Q.Figure 2.15 gives a speed-time graph of a particle in motion along a constant direction. Three equal intervals of time are shown. In which interval is the average acceleration greatest in magnitude? In which interval is the average speed greatest? Choosing the positive direction as the constant direction of motion, give the signs of $v$ and $a$ in the three intervals. What are the accelerations at the points A, B, C and D?

Concept

For a speed-time graph of motion in a fixed direction:

The three intervals have equal duration, so comparisons reduce to comparing the change in speed (for acceleration) and the typical height (for speed).

Greatest average acceleration

• Interval 1: speed rises A → B (a moderate increase).
• Interval 2: speed falls from the high peak B down to the trough C — the largest change in speed over an equal time, hence the steepest chord.
• Interval 3: speed rises C → D.

The biggest magnitude of $\Delta v/\Delta t$ occurs in interval 2.

Greatest average speed

The average speed over an interval corresponds to how high the curve lies there. The curve reaches its highest values near the tall second peak D, which is in interval 3. So the average speed is greatest in interval 3.

Signs of $v$ and $a$

The particle moves in a constant (positive) direction, so its velocity is positive at all times:

Acceleration has the sign of the slope of the speed-time curve:

• Interval 1 (speed increasing A → B): $a > 0$. …

Concept: Two Independent Readings of a Speed-Time Graph — Rise vs. Height

Method: Separate "Steepest" from "Highest" as Two Logically Distinct Criteria (they need not pick out the same interval)

It is tempting to assume that whichever interval is visually the "most dramatic" wins both comparisons — largest average acceleration and largest average speed. This method makes explicit why that assumption is wrong: average acceleration depends only on the difference between the speed at the start and end of an interval ($\bar a = \Delta v/\Delta t$), while average speed depends on how high the curve sits throughout the interval (its area, roughly its typical height). These are independent properties of a curve, and in this graph they genuinely point to two different intervals.

Steps

1. State the two formulas being compared, side by side:

Two curves can have identical endpoint heights (same $\Delta v$, hence same $\bar a$) while one sits much higher throughout the interval than the other (very different $\overline{\text{speed}}$) — the two measures simply answer different questions.

2. Apply the difference criterion (acceleration) to the three intervals. Interval 1 (A→B) is a rise to the first maximum — a moderate net change. Interval 2 (B→C) is the fall from that first maximum all the way down to the local minimum — since B is a genuine peak and C a genuine trough immediately following it, this interval captures the single sharpest reversal in the whole curve, i.e. the largest $|\Delta v|$ over an equal time window. Interval 3 (C→D) is only the early part of the climb toward the second, higher maximum D — not the whole climb — so its $|\Delta v|$ over the same short window is smaller than interval 2's sharp drop, even though D itself ends up higher than B. Interval 2 has the greatest average acceleration magnitude.

3. Apply the height criterion (speed) to the three intervals, independently of Step 2. Interval 1 sits low (climbing from a small starting value). Interval 2 sits high only briefly, near B, before dropping toward the low trough C — its typical height over the interval is pulled down by the trough end. Interval 3, by contrast, is on the approach to the taller second maximum D, which — being higher than B — keeps the curve's height elevated over that window even though the endpoint hasn't yet been reached. Interval 3 has the greatest average speed. …