Question

The position-time graph for a bug crawling along a line is shown in the figure below. Determine whether the velocity is positive, negative, or zero at each of the times marked on the graph. During which time interval(s) is the velocity decreasing and during which time interval(s) is the velocity increasing?

Answer 1

For each time marked on the graph:

• t₁ → v is negative

• t₂ → v is positive

• t₃ → v is positive

• t₄ → v is decreasing

• t₅ → v is zero

Intervals:

• Increasing: where the graph is concave up

,

• Decreasing: where the graph is concave up

Step-by-step explanation

In a position-time graph, the velocity of the object is represented by the slope of the curve at each time. If the graph is decreasing, then the velocity is negative; if the graph is increasing, then the velocity is positive; if the graph has a maximum or minimum, then the velocity is zero.

In this case, there is one local minimum and one local maximum, and at the end, the curve seems to follow a constant behavior. So at all those points, the velocity is zero.

Before the first local minimum, the function is decreasing, then it increases until the maximum and, finally, decreases again until the constant value,

Hence, for each time marked on the graph, the velocity is:

• t₁ → v is negative

• t₂ → v is positive

• t₃ → v is positive

• t₄ → v is decreasing

• t₅ → v is zero

The increasing or decreasing intervals of the velocity are given by the second derivative of the position - in other words, they are given by the sign of the acceleration.

The second derivative is represented by the concavity of the graph,

So, the velocity is increasing during the intervals where the graph is concave up - approximately from 0 to t₃, and decreases where the graph is concave down.