Calculting the slope of the tangent to find Instantaneous velocity

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Homework Help Overview

The discussion revolves around calculating instantaneous velocity using the concept of tangents to a position versus time graph. The original poster presents a problem involving the position of a pinewood derby car at various times and seeks assistance in understanding how to derive instantaneous velocity from the graph.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationship between instantaneous velocity and the slope of the tangent line on a graph. The original poster expresses confusion about how to determine the slope from the tangent and how to apply the concept of limits in this context.

Discussion Status

Some participants have offered clarifications on the definitions of instantaneous velocity and the slope of the tangent line. There is ongoing exploration of how to visually represent and measure the tangent on the graph, with various interpretations of the problem being discussed.

Contextual Notes

There are indications of uncertainty regarding the calculations of average velocity and the method for finding instantaneous velocity through graphical representation. Some participants question the accuracy of previous calculations and suggest alternative approaches for measuring slopes.

FatIndian
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EDIT: Ugh... I meant "calculating", of course.

This is an assignment that was due on the 16th. It's already been handed in, I just need some help with a question that I obviously was not able to answer in due time. I am coming back from a long, long hiatus from my academic pursuits and so some concepts that I had a firm grasp on in the past are now pretty fuzzy to me.

Homework Statement


P1. The position of a pinewood derby car was observed at various times; the results are summarized in the following table. Find the Average velocity of the car for (a) the first second, (b) the last 3s and (c) the entire period of observation.

t(s) 0-------1.0-------2.0-------3.0-------4.0-------5.0
x(m) 0-------2.3-------9.2-------20.7-------36.8-------57.5

P2. (a) Use the data in P1 to construct a smooth graph of position versus time. (b) By constructing tangents to the x(t) curve, find the instantaneous velocity of the car at several instants. (c) Plot the velocity versus time and, from this, determine the average acceleration of the car. (d) What was the initial velocity of the car?

Homework Equations


Average velocity: dx/dt
So P1 was easy enough:

(a) 2.30 m/s
(b) (57.5 - 9.20) / 3.00 = 16.1 m/s
(c) (57.5 - 0) / 5.00 = 11.5 m/s

P2:

(a) Constructing the graph was simple
(b) This is where I got stuck
(c) Couldn't calculate dv/dt because the previous part was incomplete

The Attempt at a Solution



So I know the instantaneous velocity corresponds to the limit of dx/dt as dt approaches zero, but I just didn't know how to go about doing that. The problem asks me to draw tangents and I did, I just didn't know how to derive the slope from that. I know the way to do this is pretty simple, but I'm just not getting it even after reading the relevant chapter several times.

Thanks in advance.
 
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Welcome to PF!

FatIndian said:
So I know the instantaneous velocity corresponds to the limit of dx/dt as dt approaches zero, but I just didn't know how to go about doing that. The problem asks me to draw tangents and I did, I just didn't know how to derive the slope from that. I know the way to do this is pretty simple, but I'm just not getting it even after reading the relevant chapter several times.

Hi FatIndian! Welcome to PF! :smile:

I think you're confusing two different things …

i] The instantaneous velocity is dx/dt.

ii] dx/dt is defined as the limit of {x(t + ∆t) - x(t)}/∆t as ∆t –> 0.

Forget ii]!

The velocity is dx/dt.

dx/dt is the slope of the tangent line.

The slope of the tangent line is the tangent of the angle that that line makes with the t-axis.

Any questions? :smile:
 
But how do I figure out at which point the tangent is hitting the t-axis?

Let's take t=5.0 as an example. The point is (5.0s, 57.5m). I know that when its tangent hits the t-axis, x=0, so all I have is (??s, 0m). I know I'm going about this the wrong way, I just can't figure out how.
 
FatIndian said:
But how do I figure out at which point the tangent is hitting the t-axis?

Let's take t=5.0 as an example. The point is (5.0s, 57.5m). I know that when its tangent hits the t-axis, x=0, so all I have is (??s, 0m). I know I'm going about this the wrong way, I just can't figure out how.

Hi FatIndian! :smile:

The question wants you to draw the curve, as smoothly as you can, and then draw the tangent at (5.0s, 57.5m).

You can draw it all the way to the t-axis if you like, or you can just draw enough of it to enable you to measure the slope.

But this is a drawing and measuring problem, not a calculating one. :smile:

(to calculate, you would need to use calculus … this is an introduction to how graphs and calculus are interrelated, and how you can use graphs instead of calculus. :wink:)
 
hey I had a similar question.
first I think you got wrong p1 b)
I think it suppose to be:57.5-20.7/5.0-3.0 which is 18.4..

second I am not sure myself how to find the instantenous velocity by drawing a tangent I think you need to crate a small triangle from the tangent then divide total y by total x and this is the instantenous velocity anyway I remmber I had troubles with this so I am not sure if any1 can explain better I would like to read thank you all
 
Hey Soulax! :wink:
Soulax said:
second I am not sure myself how to find the instantenous velocity by drawing a tangent I think you need to crate a small triangle from the tangent then divide total y by total x and this is the instantenous velocity

That's correct. :smile:

You can either draw a little triangle, or a great big one (usually as far as one of the axes) …

obviously, the larger the triangle, the easier it is to measure it! :biggrin:
 

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