How Is Instantaneous Acceleration Calculated in Velocity-Time Graphs?

[cFleming]

Hello, I'm trying to teach myself basic physics from a book I bought at Davis-Kidd, and I've come across a problem which puzzles me to no end.

It's quite simple: It's a problem that uses a graph of a baseball player's change in velocity with time.

It labels 3 points of the graph and asks one question about each of them. My problem is with the very first question.

The baseball player's velocity is steadily increasing, and is represented as a diagonally upward line on a velocity/time graph.

At point A his velocity is 2 meters/second after one second. now they ask you what is his instantaneous acceleration at that point. I would say to find the instantaneous acceleration I need to find the slope of an imaginary line tangent to point A. The equation would be change in velocity/change in time as time approaches zero

How did the authors get 4 m/s in the numerator? and 2s in the denominator, for a final answer of +2 m/s.

Sorry for the wordiness. I just hope I'm not ridiculed for my stupidity. I really am interested in basic physics, it's just the nuts and bolts that I need to work on. I even took two college calculus classes, but that was long ago.

 

[Rap]

If the plot of velocity on a velocity-time graph is a straight line sloping upward, then a tangent to that line at any point is the line itself. So I bet point B (or maybe C) is two seconds after A, and the velocity there is 6 m/s. So the acceleration is the change in velocity (6-2)m/s=4 m/s divided by the change in time which was 2s to give an acceleration of 2 m/s^2 (I assume you meant the final answer was 2 m/s^2, not 2 m/s as you wrote). Since its a straight line, that will also be the slope of the tangent at point A.

 

[cFleming]

Thanks Rap, I figured I would be ridiculed out of the forum. I am nowhere near as intelligent as the average physicist.

Your answer flows like a river. :PAnd you are omniscient as well! Point B IS two seconds after point A. You make this seem easy. Physics rocks!

 

[cFleming]

Ah, I was wrong.

At point B the velocity is lying on a horizontal line at 4 m/s. from seconds 2 to 3, the velocity is constant at 4 m/s. So something is not right with my calculations.

 

[Rap]

Hmm - I guess I am not omnicient. Can you describe the time-velocity graph in more detail,including where the three points are?

 

[cFleming]

Sure!

1] The graph goes: < upward to 2 m/s for the first 1 s, and that is where point A is>
Then it goes: <upward from 2 m/s to 4 m/s at 2 s for 1s from Point A to 2s (Haven't reached point B yet)>
It stays at: <steady acceleration of 4 m/s from s 2-3s and at 2 1/2 s is point B at 4 m/s>
Finally, at: <3s to 4s it goes down from 4 m/s to 2 m/s. Point C is at 3 1/2 s at 3 m/s>

Sorry, that's a lot to digest.

Thanks for the quick reply!

 

[11markus04]

The point in question is point A. The graph of the line at point A is f(x)=2x; therefore, its derivative (ie: the slope of a line tangent to point A) is f(x)'=2+C. This shows the instantanious acceleration to be 2m/(s^2). Dont over think it.

 

[11markus04]

ps: and keep up the self study man, good for you.

 

[11markus04]

Also, I just noticed your question is "why do the authors have 4m/2s," this is because they are trying to find the AVERAGE VELOCITY of the first straight portion of the graph. ie:
(4m-0m)/(2s-0s)= 4m/2s= 2m/s at point A (the half-way point of that portion of the line). Thats all this is. Then use this to find the acceleration next.

Also, be careful not to confuse velocity with acceleration, as they have different units, as well as different properties.

Hope this helps

 

[Rap]

11markus04 is right, but also, since it is a straight line, the tangent to the line at A is the line itself, so that is also the acceleration at point A, as well as the average acceleration at any point on that straight line (i.e. from 0 sec to 2 sec, where it takes a turn).