1.4 Reference Frames and Relative Motion — Worked Solutions
Work through Topic 1.4
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 bus moves +10 m/s relative to the road. A rider walks −1 m/s relative to the bus. What is the road velocity?
- +9 m/s
- −11 m/s
- +11 m/s
Step 1. Use v_rider,road = v_rider,bus + v_bus,road with the shared positive road direction.
Why this step? The bus displacement and the rider’s displacement within the bus add during the same elapsed time.
Step 2. Substitute −1 + 10 = +9 m/s.
Why this step? The rider walks backward relative to the bus, so that component reduces rather than increases the forward road velocity.
Step 3. The rider still moves forward relative to the road.
Why this step? The bus’s 10 m/s motion is larger than the rider’s 1 m/s backward walk; −11 and +11 use incorrect signs.
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Question 2
Two inertial observers move at constant velocity relative to each other. Which measurement must agree?
- Object velocity
- Object position
- Object acceleration
Step 1. Write the second observer’s velocity as v′ = v − V, where V is the constant relative frame velocity.
Why this step? Two inertial observers in this question differ by a constant-velocity motion, not an accelerating coordinate system.
Step 2. The change is v_f′ − v_i′ = (v_f − V) − (v_i − V) = v_f − v_i.
Why this step? The same constant offset appears at both endpoints and cancels rather than altering the velocity change.
Step 3. Divide by the same time interval: a′ = a.
Why this step? In the ordinary nonrelativistic model the observers share elapsed time, so equal velocity changes give equal acceleration; positions and velocities can differ.
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Question 3
A car has road velocity +6 m/s and a truck +10 m/s. What is the car velocity relative to the truck?
- +4 m/s
- −4 m/s
- +16 m/s
Step 1. The truck is the observer, so v_car,truck = v_car,road − v_truck,road.
Why this step? A truck-fixed origin advances with the truck; subtracting that motion leaves the car’s motion within that frame.
Step 2. Calculate 6 − 10 = −4 m/s.
Why this step? Both road velocities are positive, but the car is slower, so its relative coordinate decreases.
Step 3. Interpret −4 m/s as backward relative to the truck, not backward on the road.
Why this step? The car’s road velocity remains +6 m/s; changing observers can reverse a component without reversing the actual road motion.
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Question 4
A walkway moves +1.5 m/s. A traveler has +2.0 m/s relative to it. Find the traveler velocity relative to the building, then find the traveler velocity if they turn and walk at the same speed relative to the walkway.
Step 1. Keep the walkway’s +1.5 m/s velocity and add the traveler’s signed walkway-relative velocity.
Why this step? The walkway carries its own origin through the building, while the traveler also moves within that origin’s frame.
Step 2. Walking forward gives v_building = +2.0 + 1.5 = +3.5 m/s.
Why this step? The two motions share a direction, so they reinforce rather than cancel.
Step 3. Walking backward gives v_building = −2.0 + 1.5 = −0.5 m/s.
Why this step? Turning reverses only the traveler’s walkway-relative component; the walkway still moves forward at 1.5 m/s.
Step 4. The final result means 0.5 m/s backward through the building.
Why this step? The backward walking magnitude exceeds the walkway’s forward speed, making the building-relative resultant negative.
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Question 5
A ground velocity graph rises from 4 to 10 m/s in 3 s. Sketch the velocity graph measured from a frame moving +2 m/s. Compare slopes and accelerations.
Step 1. Use v′(t) = v_ground(t) − 2 m/s at every plotted time.
Why this step? The moving observer has a constant +2 m/s velocity, so its frame transformation is one fixed vertical graph shift.
Step 2. The endpoints become 4 − 2 = 2 m/s and 10 − 2 = 8 m/s over the same 3 s.
Why this step? Each endpoint must be transformed with the same observer velocity; changing only one would invent a different acceleration.
Step 3. Both slopes equal (10 − 4)/3 = (8 − 2)/3 = 2 m/s².
Why this step? Subtracting equal offsets leaves the numerator unchanged, and inertial frames share the interval in this model.
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Question 6
Cars A and B have x_A = 2 + 4t and x_B = 14 + 2t, with x in meters and t in seconds. Use relative motion to find when they meet.
Step 1. At t = 0, B is 14 − 2 = 12 m ahead. A gains on B at 4 − 2 = 2 m/s.
Why this step? The constant coefficients of t are velocities; subtracting them gives the rate at which their separation closes.
Step 2. Set the remaining separation to zero: 12 − 2t = 0, giving t = 6 s.
Why this step? A meeting occurs when the initially positive separation has been fully removed, assuming the stated constant velocities continue.
Step 3. Substitute into both positions: x_A = 2 + 4·6 = 26 m and x_B = 14 + 2·6 = 26 m.
Why this step? Matching absolute positions verifies that the relative-motion result identifies the same event for both cars.