Acceleration — Question 6

Question 6

Combine acceleration records through their velocity changes

A logger divides a 10 s run on an east-positive line into three consecutive intervals: The durations and average accelerations are A: $2\ \mathrm{s}$ and $+3\ \mathrm{m/s^2}$; B: $5\ \mathrm{s}$ and $-2\ \mathrm{m/s^2}$; C: $3\ \mathrm{s}$ and $+4\ \mathrm{m/s^2}$. The initial velocity is $-2\ \mathrm{m/s}$. Each average is exact and includes the full interval. Velocity is continuous throughout, but no linearity or constant-acceleration assumption is supplied.

1. Recover each interval’s velocity change and the velocities at 2, 7, and 10 s. Find the full-run average acceleration.

2. Derive a formula that combines interval-average accelerations using their durations. Compare the correct result here with the unweighted arithmetic mean of the three listed values.

3. Do the records determine acceleration at 4 s, or whether speed increases throughout B? Explain why net changes need not fix the intervening slopes or signs.

4. Do the boundary velocities nevertheless force the cart to reverse direction somewhere within each of A, B and C? Use continuity and the endpoint signs. Do they determine the number or exact times of reversals?

Question 6 - Solution

Combine acceleration records through their velocity changes

(a) Multiply each average by the duration it summarizes.

$$\bar a=\frac{3(2)-2(5)+4(3)}{2+5+3}=\frac8{10}=0.8\ \mathrm{m/s^2}.$$

The velocity changes are $\Delta v_A=+6$, $\Delta v_B=-10$ and $\Delta v_C=+12\ \mathrm{m/s}$. Starting at $-2\ \mathrm{m/s}$ gives $$\boxed{v(2)=+4,\quad v(7)=-6,\quad v(10)=+6\ \mathrm{m/s}}.$$ Thus the net change is $+8\ \mathrm{m/s}$ in 10 s, yielding $\boxed{\bar a=+0.8\ \mathrm{m/s^2}}$.

(b) Add velocity changes before dividing by total time.

For consecutive intervals, $$\boxed{\bar a_{\mathrm{all}}=\frac{\sum_i\bar a_iT_i}{\sum_iT_i}}.$$ The unweighted mean $(3-2+4)/3=5/3\ \mathrm{m/s^2}$ is incorrect for this record because the durations are unequal. Equal durations would guarantee that the arithmetic mean works for arbitrary interval-average values.

(c) Interval averages do not prescribe instantaneous derivatives.

Many continuous velocity curves can connect the same boundary values, with different slopes at 4 s; some could even have a corner there, making acceleration undefined. No unique instantaneous value follows.

In B, velocity starts positive and ends negative, so a continuous history must pass through zero. Its speed cannot increase throughout: it must drop from 4 m/s to zero at some stage before reaching the final 6 m/s. A negative average acceleration alone does not describe how the speed changes at each instant.

(d) Opposite endpoint signs force a change in motion direction.

Each interval begins and ends with opposite nonzero velocity signs. Continuity therefore requires at least one transition between westward and eastward motion within each interval, possibly with a pause at zero between them. These records do not establish exactly one reversal, a unique reversal instant, or the absence of additional excursions. The net velocity changes remain exact even when those details are unknown.