1.1 Scalars and Vectors in One Dimension — Worked Solutions
Work through Topic 1.1
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 person walks 7 m east, then 3 m west. With east positive, which pair is correct?
- Distance 10 m; displacement +4 m
- Distance 4 m; displacement +10 m
- Distance 10 m; displacement −4 m
Step 1. Set east positive: the first displacement is +7 m and the return displacement is −3 m.
Why this step? The question fixes the axis; westward travel must have the opposite component sign.
Step 2. Add signed displacements: Δx = 7 − 3 = +4 m. Add traveled lengths: distance = 7 + 3 = 10 m.
Why this step? The return leg partly cancels the position change, but it still adds 3 m to the path traveled.
Step 3. Choose distance 10 m and displacement +4 m.
Why this step? The positive resultant means the person finishes east of the start; neither 10 m of displacement nor a westward resultant matches these endpoints.
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Question 2
A velocity component is −5 m/s. What is the speed?
- −5 m/s
- 5 m/s
- It cannot be found without acceleration
Step 1. The component vₓ = −5 m/s describes motion in the negative x direction.
Why this step? The minus sign carries direction information; it is not a negative amount of speed.
Step 2. Take the magnitude: speed = |−5| = 5 m/s.
Why this step? Speed answers how fast the object moves, so the direction sign is removed rather than changing the magnitude.
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Question 3
You reverse the positive direction of an axis. What happens to a nonzero displacement component?
- Its sign reverses; its magnitude stays the same
- Its magnitude doubles
- Both its sign and physical direction reverse
Step 1. Let an unchanged displacement have component Δx in the original axis. In the reversed axis its component is Δx′ = −Δx.
Why this step? Reversing the positive direction relabels the same physical arrow, so its projection changes sign.
Step 2. Its magnitude remains |Δx′| = |Δx|.
Why this step? The coordinate choice does not move either endpoint or stretch the physical displacement arrow.
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Question 4
A robot starts at x = −2 m, moves to x = +5 m, then ends at x = +1 m. Find each displacement, the total displacement, and the distance.
Step 1. Find the first leg: Δx₁ = +5 − (−2) = +7 m.
Why this step? A displacement uses final minus initial position; subtracting the negative starting position adds 2 m.
Step 2. Find the return leg: Δx₂ = +1 − (+5) = −4 m.
Why this step? The robot ends left of its turnaround point, so this leg has a negative signed change.
Step 3. Total displacement is +7 − 4 = +3 m; distance is |7| + |−4| = 11 m.
Why this step? Vector changes combine with signs, whereas every traveled leg contributes a positive length.
Step 4. Check the endpoint result: +1 − (−2) = +3 m.
Why this step? The net displacement must depend only on the start and end; agreement catches a sign error in either leg.
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Question 5
Draw two opposite displacement arrows, 4 m and 4 m, on a scale of 1 cm per meter. What can you conclude about the final position? What cannot you conclude about the travel time?
Step 1. Draw a 4 cm arrow one way, then a 4 cm arrow in the opposite direction, head to tail.
Why this step? The stated scale turns each 4 m displacement into 4 cm; equal opposite arrows bring the final head back to the first tail.
Step 2. The resultant is zero, so final position equals initial position.
Why this step? The resultant joins the first tail to the final head, which coincide; total traveled distance is still 8 m.
Step 3. Travel time cannot be determined from these arrows.
Why this step? Their lengths represent meters, not seconds; many different speeds could produce the same out-and-back displacement.
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Question 6
A student says “negative acceleration always means slowing down.” Explain why a direction sign by itself cannot establish that claim.
Step 1. Negative acceleration means that the signed velocity is becoming more negative.
Why this step? Acceleration describes a velocity change along the chosen axis, so its sign alone does not describe the velocity already present.
Step 2. For v = +5 m/s and a = −1 m/s², one second later v = +4 m/s and speed falls from 5 to 4 m/s.
Why this step? Here acceleration opposes the initial velocity, reducing its magnitude before any reversal.
Step 3. For v = −5 m/s with the same acceleration, one second later v = −6 m/s and speed rises from 5 to 6 m/s.
Why this step? Here acceleration points with the initial motion, so the magnitude grows even though acceleration remains negative.