Relative Velocity in Two Dimensions Quiz
Test your understanding of relative-velocity vectors, river crossings, and aircraft motion in wind. Work through each question by identifying the reference frame and resolving velocity into components when needed.
Q1. For objects A and B, what does the vector equation ##\vec{v}_{AB}=\vec{v}_A-\vec{v}_B## represent?
By definition, the velocity of A as measured by B is ##\vec{v}_{AB}=\vec{v}_A-\vec{v}_B##. The result depends on both velocities being expressed in the same reference frame.
Q2. A train moves east at ##20\ \text{m/s}##. A passenger walks east at ##2\ \text{m/s}## relative to the train. What is the passenger's velocity relative to the ground?
The passenger's ground velocity is the train's velocity plus the passenger's velocity relative to the train: ##20+2=22\ \text{m/s}## east.
Q3. A boat moves east at ##8\ \text{m/s}## relative to the water, while the river flows north at ##3\ \text{m/s}##. What is the boat's velocity relative to the shore?
The perpendicular components combine to give a speed of ##\sqrt{8^2+3^2}=\sqrt{73}\ \text{m/s}##. The direction is northeast because both components are positive.
Q4. A river is ##180\ \text{m}## wide. A boat's velocity across the water is ##6\ \text{m/s}##, perpendicular to the bank, while the current is ##2\ \text{m/s}## downstream. How long does the crossing take?
The current does not change the across-river component. The crossing time is ##180/6=30\ \text{s}##.
Q5. Using the boat and river speeds in the previous type of situation, a boat travels across a ##180\ \text{m}## river at ##6\ \text{m/s}## while the current flows downstream at ##2\ \text{m/s}##. How far downstream does it drift before reaching the opposite bank?
The crossing takes ##180/6=30\ \text{s}##. During that time, the current carries the boat ##2\times30=60\ \text{m}## downstream.
Q6. A boat moves at ##5\ \text{m/s}## relative to the water and must travel directly across a river flowing at ##3\ \text{m/s}##. In what direction should the boat point relative to straight across?
To cancel the ##3\ \text{m/s}## downstream current, the boat needs an upstream component of ##3\ \text{m/s}##. Thus ##\sin\theta=3/5##, giving ##\theta\approx36.9^\circ## upstream from straight across.
Q7. A boat's speed relative to the water is ##4\ \text{m/s}## directly across a river, and the current is ##3\ \text{m/s}## downstream. What is its speed relative to the shore?
The two perpendicular components give a shore-relative speed of ##\sqrt{4^2+3^2}=5\ \text{m/s}##.
Q8. An aircraft flies north at ##150\ \text{km/h}## relative to the air. A wind blows east at ##50\ \text{km/h}##. What is the aircraft's ground speed?
The northward and eastward components are perpendicular, so the ground speed is ##\sqrt{150^2+50^2}=\sqrt{25000}\approx158\ \text{km/h}##.
Q9. An aircraft's velocity relative to the air is ##200\ \text{km/h}## due north. A wind blows east at ##40\ \text{km/h}##. What is the direction of the aircraft's ground velocity relative to north?
The eastward component is ##40\ \text{km/h}## and the northward component is ##200\ \text{km/h}##. Therefore ##\tan\theta=40/200##, so ##\theta\approx11.3^\circ## east of north.
Q10. An aircraft must travel due north relative to the ground while a wind blows east at ##60\ \text{km/h}##. Which way must its velocity relative to the air point?
The aircraft must have a westward component to cancel the wind's eastward component. Its air-relative velocity therefore points west of north.
Q11. A boat moves at ##4\ \text{m/s}## across a ##120\ \text{m}## river, while the current carries it downstream at ##3\ \text{m/s}##. What is the magnitude of its velocity relative to the shore?
The shore-relative velocity combines perpendicular across-river and downstream components, giving ##\sqrt{4^2+3^2}=5\ \text{m/s}##.
Q12. A boat travels ##120\ \text{m}## across a river with an across-water speed of ##4\ \text{m/s}##. The current flows downstream at ##3\ \text{m/s}##. What is the angle of its drift from straight across?
The across-river component is ##4\ \text{m/s}## and the downstream component is ##3\ \text{m/s}##. Thus ##\tan\theta=3/4##, giving ##\theta\approx36.9^\circ## downstream from straight across.
Q13. A person walks east at ##1.5\ \text{m/s}## relative to a moving walkway that travels east at ##2.5\ \text{m/s}## relative to the ground. What is the person's ground velocity?
The velocities point in the same direction, so they add: ##1.5+2.5=4.0\ \text{m/s}## east.
Q14. A car moves east at ##18\ \text{m/s}##, and a cyclist moves east at ##12\ \text{m/s}##. What is the car's velocity relative to the cyclist?
Subtract the cyclist's velocity from the car's: ##18-12=6\ \text{m/s}## east. The positive eastward result means the car moves east relative to the cyclist.
Q15. A boat's velocity relative to the water is ##\langle 0,5\rangle\ \text{m/s}##, and the river current is ##\langle 2,0\rangle\ \text{m/s}##, with the first component pointing downstream. What is the boat's velocity relative to the shore?
The shore-relative velocity is the boat's velocity relative to the water plus the water's velocity relative to the shore: ##\langle 0,5\rangle+\langle 2,0\rangle=\langle 2,5\rangle\ \text{m/s}##.
Q16. A river is ##240\ \text{m}## wide, and a boat's across-river speed is ##8\ \text{m/s}##. The current flows downstream at ##3\ \text{m/s}##. How long does the boat take to cross?
The across-river component determines crossing time: ##t=240/8=30\ \text{s}##. The downstream current changes the landing position, not the crossing time.
Q17. A boat crosses a ##240\ \text{m}## river in ##30\ \text{s}## while the current flows downstream at ##3\ \text{m/s}##. The boat heads directly across the river, perpendicular to the banks. How far downstream does it land?
Because the boat heads directly across, its velocity relative to the water has no downstream component. The current carries it downstream for the full crossing time, so the drift is ##3\times30=90\ \text{m}##.
Q18. In a graphical solution to a two-dimensional relative-velocity problem, what does placing velocity vectors head-to-tail allow you to determine?
A head-to-tail vector diagram shows the vector sum or difference needed for the chosen relative-velocity relationship. Its resultant gives the relevant velocity's magnitude and direction.
Q19. A frame ##S'## moves at constant velocity ##\vec{u}## relative to frame ##S##. According to the Galilean velocity transformation, how are an object's velocities related?
For a frame moving with velocity ##\vec{u}## relative to ##S##, the position relation is ##\vec{r}'=\vec{r}-\vec{u}t##. Differentiating with respect to time gives ##\vec{v}'=\vec{v}-\vec{u}##.
Q20. An object and an observer have the same velocity vector in the same reference frame. What is the object's velocity relative to the observer?
Using ##\vec{v}_{AB}=\vec{v}_A-\vec{v}_B##, equal velocity vectors cancel, so the relative velocity is zero.