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.