Question at position 3 In Example 16 (p. 49), what is the cyclist's average acceleration for the entire trip (i.e., from 0 s to 1800 s)? -2 ms2\frac{m}{s^2}-1 ms2\frac{m}{s^2}-0.0006 ms2\frac{m}{s^2}0 ms2\frac{m}{s^2}0.0006 ms2\frac{m}{s^2}1 ms2\frac{m}{s^2}2 ms2\frac{m}{s^2}Clear my selection

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Question at position 3  In Example 16 (p. 49), what is the cyclist's average acceleration for the entire trip (i.e., from 0 s to 1800 s)? -2 ms2\frac{m}{s^2}-1 ms2\frac{m}{s^2}-0.0006 ms2\frac{m}{s^2}0 ms2\frac{m}{s^2}0.0006 ms2\frac{m}{s^2}1 ms2\frac{m}{s^2}2 ms2\frac{m}{s^2}Clear my selection

BlackTom AI · Accepted Answer

Question being analyzed: 'Question at position 3 In Example 16 (p. 49), what is the cyclist's average acceleration for the entire trip (i.e., from 0 s to 1800 s)?'

Option 1: '-2 m/s^2' — This would imply a fairly strong negative change in velocity over the 1800 s interval. To obtain such a value, the net change in velocity would have to be -2 m/s every second on average, which is substantial. Unless the cyclist ends up moving in the opposite direction or slows down dramatically, this seems unlikely when considering typical cycling trips that aim for moderate speeds over 1800 s.

Option 2: '-1 m/s^2' — A still noticeable deceleration. While possible in some scenarios, a single-step average of -1 m/s^2 over 1800 seconds would indicate a total velocity change of -1800 m/s, which is not realistic for a cyclist in a standard example unless there is a dramatic sustained braking phase. This makes it questionable as the average unless the data specifically show a large negative velocity change.

Option 3: '-0.0006 m/s^2' — This is a very small magnitude, suggesting almost no net change in velocity over 1800 seconds. If the cyclist starts and ends with nearly the same speed, the average acceleration would be close to zero, making such a tiny negative value plausible only if there was a negligible total velocity change.

Option 4: '0 m/s^2' — This represents no net change in velocity over the entire trip. If the cyclist finishes at the same speed they started with, the difference v_f − v_i would be zero, yielding an average acceleration of zero regardless of how speed fluctuated in between. This scenario often appears in textbook problems where the acceleration profile is variable but ends up with no overall velocity change.

Option 5: '0.0006 m/s^2' — A very small positive average acceleration. This would imply a tiny net gain in velocity over 1800 s. For most questions designed to be clean, this is less likely unless the data show a very slight overall speeding up from start to finish.

Option 6: '1 m/s^2' — A moderate positive average acceleration would mean a substantial net velocity increase across 1800 s. This is large for an average over a long interval and would require a sustained acceleration pattern or a higher ending speed, which may not align with typical example data unless explicitly described.

Option 7: '2 m/s^2' — A fairly large positive average acceleration. Similar to the previous option, this would require a sizable net velocity gain across the trip, which again would need specific conditions in the data. Without those, this option seems unlikely.

General reasoning to compare options: To compute average acceleration over the interval, you would use a_f_avg = (v_f − v_i) / Δt, where Δt = 1800 s. If the problem data indicate the cyclist starts at some speed and ends at the same speed, v_f = v_i, leading directly to a_f_avg = 0 m/s^2. If the velocity change is very small, the result would be a tiny magnitude near zero (positive or negative). If there is a clearly large increase or decrease in velocity from start to finish, the average would reflect that magnitude divided by 1800 s, yielding the larger values (on the order of 0.5–2 m/s^2 would require substantial velocity changes). In many standard textbook examples, the most common clean result for an average over a full trip with balanced speeding and slowing phases is zero, unless the data explicitly indicate a net change in velocity.

Summary of option-by-option assessment: 
- Option 1 (-2 m/s^2): high magnitude negative; requires large net velocity change; unlikely without explicit data supporting it. 
- Option 2 (-1 m/s^2): negative, with substantial net change; possible only if data show a clear large slowdown. 
- Option 3 (-0.0006 m/s^2): tiny negative; plausible if net velocity change is extremely small. 
- Option 4 (0 m/s^2): zero net change in velocity; common when start and end speeds are equal. 
- Option 5 (0.0006 m/s^2): tiny positive; plausible only if there is a small net speeding up. 
- Option 6 (1 m/s^2) and Option 7 (2 m/s^2): sizable positive accelerations; would require notable net velocity gain across 1800 s, which is less typical unless the data specify such a trend.

Note: Without the underlying velocity data (v_i and v_f) from Example 16, you cannot compute the exact value here. The reasoning above focuses on how average acceleration relates to net velocity change over the interval and what each magnitude implies about that change.

Question at position 3 In Example 16 (p. 49), what is the cyclist's average acceleration for the entire trip (i.e., from 0 s to 1800 s)? -2 ms2\frac{m}{s^2}-1 ms2\frac{m}{s^2}-0.0006 ms2\frac{m}{s^2}0 ms2\frac{m}{s^2}0.0006 ms2\frac{m}{s^2}1 ms2\frac{m}{s^2}2 ms2\frac{m}{s^2}Clear my selection

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