301. Car A is initially \(200\,\text{m}\) west of car B on a straight road. A moves east at \(15\,\text{m s}^{-1}\), while B moves west at \(10\,\text{m s}^{-1}\). If both velocities remain constant, after how much time do they meet?
ⓐ. \(5\,\text{s}\)
ⓑ. \(8\,\text{s}\)
ⓒ. \(13.3\,\text{s}\)
ⓓ. \(20\,\text{s}\)
Correct Answer: \(8\,\text{s}\)
Explanation: Take east as positive. Then
\[
v_A=+15\,\text{m s}^{-1}
\]
and
\[
v_B=-10\,\text{m s}^{-1}.
\]
The velocity of A relative to B is
\[
v_{A/B}=v_A-v_B.
\]
Therefore
\[
v_{A/B}=15-(-10)=25\,\text{m s}^{-1}.
\]
This is the rate at which the \(200\,\text{m}\) separation closes. Hence
\[
t=\frac{200}{25}=8\,\text{s}.
\]
The addition of the two speed magnitudes is valid here because the vehicles move directly toward one another along the same line. Relative-velocity subtraction automatically produces that addition when the signed velocities are used correctly. The meeting condition concerns the reduction of relative position to zero.
302. Two particles have ground-frame velocities
\[
\vec v_A=6\hat i+8\hat j\ \text{m s}^{-1},
\qquad
\vec v_B=-2\hat i+4\hat j\ \text{m s}^{-1}.
\]
Which statement correctly describes the velocity of A relative to B?
ⓐ. \(4\hat i+12\hat j\), magnitude \(4\sqrt{10}\,\text{m s}^{-1}\)
ⓑ. \(8\hat i+12\hat j\), magnitude \(4\sqrt{13}\,\text{m s}^{-1}\)
ⓒ. \(-8\hat i-4\hat j\), magnitude \(4\sqrt5\,\text{m s}^{-1}\), north of east
ⓓ. \(8\hat i+4\hat j\), magnitude \(4\sqrt5\,\text{m s}^{-1}\), with \(\tan\theta=\dfrac12\) north of east
Correct Answer: \(8\hat i+4\hat j\), magnitude \(4\sqrt5\,\text{m s}^{-1}\), with \(\tan\theta=\dfrac12\) north of east
Explanation: Relative velocity is the vector difference
\[
\vec v_{A/B}=\vec v_A-\vec v_B.
\]
Therefore
\[
\vec v_{A/B}
=(6-(-2))\hat i+(8-4)\hat j.
\]
Hence
\[
\vec v_{A/B}=8\hat i+4\hat j.
\]
Its magnitude is
\[
|\vec v_{A/B}|=\sqrt{8^2+4^2}
=\sqrt{80}
=4\sqrt5\,\text{m s}^{-1}.
\]
Both components are positive, so the relative motion is north of east. If \(\theta\) is measured from east,
\[
\tan\theta=\frac{4}{8}=\frac12.
\]
The reverse relative velocity satisfies
\[
\vec v_{B/A}=-(8\hat i+4\hat j),
\]
so the two observers assign opposite relative-velocity vectors to one another.
303. Three particles have velocities
\[
\vec v_P=3\hat i+4\hat j,\qquad
\vec v_Q=3\hat i+4\hat j,\qquad
\vec v_R=-4\hat i+3\hat j
\]
in \(\text{m s}^{-1}\). Which statement is correct?
ⓐ. P and Q move relative to each other with speed \(5\,\text{m s}^{-1}\).
ⓑ. P is at rest relative to Q, and both have relative speed \(5\sqrt2\,\text{m s}^{-1}\) with respect to R.
ⓒ. All three are mutually at rest because each has speed \(5\,\text{m s}^{-1}\).
ⓓ. R is at rest relative to P because their ground-frame speeds are equal.
Correct Answer: P is at rest relative to Q, and both have relative speed \(5\sqrt2\,\text{m s}^{-1}\) with respect to R.
Explanation: Since
\[
\vec v_P=\vec v_Q,
\]
their relative velocity is
\[
\vec v_{P/Q}=\vec v_P-\vec v_Q=\vec0.
\]
Thus P and Q remain mutually at rest even though both may be moving relative to the ground. Now compare P with R:
\[
\vec v_{P/R}
=(3-(-4))\hat i+(4-3)\hat j.
\]
Hence
\[
\vec v_{P/R}=7\hat i+\hat j.
\]
Its magnitude is
\[
\sqrt{7^2+1^2}
=\sqrt{50}
=5\sqrt2\,\text{m s}^{-1}.
\]
Since Q has the same ground-frame velocity as P, Q has exactly the same relative velocity with respect to R. Equal ground-frame speeds alone are not enough for mutual rest; the velocity vectors themselves must be equal.
304. Rain falls vertically downward with velocity \(10\,\text{m s}^{-1}\) relative to the ground. A person runs east with speed \(6\,\text{m s}^{-1}\). Which statement describes the velocity of rain relative to the person?
ⓐ. Apparent speed \(=4\,\text{m s}^{-1}\), vertically downward with \(\tan\theta=0\), where \(\theta\) is measured from the downward vertical
ⓑ. Apparent speed \(=2\sqrt{34}\,\text{m s}^{-1}\), directed downward and east with \(\tan\theta=\dfrac53\), where \(\theta\) is measured from the downward vertical
ⓒ. Apparent speed \(=16\,\text{m s}^{-1}\), directed downward and east with \(\tan\theta=\dfrac35\), where \(\theta\) is measured from the downward vertical
ⓓ. Apparent speed \(=2\sqrt{34}\,\text{m s}^{-1}\), directed downward and west with \(\tan\theta=\dfrac35\), where \(\theta\) is measured from the downward vertical
Correct Answer: Apparent speed \(=2\sqrt{34}\,\text{m s}^{-1}\), directed downward and west with \(\tan\theta=\dfrac35\), where \(\theta\) is measured from the downward vertical
Explanation: Choose east as \(+\hat i\) and upward as \(+\hat j\). The rain velocity relative to the ground is
\[
\vec v_R=-10\hat j.
\]
The person's velocity is
\[
\vec v_P=6\hat i.
\]
Therefore the rain velocity relative to the person is
\[
\vec v_{R/P}=\vec v_R-\vec v_P.
\]
Hence
\[
\vec v_{R/P}=-6\hat i-10\hat j.
\]
The negative horizontal component means that the rain appears to move toward the west relative to the runner. Its apparent speed is
\[
|\vec v_{R/P}|=\sqrt{6^2+10^2}
=\sqrt{136}
=2\sqrt{34}\,\text{m s}^{-1}.
\]
If \(\theta\) is measured from the downward vertical,
\[
\tan\theta=\frac{6}{10}=\frac35.
\]
Thus motion of the observer changes both the apparent speed and apparent direction of the rain.
305. At \(t=0\), the position of particle B relative to particle A is
\[
\vec r_{B/A}(0)=12\hat i+5\hat j\ \text{m},
\]
and their relative velocity remains constant at
\[
\vec v_{B/A}=-4\hat i-\hat j\ \text{m s}^{-1}.
\]
Which statement correctly describes their closest approach for \(t\ge0\)?
ⓐ. Minimum separation \(=0\) at \(t=3\,\text{s}\), with relative speed \(\sqrt{17}\,\text{m s}^{-1}\), so the particles collide.
ⓑ. Minimum separation \(=\dfrac{8}{\sqrt{17}}\,\text{m}\) at \(t=\dfrac{53}{17}\,\text{s}\), with relative speed \(\sqrt{17}\,\text{m s}^{-1}\), so they do not collide.
ⓒ. Minimum separation \(=0\) at \(t=\dfrac{13}{\sqrt{17}}\,\text{s}\), with relative speed \(\sqrt{17}\,\text{m s}^{-1}\), so they collide.
ⓓ. Minimum separation \(=5\,\text{m}\) at \(t=3\,\text{s}\), with relative speed zero at that instant, so they do not collide.
Correct Answer: Minimum separation \(=\dfrac{8}{\sqrt{17}}\,\text{m}\) at \(t=\dfrac{53}{17}\,\text{s}\), with relative speed \(\sqrt{17}\,\text{m s}^{-1}\), so they do not collide.
Explanation: The relative position at time \(t\) is
\[
\vec r(t)=\vec r_0+\vec v_{\rm rel}t.
\]
For constant relative velocity, closest approach occurs when the separation vector is perpendicular to the relative velocity:
\[
\vec r(t)\cdot\vec v_{\rm rel}=0.
\]
Here
\[
\vec r_0\cdot\vec v_{\rm rel}
=(12)(-4)+(5)(-1)=-53,
\]
and
\[
v_{\rm rel}^2=(-4)^2+(-1)^2=17.
\]
Thus
\[
t_{\min}=-\frac{\vec r_0\cdot\vec v_{\rm rel}}{v_{\rm rel}^2}
=\frac{53}{17}\,\text{s}.
\]
At that instant,
\[
\vec r_{\min}
=(12\hat i+5\hat j)+\frac{53}{17}(-4\hat i-\hat j)
=\frac{-8\hat i+32\hat j}{17}\,\text{m}.
\]
Hence
\[
r_{\min}=\frac{\sqrt{(-8)^2+32^2}}{17}
=\frac{8}{\sqrt{17}}\,\text{m}.
\]
The relative speed is constant:
\[
|\vec v_{\rm rel}|=\sqrt{17}\,\text{m s}^{-1}.
\]
Because the minimum separation is non-zero, the particles never occupy the same point and therefore do not collide.
306. A platform moves east with constant velocity
\[
5\hat i\ \text{m s}^{-1}
\]
relative to the ground. A passenger walks north on the platform with velocity
\[
3\hat j\ \text{m s}^{-1}
\]
relative to the platform. Which statement is correct?
ⓐ. Ground-frame velocity \(=5\hat i-3\hat j\), ground-frame speed \(=\sqrt{34}\,\text{m s}^{-1}\), and platform velocity relative to the passenger \(=\vec0\).
ⓑ. Ground-frame velocity \(=5\hat i+3\hat j\), ground-frame speed \(=\sqrt{34}\,\text{m s}^{-1}\), and platform velocity relative to the passenger \(=-3\hat j\,\text{m s}^{-1}\).
ⓒ. Ground-frame velocity \(=5\hat i+3\hat j\), ground-frame speed \(=8\,\text{m s}^{-1}\), and platform velocity relative to the passenger \(=-3\hat j\,\text{m s}^{-1}\).
ⓓ. Ground-frame velocity \(=3\hat j\), ground-frame speed \(=3\,\text{m s}^{-1}\), and platform velocity relative to the passenger \(=5\hat i\,\text{m s}^{-1}\).
Correct Answer: Ground-frame velocity \(=5\hat i+3\hat j\), ground-frame speed \(=\sqrt{34}\,\text{m s}^{-1}\), and platform velocity relative to the passenger \(=-3\hat j\,\text{m s}^{-1}\).
Explanation: The velocity-addition relation is
\[
\vec v_{P/G}
=\vec v_{P/\text{platform}}
+\vec v_{\text{platform}/G}.
\]
Therefore
\[
\vec v_{P/G}=3\hat j+5\hat i
=5\hat i+3\hat j.
\]
Its magnitude is
\[
|\vec v_{P/G}|
=\sqrt{5^2+3^2}
=\sqrt{34}\,\text{m s}^{-1}.
\]
The perpendicular velocity components do not add directly as \(5+3\). To determine how the platform appears to the passenger,
\[
\vec v_{\text{platform}/P}
=\vec v_{\text{platform}/G}-\vec v_{P/G}.
\]
Thus
\[
\vec v_{\text{platform}/P}
=5\hat i-(5\hat i+3\hat j)
=-3\hat j.
\]
Hence the passenger sees the platform moving south at \(3\,\text{m s}^{-1}\), exactly opposite to the passenger's walking velocity relative to the platform.
307. The position vector of a particle moving in a plane is
\[
\vec r(t)=2t\,\hat i+t^2\,\hat j
\]
in metres, where \(t\) is in seconds. Which statement is correct at \(t=2\,\text{s}\)?
ⓐ. \(\vec v=4\hat j\,\text{m s}^{-1}\), \(\vec a=2\hat j\,\text{m s}^{-2}\), and the particle has no \(x\)-motion
ⓑ. \(\vec v=4\hat i+2\hat j\,\text{m s}^{-1}\), speed \(=2\sqrt5\,\text{m s}^{-1}\), and \(\vec a=2\hat i\,\text{m s}^{-2}\)
ⓒ. \(\vec v=2\hat i+2\hat j\,\text{m s}^{-1}\), speed \(=2\sqrt2\,\text{m s}^{-1}\), and the path is a straight line
ⓓ. \(\vec v=2\hat i+4\hat j\,\text{m s}^{-1}\), speed \(=2\sqrt5\,\text{m s}^{-1}\), \(\vec a=2\hat j\,\text{m s}^{-2}\), and the trajectory satisfies \(y=\dfrac{x^2}{4}\)
Correct Answer: \(\vec v=2\hat i+4\hat j\,\text{m s}^{-1}\), speed \(=2\sqrt5\,\text{m s}^{-1}\), \(\vec a=2\hat j\,\text{m s}^{-2}\), and the trajectory satisfies \(y=\dfrac{x^2}{4}\)
Explanation: The instantaneous velocity is the time derivative of the position vector:
\[
\vec v=\frac{d\vec r}{dt}.
\]
Therefore
\[
\vec v=2\hat i+2t\,\hat j.
\]
At \(t=2\,\text{s}\),
\[
\vec v=2\hat i+4\hat j\,\text{m s}^{-1}.
\]
Its speed is the magnitude
\[
v=\sqrt{2^2+4^2}=2\sqrt5\,\text{m s}^{-1}.
\]
The acceleration is
\[
\vec a=\frac{d\vec v}{dt}=2\hat j\,\text{m s}^{-2}.
\]
To obtain the trajectory, use
\[
x=2t,\qquad y=t^2.
\]
Hence
\[
t=\frac{x}{2}.
\]
Substitution into \(y=t^2\) gives
\[
y=\frac{x^2}{4}.
\]
Thus the particle has constant \(x\)-component of velocity, increasing \(y\)-component of velocity and a parabolic trajectory in the plane.
308. A river is \(120\,\text{m}\) wide and flows east at \(3\,\text{m s}^{-1}\). A boat is pointed perpendicular to the banks and moves north at \(4\,\text{m s}^{-1}\) relative to the water. Which set of results is correct?
ⓐ. Crossing time \(=30\,\text{s}\), downstream drift \(=120\,\text{m}\), ground-frame path length \(=90\,\text{m}\)
ⓑ. Crossing time \(=24\,\text{s}\), downstream drift \(=72\,\text{m}\), ground-frame path length \(=120\,\text{m}\)
ⓒ. Crossing time \(=30\,\text{s}\), downstream drift \(=90\,\text{m}\), ground-frame path length \(=150\,\text{m}\)
ⓓ. Crossing time \(=40\,\text{s}\), downstream drift \(=120\,\text{m}\), ground-frame path length \(=200\,\text{m}\)
Correct Answer: Crossing time \(=30\,\text{s}\), downstream drift \(=90\,\text{m}\), ground-frame path length \(=150\,\text{m}\)
Explanation: Take east as \(+\hat i\) and the direction directly across the river as \(+\hat j\). The boat's velocity relative to water is
\[
\vec v_{B/W}=4\hat j.
\]
The water velocity relative to ground is
\[
\vec v_{W/G}=3\hat i.
\]
Hence the boat's ground velocity is
\[
\vec v_{B/G}=\vec v_{B/W}+\vec v_{W/G}=3\hat i+4\hat j.
\]
Only the northward component carries the boat across the \(120\,\text{m}\) width, so
\[
t=\frac{120}{4}=30\,\text{s}.
\]
During this time the current produces an eastward displacement
\[
x=(3)(30)=90\,\text{m}.
\]
The ground-frame speed is
\[
|\vec v_{B/G}|=\sqrt{3^2+4^2}=5\,\text{m s}^{-1}.
\]
Therefore the actual ground-frame path length is
\[
s=(5)(30)=150\,\text{m}.
\]
The current changes the ground path and drift but does not alter the \(4\,\text{m s}^{-1}\) component perpendicular to the banks.
309. A river flows east at \(3\,\text{m s}^{-1}\) and is \(120\,\text{m}\) wide. A boat can move at \(5\,\text{m s}^{-1}\) relative to the water. The boat is to land exactly at the point directly opposite its starting point. Which statement is correct?
ⓐ. The boat must point due north and crosses in \(24\,\text{s}\).
ⓑ. The boat must have a \(3\,\text{m s}^{-1}\) westward and \(4\,\text{m s}^{-1}\) northward velocity relative to water, and crosses in \(30\,\text{s}\).
ⓒ. The boat must have a \(4\,\text{m s}^{-1}\) westward and \(3\,\text{m s}^{-1}\) northward velocity relative to water, and crosses in \(40\,\text{s}\).
ⓓ. The boat must point due west until the current is cancelled and then turn north.
Correct Answer: The boat must have a \(3\,\text{m s}^{-1}\) westward and \(4\,\text{m s}^{-1}\) northward velocity relative to water, and crosses in \(30\,\text{s}\).
Explanation: To land directly opposite, the boat's ground-frame east-west velocity must be zero. The river contributes
\[
+3\,\text{m s}^{-1}
\]
eastward. Therefore the boat must supply a water-relative component
\[
v_{x,B/W}=-3\,\text{m s}^{-1}.
\]
Its total speed relative to water is \(5\,\text{m s}^{-1}\), so its northward component satisfies
\[
5^2=3^2+v_y^2.
\]
Hence
\[
v_y=4\,\text{m s}^{-1}.
\]
The water-relative velocity is therefore
\[
\vec v_{B/W}=-3\hat i+4\hat j.
\]
After adding the river velocity \(3\hat i\),
\[
\vec v_{B/G}=4\hat j.
\]
Thus the boat moves directly across relative to the ground and takes
\[
t=\frac{120}{4}=30\,\text{s}.
\]
The upstream component cancels the current but reduces the component available for crossing.
310. A river is \(120\,\text{m}\) wide, its current is \(3\,\text{m s}^{-1}\) eastward, and a boat moves at \(5\,\text{m s}^{-1}\) relative to water. Boat P points perpendicular to the banks. Boat Q chooses its direction so that it lands exactly opposite its starting point. Let their crossing times be \(T_P,T_Q\) and their ground-frame path lengths be \(d_P,d_Q\). Which statement is correct?
ⓐ. \(\dfrac{T_Q}{T_P}=\dfrac45,\ \dfrac{d_P}{d_Q}=\dfrac5{\sqrt{34}}\); Q crosses sooner and P follows the shorter ground path
ⓑ. \(\dfrac{T_Q}{T_P}=1,\ \dfrac{d_P}{d_Q}=\dfrac{\sqrt{34}}5\); both boats take the same crossing time
ⓒ. \(\dfrac{T_Q}{T_P}=\dfrac54,\ \dfrac{d_P}{d_Q}=\dfrac{\sqrt{34}}5\); P crosses sooner and Q follows the shorter ground path
ⓓ. \(\dfrac{T_Q}{T_P}=\dfrac45,\ \dfrac{d_P}{d_Q}=\dfrac5{\sqrt{34}}\); P crosses sooner and Q follows the longer ground path
Correct Answer: \(\dfrac{T_Q}{T_P}=\dfrac54,\ \dfrac{d_P}{d_Q}=\dfrac{\sqrt{34}}5\); P crosses sooner and Q follows the shorter ground path
Explanation: For P, the full \(5\,\text{m s}^{-1}\) water-relative speed is directed across the river. Therefore
\[
T_P=\frac{120}{5}=24\,\text{s}.
\]
Its ground velocity is
\[
\vec v_{P/G}=3\hat i+5\hat j,
\]
whose magnitude is
\[
\sqrt{34}\,\text{m s}^{-1}.
\]
Hence
\[
d_P=24\sqrt{34}\,\text{m}.
\]
For Q, cancellation of the \(3\,\text{m s}^{-1}\) current requires a \(3\,\text{m s}^{-1}\) upstream component. The remaining across-river component is
\[
\sqrt{5^2-3^2}=4\,\text{m s}^{-1}.
\]
Thus
\[
T_Q=\frac{120}{4}=30\,\text{s}
\]
and its ground path is directly across, so
\[
d_Q=120\,\text{m}.
\]
Therefore
\[
\frac{T_Q}{T_P}=\frac{30}{24}=\frac54
\]
and
\[
\frac{d_P}{d_Q}=\frac{24\sqrt{34}}{120}=\frac{\sqrt{34}}5.
\]
The minimum-time strategy and the directly-opposite, shortest-ground-path strategy are not the same.
311. A river flows east at \(6\,\text{m s}^{-1}\). A boat has speed \(5\,\text{m s}^{-1}\) relative to the water. During one crossing it takes \(12\,\text{s}\) to cross a \(48\,\text{m}\) wide river and lands \(36\,\text{m}\) downstream from the point directly opposite the start. Which statement correctly describes the boat's velocity relative to the water?
ⓐ. \(3\hat i+4\hat j\,\text{m s}^{-1}\)
ⓑ. \(-4\hat i+3\hat j\,\text{m s}^{-1}\)
ⓒ. \(-3\hat i+4\hat j\,\text{m s}^{-1}\)
ⓓ. \(5\hat j\,\text{m s}^{-1}\)
Correct Answer: \(-3\hat i+4\hat j\,\text{m s}^{-1}\)
Explanation: Take east as \(+\hat i\) and directly across the river as \(+\hat j\). The observed ground displacement is
\[
\Delta\vec r=36\hat i+48\hat j.
\]
Since the crossing takes \(12\,\text{s}\), the boat's ground velocity is
\[
\vec v_{B/G}=\frac{\Delta\vec r}{12}.
\]
Therefore
\[
\vec v_{B/G}=3\hat i+4\hat j\,\text{m s}^{-1}.
\]
The current velocity is
\[
\vec v_{W/G}=6\hat i\,\text{m s}^{-1}.
\]
Using
\[
\vec v_{B/G}=\vec v_{B/W}+\vec v_{W/G},
\]
we obtain
\[
\vec v_{B/W}=\vec v_{B/G}-\vec v_{W/G}.
\]
Hence
\[
\vec v_{B/W}=-3\hat i+4\hat j\,\text{m s}^{-1}.
\]
Its magnitude is
\[
\sqrt{(-3)^2+4^2}=5\,\text{m s}^{-1},
\]
which agrees with the given speed and confirms the consistency of the result.
312. Rain has speed \(13\,\text{m s}^{-1}\) relative to the ground. A cyclist moving east at \(5\,\text{m s}^{-1}\) observes the rain to fall vertically downward. The cyclist then turns around and moves west at \(5\,\text{m s}^{-1}\). What is the rain velocity relative to the cyclist after the turn?
ⓐ. \(-5\hat i-8\hat j\,\text{m s}^{-1}\), with apparent speed \(\sqrt{89}\,\text{m s}^{-1}\)
ⓑ. \(-10\hat i-12\hat j\,\text{m s}^{-1}\), with apparent speed \(22\,\text{m s}^{-1}\)
ⓒ. \(5\hat i-12\hat j\,\text{m s}^{-1}\), with apparent speed \(13\,\text{m s}^{-1}\)
ⓓ. \(10\hat i-12\hat j\,\text{m s}^{-1}\), with apparent speed \(2\sqrt{61}\,\text{m s}^{-1}\)
Correct Answer: \(10\hat i-12\hat j\,\text{m s}^{-1}\), with apparent speed \(2\sqrt{61}\,\text{m s}^{-1}\)
Explanation: Let east be \(+\hat i\) and upward be \(+\hat j\). Initially the cyclist velocity is
\[
\vec v_C=5\hat i.
\]
The rain appears vertically downward, so the horizontal component of
\[
\vec v_{R/C}=\vec v_R-\vec v_C
\]
must be zero. Therefore the rain's ground-frame horizontal component is \(+5\,\text{m s}^{-1}\). Since its ground-frame speed is \(13\,\text{m s}^{-1}\), its downward component has magnitude
\[
\sqrt{13^2-5^2}=12\,\text{m s}^{-1}.
\]
Thus
\[
\vec v_R=5\hat i-12\hat j.
\]
After turning around, the cyclist has
\[
\vec v'_C=-5\hat i.
\]
Hence
\[
\vec v_{R/C'}=\vec v_R-\vec v'_C
=(5\hat i-12\hat j)-(-5\hat i).
\]
Therefore
\[
\vec v_{R/C'}=10\hat i-12\hat j.
\]
Its magnitude is
\[
\sqrt{10^2+12^2}=2\sqrt{61}\,\text{m s}^{-1}.
\]
The observer's change of motion changes the apparent rain velocity even though the rain's ground velocity is unchanged.
313. The current of a river is uniform but initially unknown. Boat P is directed due north at \(5\,\text{m s}^{-1}\) relative to the water. It crosses a \(100\,\text{m}\) wide river in \(20\,\text{s}\) and lands \(60\,\text{m}\) east of the point directly opposite its start. Boat Q then moves in the same river with water-relative velocity
\[
-3\hat i+4\hat j\ \text{m s}^{-1}.
\]
Which statement is correct?
ⓐ. The current is \(5\,\text{m s}^{-1}\) east; Q drifts \(50\,\text{m}\) downstream.
ⓑ. The current is \(3\,\text{m s}^{-1}\) east; Q lands directly opposite after \(25\,\text{s}\).
ⓒ. The current is \(3\,\text{m s}^{-1}\) west; Q crosses in \(20\,\text{s}\).
ⓓ. The current is \(4\,\text{m s}^{-1}\) east; Q cannot reach the opposite bank.
Correct Answer: The current is \(3\,\text{m s}^{-1}\) east; Q lands directly opposite after \(25\,\text{s}\).
Explanation: Boat P's northward component is \(5\,\text{m s}^{-1}\), so crossing \(100\,\text{m}\) in
\[
\frac{100}{5}=20\,\text{s}
\]
is consistent with the data. During those \(20\,\text{s}\), the river carries P \(60\,\text{m}\) east. Therefore the current speed is
\[
v_c=\frac{60}{20}=3\,\text{m s}^{-1}.
\]
Hence
\[
\vec v_{W/G}=3\hat i.
\]
For Q,
\[
\vec v_{Q/W}=-3\hat i+4\hat j.
\]
Its ground velocity is
\[
\vec v_{Q/G}
=\vec v_{Q/W}+\vec v_{W/G}.
\]
Therefore
\[
\vec v_{Q/G}=4\hat j.
\]
The horizontal components cancel exactly, so Q lands directly opposite its starting point. Its crossing time is
\[
t=\frac{100}{4}=25\,\text{s}.
\]
Thus observations of one boat can first determine the medium velocity and then predict another boat's ground motion.
314. At \(t=0\), particle A is at the origin and particle B is at
\[
15\hat i+20\hat j\ \text{m}.
\]
Their constant velocities are
\[
\vec v_A=4\hat i+5\hat j\ \text{m s}^{-1},
\qquad
\vec v_B=\hat i+\hat j\ \text{m s}^{-1}.
\]
Which statement is correct?
ⓐ. They meet after \(4\,\text{s}\) at \(16\hat i+20\hat j\,\text{m}\).
ⓑ. They never meet because their speeds are unequal.
ⓒ. They meet at \(t=5\,\text{s}\) at \(20\hat i+25\hat j\,\text{m}\); relative speed \(=5\,\text{m s}^{-1}\).
ⓓ. They meet after \(5\,\text{s}\), but their relative velocity is zero at the meeting.
Correct Answer: They meet at \(t=5\,\text{s}\) at \(20\hat i+25\hat j\,\text{m}\); relative speed \(=5\,\text{m s}^{-1}\).
Explanation: The initial position of B relative to A is
\[
\vec r_{B/A}(0)=15\hat i+20\hat j.
\]
The relative velocity is
\[
\vec v_{B/A}=\vec v_B-\vec v_A.
\]
Thus
\[
\vec v_{B/A}=-3\hat i-4\hat j\,\text{m s}^{-1}.
\]
The relative position at time \(t\) is
\[
\vec r_{B/A}(t)
=(15-3t)\hat i+(20-4t)\hat j.
\]
Both components vanish simultaneously for
\[
t=5\,\text{s}.
\]
Hence the particles meet. Using A's motion, their meeting position is
\[
\vec r_A(5)
=(4\hat i+5\hat j)(5)
=20\hat i+25\hat j.
\]
The relative speed is
\[
|\vec v_{B/A}|=\sqrt{3^2+4^2}=5\,\text{m s}^{-1}.
\]
The relative velocity need not vanish when two particles meet; it determines how rapidly they pass one another at that event.
315. Particle A starts from the origin with constant velocity
\[
\vec v_A=5\hat i+3\hat j\ \text{m s}^{-1}.
\]
Particle B starts from
\[
12\hat i+8\hat j\ \text{m}.
\]
Which one of the following constant velocities of B makes the particles meet, and when does the meeting occur?
ⓐ. \(\vec v_B=2\hat i+\hat j\,\text{m s}^{-1}\); meeting after \(4\,\text{s}\)
ⓑ. \(\vec v_B=\hat i+\hat j\,\text{m s}^{-1}\); meeting after \(3\,\text{s}\)
ⓒ. \(\vec v_B=2\hat i+2\hat j\,\text{m s}^{-1}\); meeting after \(4\,\text{s}\)
ⓓ. \(\vec v_B=8\hat i+5\hat j\,\text{m s}^{-1}\); meeting after \(4\,\text{s}\)
Correct Answer: \(\vec v_B=2\hat i+\hat j\,\text{m s}^{-1}\); meeting after \(4\,\text{s}\)
Explanation: For the particles to meet, B's initial relative-position vector must be cancelled by its relative displacement. The initial vector from A to B is
\[
\vec r_{B/A}(0)=12\hat i+8\hat j.
\]
For option A,
\[
\vec v_{B/A}
=(2-5)\hat i+(1-3)\hat j
=-3\hat i-2\hat j.
\]
Notice that
\[
12\hat i+8\hat j
=4(3\hat i+2\hat j).
\]
Therefore
\[
\vec r_{B/A}(0)+4\vec v_{B/A}=\vec0.
\]
So the meeting time is \(4\,\text{s}\). At that instant A is at
\[
\vec r_A=(5\hat i+3\hat j)(4)
=20\hat i+12\hat j.
\]
For option B, the \(x\)-separation and \(y\)-separation would vanish at different times. The same incompatibility occurs for option C, while option D gives a relative velocity with components pointing generally away from the initial separation. A meeting requires the relative velocity to cancel the entire relative-position vector at one common time, not merely one component.
316. Consider the following Assertion and Reason.
Assertion: If two particles have unequal velocity vectors in a ground frame, no observer moving with a single constant velocity relative to the ground can make the two particles mutually at rest.
Reason: Subtracting the same observer velocity from both particle velocities leaves their relative velocity unchanged.
ⓐ. Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
ⓑ. Both Assertion and Reason are true, but Reason is not the correct explanation of Assertion.
ⓒ. Assertion is true, but Reason is false.
ⓓ. Assertion is false, but Reason is true.
Correct Answer: Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
Explanation: Let the ground-frame particle velocities be \(\vec v_A\) and \(\vec v_B\). Let an observer move with constant velocity \(\vec U\) relative to the ground. That observer measures
\[
\vec v'_A=\vec v_A-\vec U
\]
and
\[
\vec v'_B=\vec v_B-\vec U.
\]
Their relative velocity in the observer's frame is therefore
\[
\vec v'_A-\vec v'_B
=(\vec v_A-\vec U)-(\vec v_B-\vec U).
\]
The observer velocity cancels:
\[
\vec v'_A-\vec v'_B=\vec v_A-\vec v_B.
\]
Thus a common change of uniformly moving observer does not alter the relative velocity between the two particles. If \(\vec v_A\ne\vec v_B\), their relative velocity is non-zero in every such frame, so they cannot be mutually at rest. Hence the Assertion is true and the Reason directly explains it.
317. In the ground frame,
\[
\vec v_A=7\hat i+2\hat j\ \text{m s}^{-1},
\qquad
\vec v_B=\hat i+5\hat j\ \text{m s}^{-1}.
\]
An observer moves with constant velocity \(\vec U\) such that A is seen to move due east at \(4\,\text{m s}^{-1}\). Which statement is correct?
ⓐ. \(\vec U=4\hat i\); B is seen with \(-3\hat i+5\hat j\,\text{m s}^{-1}\) and speed \(\sqrt{34}\,\text{m s}^{-1}\)
ⓑ. \(\vec U=11\hat i+2\hat j\); B is seen with \(-10\hat i+3\hat j\,\text{m s}^{-1}\) and speed \(\sqrt{109}\,\text{m s}^{-1}\)
ⓒ. \(\vec U=3\hat i-2\hat j\); B is seen with \(-2\hat i+7\hat j\,\text{m s}^{-1}\) and speed \(\sqrt{53}\,\text{m s}^{-1}\)
ⓓ. \(\vec U=3\hat i+2\hat j\); B is seen with \(-2\hat i+3\hat j\,\text{m s}^{-1}\) and speed \(\sqrt{13}\,\text{m s}^{-1}\)
Correct Answer: \(\vec U=3\hat i+2\hat j\); B is seen with \(-2\hat i+3\hat j\,\text{m s}^{-1}\) and speed \(\sqrt{13}\,\text{m s}^{-1}\)
Explanation: The observer is required to measure A's velocity as
\[
\vec v'_A=4\hat i.
\]
The transformation is
\[
\vec v'_A=\vec v_A-\vec U.
\]
Therefore
\[
\vec U=\vec v_A-\vec v'_A.
\]
Substituting,
\[
\vec U=(7\hat i+2\hat j)-4\hat i
=3\hat i+2\hat j.
\]
The same observer measures B as
\[
\vec v'_B=\vec v_B-\vec U.
\]
Hence
\[
\vec v'_B
=(\hat i+5\hat j)-(3\hat i+2\hat j)
=-2\hat i+3\hat j.
\]
Its speed is
\[
|\vec v'_B|=\sqrt{(-2)^2+3^2}=\sqrt{13}\,\text{m s}^{-1}.
\]
The relative velocity of A and B is unchanged by this observer transformation because the same \(\vec U\) is subtracted from both.
318. An observer O measures particle A to have velocity
\[
3\hat i+4\hat j\ \text{m s}^{-1}.
\]
In the ground frame A actually has velocity
\[
8\hat i+2\hat j\ \text{m s}^{-1}.
\]
The same observer measures particle B to have velocity
\[
-2\hat i+\hat j\ \text{m s}^{-1}.
\]
Which statement is correct?
ⓐ. O moves with \(11\hat i+6\hat j\,\text{m s}^{-1}\); B's ground velocity is \(9\hat i+7\hat j\,\text{m s}^{-1}\)
ⓑ. O moves with \(5\hat i-2\hat j\,\text{m s}^{-1}\); B's ground velocity is \(3\hat i-\hat j\,\text{m s}^{-1}\); A relative to B has velocity \(5\hat i+3\hat j\,\text{m s}^{-1}\)
ⓒ. O moves with \(-5\hat i+2\hat j\,\text{m s}^{-1}\); B's ground velocity is \(-7\hat i+3\hat j\,\text{m s}^{-1}\)
ⓓ. O moves with \(5\hat i-2\hat j\,\text{m s}^{-1}\), but the relative velocity of A and B depends on whether it is calculated in O's frame or the ground frame
Correct Answer: O moves with \(5\hat i-2\hat j\,\text{m s}^{-1}\); B's ground velocity is \(3\hat i-\hat j\,\text{m s}^{-1}\); A relative to B has velocity \(5\hat i+3\hat j\,\text{m s}^{-1}\)
Explanation: The observer-frame relation for A is
\[
\vec v_{A/O}=\vec v_{A/G}-\vec v_{O/G}.
\]
Therefore
\[
\vec v_{O/G}
=\vec v_{A/G}-\vec v_{A/O}.
\]
Substitution gives
\[
\vec v_{O/G}
=(8\hat i+2\hat j)-(3\hat i+4\hat j)
=5\hat i-2\hat j.
\]
For B,
\[
\vec v_{B/G}
=\vec v_{B/O}+\vec v_{O/G}.
\]
Thus
\[
\vec v_{B/G}
=(-2\hat i+\hat j)+(5\hat i-2\hat j)
=3\hat i-\hat j.
\]
The ground-frame relative velocity of A with respect to B is
\[
\vec v_{A/B}
=\vec v_{A/G}-\vec v_{B/G}
=5\hat i+3\hat j.
\]
Using observer-frame velocities gives the same result:
\[
(3\hat i+4\hat j)-(-2\hat i+\hat j)=5\hat i+3\hat j.
\]
Thus the relative velocity is invariant under this common constant-velocity change of observer.
319. A river of width \(50\,\text{m}\) flows east at \(6\,\text{m s}^{-1}\). A boat can move at only \(5\,\text{m s}^{-1}\) relative to the water. Which statement is correct?
ⓐ. The boat can land directly opposite by pointing sufficiently far upstream, and the crossing time is \(10\,\text{s}\).
ⓑ. The boat cannot cross the river because the current speed exceeds the boat speed.
ⓒ. Landing directly opposite is impossible; if the boat points perpendicular to the banks, it crosses in \(10\,\text{s}\) and drifts \(60\,\text{m}\) downstream.
ⓓ. The minimum crossing time is \(\dfrac{50}{11}\,\text{s}\) because the two speeds must be added.
Correct Answer: Landing directly opposite is impossible; if the boat points perpendicular to the banks, it crosses in \(10\,\text{s}\) and drifts \(60\,\text{m}\) downstream.
Explanation: To land directly opposite, the boat would need a water-relative upstream component equal in magnitude to the current:
\[
v_{\rm upstream}=6\,\text{m s}^{-1}.
\]
But the boat's entire speed relative to the water is only
\[
5\,\text{m s}^{-1}.
\]
Therefore it cannot supply a \(6\,\text{m s}^{-1}\) upstream component, so exact current cancellation is impossible. This does not mean that the boat cannot cross. To minimise crossing time, the boat should maximise its component perpendicular to the banks. Pointing directly across gives the full
\[
5\,\text{m s}^{-1}
\]
across-river component. Hence
\[
t_{\min}=\frac{50}{5}=10\,\text{s}.
\]
During these \(10\,\text{s}\), the current produces downstream drift
\[
x=(6)(10)=60\,\text{m}.
\]
A current faster than the boat prevents a directly-opposite landing but does not prevent crossing when the boat has a non-zero across-river component.
320. A boat can move at \(5\,\text{m s}^{-1}\) relative to water across a river \(100\,\text{m}\) wide. Initially the current is \(3\,\text{m s}^{-1}\) eastward; later it increases to \(4\,\text{m s}^{-1}\). Consider two strategies: P always points the boat perpendicular to the banks, while Q always adjusts its heading so as to land directly opposite whenever possible. Which statement is correct?
ⓐ. P's crossing time increases from \(20\,\text{s}\) to \(25\,\text{s}\), while Q's crossing time remains fixed.
ⓑ. Both P and Q cross in \(20\,\text{s}\) for either current because the boat speed is unchanged.
ⓒ. For P, \(T=20\,\text{s}\) in both cases while drift changes \(60\to80\,\text{m}\); for Q, \(T\) changes \(25\to\dfrac{100}{3}\,\text{s}\).
ⓓ. Q becomes unable to cross directly opposite when the current becomes \(4\,\text{m s}^{-1}\).
Correct Answer: For P, \(T=20\,\text{s}\) in both cases while drift changes \(60\to80\,\text{m}\); for Q, \(T\) changes \(25\to\dfrac{100}{3}\,\text{s}\).
Explanation: For strategy P, the full water-relative speed \(5\,\text{m s}^{-1}\) is directed across the river. Therefore its across-river component remains
\[
5\,\text{m s}^{-1}
\]
regardless of the current parallel to the banks. Hence
\[
T_P=\frac{100}{5}=20\,\text{s}
\]
in both cases. The drift is current speed multiplied by this time, so it changes from
\[
3(20)=60\,\text{m}
\]
to
\[
4(20)=80\,\text{m}.
\]
For Q with current \(3\,\text{m s}^{-1}\), the upstream component must be \(3\), leaving across component
\[
\sqrt{25-9}=4\,\text{m s}^{-1}.
\]
Thus
\[
T_Q=\frac{100}{4}=25\,\text{s}.
\]
With current \(4\,\text{m s}^{-1}\), the remaining across component is
\[
\sqrt{25-16}=3\,\text{m s}^{-1},
\]
giving
\[
T_Q=\frac{100}{3}\,\text{s}.
\]
Thus a stronger current does not affect P's crossing component but forces Q to sacrifice more of its own speed to current cancellation.
