# Homework Help: Calculting the slope of the tangent to find Instantaneous velocity

1. Sep 18, 2008

### FatIndian

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.

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

2. Relevant 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

3. 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.

2. Sep 18, 2008

### tiny-tim

Welcome to PF!

Hi FatIndian! Welcome to PF!

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?

3. Sep 19, 2008

### FatIndian

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.

4. Sep 20, 2008

### tiny-tim

Hi FatIndian!

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.

(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. )

5. Jan 12, 2011

### Soulax

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

6. Jan 13, 2011

### tiny-tim

Hey Soulax!
That's correct.

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!