: 10 / 1.65-0.50 = 5.5
1.50: 10 / 1.50-0.75 = 6.2
A tangent line at a certain point on the graph is a straight line that touches the graph only at that point.

What does the slope of the tangent line on a distance-time graph represent? Does that help you determine points for a velocity-time graph?

AM

Chestermiller
Mentor

## Homework Statement

I need to use the following data table to:
1. make a position-time graph
2. draw tangents (it isn't specified how many, but a previous practice question only drew 3 out of a possible 5)
3. create a time-velocity table.

0 --- 0
0.25 --- 0.29
0.50 --- 1.15
0.75 --- 2.59
1.00 --- 4.60
1.25 --- 7.19
1.50 --- 10.35

vinst = d/t

## The Attempt at a Solution

The position-time graph is attached. I think that's right, but what I'm really having trouble with is drawing the tangent lines and calculating the instantaneous velocity. I feel like I didn't draw the tangents right but I'm not sure how to do it differently.
Here's my calculations:
0.50: 7 / 1.75-0.25 = 3.7
1.00: 10 / 1.65-0.50 = 5.5
1.50: 10 / 1.50-0.75 = 6.2
It seems to me you did an excellent job, and have the right idea. Nice work.

Andrew Mason
Homework Helper
Your tangent lines (supposedly tangent at .5, 1.0 and 1.5) are not touching at those points. This affects the calculation of velocity at those points. If you plot the velocity vs. time you will see this. I would choose a tangent at t=1.25 since you do not have any points after t=1.5.

AM

Your tangent lines (supposedly tangent at .5, 1.0 and 1.5) are not touching at those points. This affects the calculation of velocity at those points. If you plot the velocity vs. time you will see this. I would choose a tangent at t=1.25 since you do not have any points after t=1.5.

AM
Thanks for the help.
Do you mean do a tangent at 1.25 instead of at 1.50 or do both?

Chestermiller
Mentor
Your tangent lines (supposedly tangent at .5, 1.0 and 1.5) are not touching at those points. This affects the calculation of velocity at those points. If you plot the velocity vs. time you will see this. I would choose a tangent at t=1.25 since you do not have any points after t=1.5.

AM
Gee. It looks to me like they are pretty tangent at these points. The tangent lines aren't perfect, but they look pretty good. Yet, when I calculated the velocities using central finite difference approximations, I got a straight line as a function of time, with all the velocities calculated from the drawn tangents lying below the (more accurate) numerically calculated velocities.

Chet

Andrew Mason
Homework Helper
Thanks for the help.
Do you mean do a tangent at 1.25 instead of at 1.50 or do both?
The problem with drawing a tangent at 1.5 is that you don't know what the line looks like after. I would use 1.25 but you can use as many as you like: you can use any points in between 0 and 1.5.

AM

Andrew Mason
Homework Helper
pebbles3:

Actually, the real problem is with your calculations. The slope of the tangent you have drawn at t=1.0 is 10/(1.65-.50) = 9.1 not 5.5. The slope of the tangent as you have drawn it at 1.5 is 10.35/(1.5-.75) = 13.8 (the point is (10.35, 1.5)). Check the slope calculation for t=.5. That should give you an accurate v-t graph.

You seem to be dividing by the first term in the denominator and then subtracting the second!! Slope is rise/run

AM

Last edited:
pebbles3:

Actually, the real problem is with your calculations. The slope of the tangent you have drawn at t=1.0 is 10/(1.65-.50) = 9.1 not 5.5. The slope of the tangent as you have drawn it at 1.5 is 10.35/(1.5-.75) = 13.8 (the point is (10.35, 1.5)). Check the slope calculation for t=.5. That should give you an accurate v-t graph.

You seem to be dividing by the first term in the denominator and then subtracting the second!! Slope is rise/run

AM
Ah I see, thank you!