Why would this be accelerating positively?

[helloword365]

Homework Statement

Check graph below (it'll make more sense)

Relevant Equations

No relevant equations. This is more like a concept/understanding question.

Above is a graph of the distance vs. time for car moving along a road. According the graph, at which of the following times would the automobile have been accelerating positively?

OK, so I was able to get that at t=5 and t=29, it's accelerating positively (I graphed the velocity vs time graph, and then got the acceleration graph (which is some constant c)), but I don't really get why at t=57 it also has a positive acceleration. After all, wouldn't the velocity be moving against the positive direction (since the displacement is negative, it's returning to the original location)? Hence, the acceleration should be negative since the velocity is moving against the positive direction.

 

[haruspex]

Homework Statement: Check graph below (it'll make more sense)
Relevant Equations: No relevant equations. This is more like a concept/understanding question.

View attachment 349099
Above is a graph of the distance vs. time for car moving along a road. According the graph, at which of the following times would the automobile have been accelerating positively?

OK, so I was able to get that at t=5 and t=29, it's accelerating positively (I graphed the velocity vs time graph, and then got the acceleration graph (which is some constant c)), but I don't really get why at t=57 it also has a positive acceleration. After all, wouldn't the velocity be moving against the positive direction (since the displacement is negative, it's returning to the original location)? Hence, the acceleration should be negative since the velocity is moving against the positive direction.

At t=59s, the velocity is negative (graph slopes down) but becoming less negative (slope levelling out). The change in velocity is positive, so the acceleration is positive.

 

[robphy]

This is a useful mnemonic:
$$a>0: \stackrel{++}{\cup} \qquad a=0: \stackrel{oo}{-} \qquad a<0: \stackrel{--}{\cap}$$

The acceleration (the second-derivative of displacement with respect to time)
is positive (a smiley face) at a point
when the displacement-vs-time graph near that point is a portion of a cup (a smile).
Note: for a cup, the velocity (i.e. the slope of x-vs-t) is increasing.

You can interpret the other cases similarly.

 

[haruspex]

This is a useful mnemonic:
$$a>0: \stackrel{++}{\cup} \qquad a=0: \stackrel{oo}{-} \qquad a<0: \stackrel{--}{\cap}$$

The acceleration (the second-derivative of displacement with respect to time)
is positive (a smiley face) at a point
when the displacement-vs-time graph near that point is a portion of a cup (a smile).
Note: for a cup, the velocity (i.e. the slope of x-vs-t) is increasing.

You can interpret the other cases similarly.

Just to add that while a positive acceleration implies a cup and a negative acceleration implies a cap, it doesn’t quite work the other way. If the first three derivatives are all zero and the fourth is positive, for example, you will also see a cup.

I would discard the horizontal line image. A sustained clearly straight line, at any angle, means zero acceleration.

 

[kuruman]

I have two one comments on this.

First, there is an additional point at which the acceleration is positive. The car comes to rest at about 18 min and then starts moving again in the positive x-direction at about t = 28 min.

Second, t Talking about positive acceleration is ambiguous unless one defines which way is "positive" which is not explicitly stated here. I prefer to think of the direction of the velocity relative to the direction of the acceleration.

• If the velocity and the acceleration point in the same direction, the speed is increasing.
• If the velocity and the acceleration point in opposite directions, the speed is decreasing.

So I look at the graph and I see that

• At t = 0 min. the second derivative (acceleration) is positive because the graph curves away from the time axis. The first derivative (velocity) is also positive because $x$ increases with time. Therefore the speed is increasing.
• At t = 12 min. the second derivative (acceleration) is negative because the graph curves towards the time axis. The first derivative (velocity) is positive because $x$ increases with time. Therefore the speed is decreasing.
• And so on.

I think what's important is not the arbitrary decision of whether the vectors involved are positive or negative, but what the particular shape of the graph means in terms of objects speeding up or slowing down.

 

[pbuk]

First, there is an additional point at which the acceleration is positive. The car comes to rest at about 18 min and then starts moving again in the positive x-direction at about t = 28 min.

That is not an 'additional' point - it was already mentioned in the OP:

OK, so I was able to get that at t=5 and t=29, it's accelerating positively

Second, talking about positive acceleration is ambiguous unless one defines which way is "positive" which is not explicitly stated here.

Yes it is: the question defines it by reference to the graph ("according to the graph..."). The y-axis is labelled with distance increasing upwards and so acceleration is positive when the second derivative of distance is positive.

I prefer to think of the direction of the velocity relative to the direction of the acceleration.

Your personal preference is not relevant here.

I think what's important is not the arbitrary decision of whether the vectors involved are positive or negative, but what the particular shape of the graph means in terms of objects speeding up or slowing down.

The question does not ask about objects speeding up or slowing down, it asks when acceleration is positive.

This is why the answer includes an interval near t=57 when the car is clearly slowing down, yet in the framework of the question acceleration, which is the second derivative of distance, is positive.

 

[kuruman]

That is not an 'additional' point - it was already mentioned in the OP

I agree. Somehow I missed it. Point crossed out. Thanks.

Yes it is: the question defines it by reference to the graph ("according to the graph..."). The y-axis is labelled with distance increasing upwards and so acceleration is positive when the second derivative of distance is positive.

Here I disagree. Acceleration is a vector and as such it requires a coordinate system to be expressed in. Distance is a scalar magnitude and as such is independent of coordinate system and never negative. If the ordinate were labeled "Position", I would agree with you.

Your personal preference is not relevant here.

Perhaps. However in cases where the sign of accelerations and velocities needs to be determined without an explicit definition of coordinate axes, it is a good idea to define as positive the direction of the velocity in which case the speed is always increasing for positive acceleration and decreasing for negative acceleration. It is a good idea because when novices consider these quantities, they attach disproportionate importance to their sign, which they believe is intrinsic property, and do not realize it is only a choice of convention. Shown below are only a few examples of the perennial question on this homework forum "Is the acceleration of gravity positive or negative?"

https://www.physicsforums.com/threa...ity-negative-and-when-is-it-positive.1054243/
https://www.physicsforums.com/threa...ty-as-positive-negative.1064624/#post-7108070
https://www.physicsforums.com/threads/question-about-acceleration-of-free-fall.1061243/#post-7071840
https://www.physicsforums.com/threads/is-the-value-of-g-negative-or-positive.896867/

And we have not seen the end. I think a departure from this rabbit hole can be achieved if, as homework helpers, we minimize the importance of the arbitrary sign of the acceleration vector and emphasize the importance of the relative sign between acceleration and velocity. The former is not as informative about the motion of the object as the latter.

So this question can be criticized in terms of what it teaches the student. Instead of asking "According the graph, at which of the following times would the automobile have been accelerating positively?", it could have asked "According the graph, at which times is the speed of the automobile (a) increasing and (b) decreasing?"