Reference Frames and Relative Motion Practice

Sample questions

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1. Sample 1difficulty 2/5

   Car A drives east at $25~\text{m/s}$; car B drives west at $15~\text{m/s}$ on the same road. What is the speed of car A as observed from car B?

   • A

     $375~\text{m/s}$

   • B

     $10~\text{m/s}$

   • C

     $40~\text{m/s}$

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   • D

     $20~\text{m/s}$

   Why

   Relative velocity: $v_{A/B} = v_A - v_B = 25 - (-15) = 40~\text{m/s}$ (toward car B).

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2. Sample 2difficulty 2/5

   Two cars travel in the same direction. Car A is at $25~\text{m/s}$, car B is $40~\text{m}$ ahead at $20~\text{m/s}$. How long until A catches B?

   • A

     $2~\text{s}$

   • B

     $5~\text{s}$

   • C

     $8~\text{s}$

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   • D

     $10~\text{s}$

   Why

   Closing speed $5~\text{m/s}$. Time = $40/5 = 8~\text{s}$.

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3. Sample 3difficulty 3/5

   A boat that can move at $5~\text{m/s}$ relative to water aims directly across a river that flows at $3~\text{m/s}$. The river is $40~\text{m}$ wide. How far downstream does the boat drift by the time it reaches the other bank?

   • A

     $8~\text{m}$

   • B

     $40~\text{m}$

   • C

     $24~\text{m}$

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   • D

     $12~\text{m}$

   Why

   Crossing time: $40/5 = 8~\text{s}$. Drift: $3 \times 8 = 24~\text{m}$.

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4. Sample 4difficulty 4/5

   A boat travels at $v_b = 4.0\text{ m/s}$ relative to water, aimed straight across a $w = 120\text{ m}$ river. The current flows at $v_c = 3.0\text{ m/s}$. How far downstream does the boat land?

   • A

     90 m

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   • B

     60 m

   • C

     120 m

   • D

     75 m

   Why

   Crossing time: $t = w/v_b = 120/4 = 30\text{ s}$. Downstream drift: $\Delta x = v_c t = 3 \times 30 = 90\text{ m}$.

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5. Sample 5difficulty 4/5

   A river of width $W$ flows with speed $v_r$. A boat moves at speed $v_b$ relative to water. To minimize the time to reach the far bank, the boat must aim straight across (perpendicular to the banks). What is the minimum crossing time and how far downstream does the boat land?

   • A

     time $W/(v_b + v_r)$, drift $W v_r/(v_b+v_r)$

   • B

     time $W/v_b$, drift $W v_r / v_b$

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   • C

     time $W/(v_b - v_r)$, drift $W$

   • D

     time $W/\sqrt{v_b^2 - v_r^2}$, drift $0$

   Why

   Aiming perpendicular gives the largest cross-stream component $v_b$, so the time is $W/v_b$. During that time the current carries the boat a distance $v_r \cdot (W/v_b) = W v_r/v_b$ downstream.

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