# Instantaneous Acceleration from a Velocity-Time graph

1. Feb 27, 2015

### physxaffinity

1. The problem statement, all variables and given/known data
A test driver is testing a new model car with a speedometer calibrated to read m/s rather than mi/h. The following series of speedometer readings were obtained during a test run along a long, straight road:

Time (s): 0 2 4 6 8 10 12 14 16
Speed (m/s): 0 0 2 6 10 16 19 22 22

I.e.,
(0 s, 0 m/s)
(2, 0)
(4, 2)
(6, 6)
(8, 10)
(10, 16)
(12, 19)
(14, 22)
(16, 22)

(a) Compute the average acceleration during each 2-s interval. Is the acceleration constant? Is it constant during any part of the test run?
(b) Make a velocityx-time graph of the data above, using scales of 1 cm = 1 s horizontally and 1 cm = 2 m/s vertically. Draw a smooth curve through the plotted points. By measuring the slope of your curve, find the instantaneous acceleration at t = 9 s, 13 s, and 15 s.

2. Relevant equations
Average Acceleration
a = (Δvx)/Δt

Instantaneous Acceleration
a = dvx/dt

3. The attempt at a solution

I know how to get the answers for (a) and these are 0m/s/s, 1 m/s/s, 2 m/s/s, 2 m/s/s, 3 m/s/s, 1.5 m/s/s, 1.5 m/s/s, and 0 m/s/s. Acceleration is constant at 2 m/s/s and 1.5 m/s/s.

For (b), I made a graph as specified and tried finding the slopes of the tangent lines at t = 9 s, 13 s, and 15 s. My answers have not matched the book's, which are 2.5 m/s/s, 1.5 m/s/s, and 0 m/s/s, respectively. It makes sense that the instantaneous acceleration at 15 s is 0 m/s/s as there is no rise in the tangent line at t = 15 s and so the slope equals 0.

2. Feb 27, 2015

### BvU

Hello Aff, and (a belated) welcome to PF

I looked at the plot and I can imagine that smoothing would move the point (10, 16) a bit down, e.g. to (10,15). Then the slope would match the 2.5 m/s2.

For the other two I wouldn't expeect your answers to deviate very much. am I right ?

3. Feb 27, 2015

### Staff: Mentor

Your calculated accelerations are correct, and match the book answers at 13 and 15 s. The book answer is wrong at 9 s, but yours is correct there too.

Chet

4. Feb 27, 2015

### physxaffinity

Hey BvU,

Thanks for the welcome!

My answers varied as I tried quite a few different tangent lines for each point. I was able to get an instantaneous accelerations of 3.5 m/s/s and ~3.0 at 9 s, 1.5 m/s/s and ~1.33 m/s/s at 13 s, and of course 0 m/s/s at 15 s.

@Chestermiller

Is finding the slope of the tangent lines at each point a correct way of finding the instantaneous acceleration at those points?

5. Feb 27, 2015

### Staff: Mentor

It's a very accurate approximation for the center of each time interval. With discrete data like this, it's the best you can do.

6. Feb 28, 2015

### BvU

Hi folks,

This is pretty pedantic, but I liked this exercise, so there. Some remarks:

The story is about an accelerating car, so gear shifts can be a reason for a not-all-that-smooth acceleration versus time.
That aside, the exercise tekst wants a smooth curve drawn. I'm no artist, so I let a program (Excel ) do the work: a third order polynomial does a reasonable job. Again, it doesn't know the driver takes his foot off the pedal at 14 s, so it overshoots, but never mind. And the four-digits are way too much accuracy (And yes, I did find the button to reduce that, but too late). See the red line.

I am pleased to see it indeed draws the line through (10, 15) -- but, again, it doesn't know about possible gear shifts.
Slope of line = derivative = instantaneous acceleration at t = 9, 13, 15s: 2.4, 1.25 and 0.04, respectively.
In view of the accuracy of the data, 2.5, 1.5 and 0 are adequate.

3.5 or 3 m/s2 at t = 9s to me seems on the high side. Unless it's a pretty bad driver, fumbling that gear shift -- but in that case a smooth line isn't justified.

7. Feb 28, 2015

### Staff: Mentor

Ah. I didn't notice the part about putting a smooth curve through the data.

Chet