1.3 Representing Motion — Worked Solutions
Work through Topic 1.3
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 velocity–time graph lies below zero and slopes upward. What does it describe?
- Moving negative and initially slowing down
- Moving positive and speeding up
- No motion because velocity is negative
Step 1. The graph below zero means v is negative; its upward slope means a is positive.
Why this step? Graph height gives the signed velocity, while the rate that height changes gives acceleration; these are separate readings.
Step 2. While v stays negative, positive acceleration moves it toward zero and reduces |v|.
Why this step? The acceleration is opposite the motion in this interval, so the object initially slows rather than speeding up.
Step 3. Choose moving negative and initially slowing down.
Why this step? A negative velocity is still motion, and an upward graph slope does not make that velocity positive.
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Question 2
What does the area under an acceleration–time graph give?
- Position
- Velocity change
- Acceleration change
Step 1. For a short constant-acceleration interval, area = aΔt and Δv = aΔt.
Why this step? The acceleration definition links a rectangular graph area directly to the change in velocity over that same interval.
Step 2. Add signed areas for successive intervals to obtain the total Δv.
Why this step? Below-axis intervals subtract velocity change; summing the pieces keeps their direction information.
Step 3. Final velocity is v_f = v_i + area, so area alone is velocity change.
Why this step? A nonzero initial velocity must still be added; Position information cannot be obtained from this area directly.
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Question 3
At the highest point of a vertical toss without air resistance, which statement is correct?
- Velocity and acceleration are both zero
- Velocity is zero; acceleration is downward
- Acceleration changes from upward to downward
Step 1. At the highest point of a vertical toss, the upward velocity has reached zero before reversing downward.
Why this step? A turning point is where the one-dimensional position slope changes sign, so the instantaneous vertical velocity vanishes.
Step 2. Gravity is still downward, giving a = −g if up is positive.
Why this step? With air resistance excluded, Earth is still the only force source; being momentarily at rest does not remove its force.
Step 3. Choose velocity zero and acceleration downward.
Why this step? Confusing a zero value of velocity with a zero rate of velocity change would incorrectly predict that the ball stays at the peak.
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Question 4
A cart’s velocity is +4 m/s for 2 s, then falls linearly to zero during the next 2 s. Sketch v versus t, and find displacement and the acceleration in each interval.
Step 1. Draw a 4 m/s horizontal segment from 0 to 2 s, then a straight line reaching 0 m/s at 4 s.
Why this step? The first interval has fixed velocity; the stated linear fall in the next interval sets a constant negative slope.
Step 2. The first displacement is the rectangle area: (4 m/s)(2 s) = 8 m.
Why this step? Velocity is positive throughout this interval, so its area contributes positive displacement.
Step 3. The second displacement is the triangle area: ½(2 s)(4 m/s) = 4 m. Total displacement is 12 m.
Why this step? The velocity falls but remains nonnegative, so this area adds to rather than subtracts from the first displacement.
Step 4. Acceleration is 0 for 0–2 s and (0 − 4)/(4 − 2) = −2 m/s² for 2–4 s.
Why this step? Acceleration comes from each segment’s slope; the first is horizontal and the second loses 4 m/s in 2 s.
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Question 5
Derive Δx = ½(v_i + v_f)Δt from a straight velocity–time graph. State the condition needed for that graph.
Step 1. For a straight v–t line, the signed area over Δt is a rectangle v_iΔt plus a triangle ½(v_f − v_i)Δt.
Why this step? Straight-line interpolation makes the excess velocity above v_i a triangle; a curved graph would not have that triangular area.
Step 2. Combine terms: Δx = [v_i + ½(v_f − v_i)]Δt = ½(v_i + v_f)Δt.
Why this step? Both geometric pieces span the same interval, so their algebra reduces to the endpoint average velocity times duration.
Step 3. The condition is constant acceleration during this interval.
Why this step? A fixed slope is exactly a fixed acceleration; endpoint averaging is not generally valid for a curved velocity graph.
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Question 6
A falling ball experiences increasing air resistance. Sketch the qualitative velocity and acceleration graphs with downward positive. Explain why a constant downward acceleration graph would fail.
Step 1. With downward positive, the net force is mg − F_drag and a = g − F_drag/m.
Why this step? For a downward-falling ball, drag points upward, so increasing drag subtracts from the gravitational force.
Step 2. Sketch v positive and increasing with a progressively smaller slope.
Why this step? As long as mg exceeds drag, acceleration remains downward, but its declining magnitude makes velocity grow more slowly.
Step 3. Sketch a starting near +g and decreasing toward zero as drag approaches mg.
Why this step? At terminal motion the opposing forces balance, giving zero acceleration while downward velocity stays nonzero.
Step 4. A constant +g line would omit the increasing drag.
Why this step? The net force changes in the stated situation; only a qualitative variable-acceleration description is needed for this AP task.