1.2 Displacement, Velocity, and Acceleration — Worked Solutions
Work through Topic 1.2
These are the full worked solutions to the lesson questions. Try the questions first, then check each step and the reason for it.
Question 1
A runner returns to the start after 20 s. Which statement must be true?
- Average velocity is zero
- Average speed is zero
- Instantaneous velocity was always zero
Step 1. A return to the starting point gives Δx = 0.
Why this step? The initial and final positions match, whatever route the runner took between them.
Step 2. Average velocity is Δx/Δt = 0/20 = 0 m/s.
Why this step? The full-interval definition uses net displacement; a positive traveled distance cannot replace that numerator.
Step 3. Average speed and instantaneous velocity need not be zero.
Why this step? Running out and back includes motion even though the endpoint changes cancel, so only average velocity is guaranteed.
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Question 2
Velocity changes from −3 m/s to −7 m/s in 2 s. What is the average acceleration?
- +2 m/s²
- −2 m/s²
- −5 m/s²
Step 1. Compute the signed change: Δv = −7 − (−3) = −4 m/s.
Why this step? Use final minus initial; subtracting the initial negative velocity prevents treating −7 and −3 as unsigned speeds.
Step 2. Divide by elapsed time: a_avg = (−4 m/s)/(2 s) = −2 m/s².
Why this step? The 2 s interval measures how quickly the velocity changed; the unit becomes velocity per second.
Step 3. The motion is speeding up in the negative direction.
Why this step? The speed magnitude grows from 3 to 7 m/s, so the negative acceleration does not mean slowing down.
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Question 3
A car travels around a circular bend at fixed speed. Is it accelerating?
- No; only changing speed causes acceleration
- Yes; the velocity direction changes
- Only if its position is negative
Step 1. Compare the car’s velocity arrows at two different points on the bend. They have equal lengths but different directions.
Why this step? Fixed speed fixes arrow length, while the bend changes the direction in which the car moves.
Step 2. The velocity change is nonzero, so the car accelerates.
Why this step? Acceleration depends on the vector change; keeping its magnitude constant does not keep the vector constant.
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Question 4
A tracker gives positions 0 m, 2 m, and 8 m at times 0 s, 1 s, and 3 s. Calculate average velocity during each interval and during the full 3 s.
Step 1. From 0 to 1 s: v_avg = (2 − 0)/(1 − 0) = 2 m/s.
Why this step? These are the position change and duration of the first interval, not of the entire record.
Step 2. From 1 to 3 s: v_avg = (8 − 2)/(3 − 1) = 3 m/s.
Why this step? The second displacement is 6 m over 2 s; using 3 s here would incorrectly include the earlier interval.
Step 3. For the full trip: v_avg = (8 − 0)/(3 − 0) = 8/3 ≈ 2.67 m/s.
Why this step? A whole-trip average must use whole-trip displacement and time, so the longer interval carries more weight.
Step 4. Equivalently, (2·1 + 3·2)/(1 + 2) = 8/3 m/s.
Why this step? Time weighting adds the two interval displacements; the unweighted mean 2.5 m/s assumes equal durations that were not given.
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Question 5
A cart moving right has positive acceleration. Describe what happens to its speed. Then repeat for a cart moving left with the same positive acceleration.
Step 1. With right positive, positive acceleration adds a positive change to the velocity.
Why this step? A single fixed axis lets us compare acceleration with each cart’s initial motion rather than with its position.
Step 2. For a rightward cart, positive velocity grows more positive, so speed increases.
Why this step? The velocity and acceleration point together, increasing the velocity magnitude.
Step 3. For a leftward cart, negative velocity initially moves toward zero, so speed decreases. It can later reverse and speed up rightward.
Why this step? Acceleration initially opposes the motion; after velocity passes through zero, it instead points with the new rightward motion.
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Question 6
Two position–time curves share the same endpoints. One is a straight line; the other bends upward. Explain how their full-interval average velocities can match while their instantaneous velocities differ.
Step 1. Draw the secant joining the common start and end points. Its slope is the same for both curves.
Why this step? The two records have identical full-interval displacement and duration, which are the two inputs to average velocity.
Step 2. At a chosen time, compare the tangent slopes. The straight line has one fixed slope; the upward-bending curve has changing slope.
Why this step? A local tangent uses nearby motion rather than just the endpoints, so it reveals differences that a full-interval average discards.
Step 3. Both trips can therefore have the same average velocity and different instantaneous velocities.
Why this step? Agreement of an endpoint ratio does not require agreement at every intermediate time.