201. Consider the following Assertion and Reason for one-dimensional motion with constant acceleration.
Assertion: If the signed displacements during two consecutive equal time intervals are equal, the acceleration must be zero.
Reason: For constant acceleration, the difference between the signed displacements in two consecutive equal intervals of duration \(T\) is \(aT^2\).
ⓐ. 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 first interval have duration \(T\). For constant acceleration, its displacement can be written as
\[
s_1=uT+\frac12aT^2.
\]
At the beginning of the second interval, the velocity is
\[
u+aT.
\]
Hence the second displacement is
\[
s_2=(u+aT)T+\frac12aT^2.
\]
Therefore
\[
s_2-s_1=aT^2.
\]
If the signed displacements are equal,
\[
s_2-s_1=0.
\]
Since \(T\ne0\),
\[
a=0.
\]
The use of signed displacement is important: equal path lengths in two intervals would not necessarily lead to the same conclusion if a reversal occurred. Thus both the Assertion and Reason are true, and the Reason directly proves the Assertion.
202. A vehicle moving with initial speed \(u\) is brought to rest by a constant retardation of magnitude \(a\). Let its complete stopping distance and stopping time be \(S\) and \(T\), respectively. When it has covered exactly half of \(S\), which result is correct?
ⓐ. Speed \(=\dfrac{u}{2}\), and remaining stopping time \(=\dfrac{T}{2}\)
ⓑ. Speed \(=\dfrac{u}{\sqrt2}\), and elapsed time \(=\dfrac{T}{\sqrt2}\)
ⓒ. Speed \(=\dfrac{u}{\sqrt2}\), and remaining stopping time \(=\dfrac{T}{\sqrt2}\)
ⓓ. Speed \(=\dfrac{u}{2}\), and remaining stopping time \(=\dfrac{T}{\sqrt2}\)
Correct Answer: Speed \(=\dfrac{u}{\sqrt2}\), and remaining stopping time \(=\dfrac{T}{\sqrt2}\)
Explanation: For uniform braking,
\[
S=\frac{u^2}{2a}.
\]
After the vehicle has travelled \(S/2\), use
\[
v^2=u^2-2a\left(\frac S2\right).
\]
Since
\[
aS=\frac{u^2}{2},
\]
we obtain
\[
v^2=u^2-\frac{u^2}{2}=\frac{u^2}{2}.
\]
Hence
\[
v=\frac{u}{\sqrt2}.
\]
The full stopping time is
\[
T=\frac{u}{a}.
\]
From the halfway-distance point, the vehicle still has speed \(u/\sqrt2\) and the same constant retardation \(a\). Therefore the remaining time is
\[
T_{\rm rem}=\frac{u/\sqrt2}{a}
=\frac{T}{\sqrt2}.
\]
Half the stopping distance is therefore not reached at half the stopping time because displacement depends quadratically on the changing velocity.
203. A particle starts from rest at \(x=0\). For the first \(3\,\text{s}\), it has constant acceleration \(+2\,\text{m s}^{-2}\). Immediately afterward, its acceleration changes to \(-6\,\text{m s}^{-2}\) and remains constant until the particle first returns to \(x=0\). Which set of results is correct?
ⓐ. Return occurs \(3\,\text{s}\) after the acceleration change; final velocity \(=-12\,\text{m s}^{-1}\); total distance \(=24\,\text{m}\)
ⓑ. Return occurs \(2\,\text{s}\) after the acceleration change; final velocity \(=-6\,\text{m s}^{-1}\); total distance \(=18\,\text{m}\)
ⓒ. Return occurs \(3\,\text{s}\) after the acceleration change; final velocity \(=0\); total distance \(=18\,\text{m}\)
ⓓ. Return occurs \(4\,\text{s}\) after the acceleration change; final velocity \(=-18\,\text{m s}^{-1}\); total distance \(=30\,\text{m}\)
Correct Answer: Return occurs \(3\,\text{s}\) after the acceleration change; final velocity \(=-12\,\text{m s}^{-1}\); total distance \(=24\,\text{m}\)
Explanation: During the first stage,
\[
v_1=(2)(3)=6\,\text{m s}^{-1},
\]
and
\[
s_1=\frac12(2)(3^2)=9\,\text{m}.
\]
Thus the second stage begins at \(x=9\,\text{m}\) with velocity \(+6\,\text{m s}^{-1}\). To return to \(x=0\), the second-stage displacement must be
\[
s_2=-9\,\text{m}.
\]
Using
\[
v_f^2=u^2+2as,
\]
\[
v_f^2=6^2+2(-6)(-9)=36+108=144.
\]
The particle must be moving in the negative direction on returning, so
\[
v_f=-12\,\text{m s}^{-1}.
\]
Now
\[
-12=6-6T,
\]
giving
\[
T=3\,\text{s}.
\]
During the second stage, the particle first stops after \(1\,\text{s}\), having moved another
\[
\frac{6^2}{2(6)}=3\,\text{m}
\]
forward to \(x=12\,\text{m}\). It then travels \(12\,\text{m}\) back to the origin. Hence total distance is
\[
9+3+12=24\,\text{m}.
\]
204. A particle starts at \(x=1\,\text{m}\). Its velocity-time graph is described below.
From \(t=0\) to \(3\,\text{s}\), velocity increases linearly from \(-2\,\text{m s}^{-1}\) to \(+4\,\text{m s}^{-1}\). From \(t=3\) to \(5\,\text{s}\), velocity remains constant at \(+4\,\text{m s}^{-1}\).
Which statement is correct?
ⓐ. At \(t=1\,\text{s}\), \(x=0\) and \(v=0\); the particle reverses there without entering \(x\lt 0\) and ends at \(x=12\,\text{m}\).
ⓑ. The particle crosses into negative \(x\) before reversing and finishes at \(x=10\,\text{m}\).
ⓒ. The particle reaches \(x=0\) at \(t=2\,\text{s}\) and remains there until \(t=3\,\text{s}\).
ⓓ. The particle never reaches the origin because its initial position is positive.
Correct Answer: At \(t=1\,\text{s}\), \(x=0\) and \(v=0\); the particle reverses there without entering \(x\lt 0\) and ends at \(x=12\,\text{m}\).
Explanation: The acceleration during the first stage is
\[
a=\frac{4-(-2)}{3}=2\,\text{m s}^{-2}.
\]
Thus velocity becomes zero at
\[
-2+2t=0,
\]
so
\[
t=1\,\text{s}.
\]
The displacement from \(0\) to \(1\,\text{s}\) is the negative triangular area
\[
s=-\frac12(1)(2)=-1\,\text{m}.
\]
Since the initial position is \(+1\,\text{m}\), the particle reaches
\[
x=0
\]
exactly at the zero-velocity instant. The velocity changes from negative to positive there, so the particle reverses without entering \(x\lt 0\). From \(1\) to \(3\,\text{s}\), the positive triangular area is
\[
\frac12(2)(4)=4\,\text{m}.
\]
Thus \(x(3)=4\,\text{m}\). During the final \(2\,\text{s}\),
\[
\Delta x=(4)(2)=8\,\text{m},
\]
so
\[
x(5)=12\,\text{m}.
\]
205. Match the velocity-time behaviour in List I with the corresponding position-time behaviour in List II.
| List I | \(v-t\) behaviour | List II | Corresponding \(x-t\) behaviour |
|---|
| P | Horizontal line at constant \(v\gt 0\) | 1 | Straight line with constant positive slope |
| Q | Straight line decreasing from positive \(v\) to \(0\) | 2 | Rising curve with decreasing slope, ending with a horizontal tangent |
| R | Straight line increasing from negative \(v\) to \(0\) | 3 | Falling curve with slope becoming less negative, ending with a horizontal tangent |
| S | Straight line crossing from negative \(v\) to positive \(v\) | 4 | Curve with a local minimum at the zero-velocity instant |
ⓐ. P-1, Q-3, R-2, S-4
ⓑ. P-2, Q-1, R-4, S-3
ⓒ. P-1, Q-2, R-3, S-4
ⓓ. P-4, Q-2, R-3, S-1
Correct Answer: P-1, Q-2, R-3, S-4
Explanation: For P, constant positive velocity means the \(x-t\) slope is constant and positive, so P matches 1. For Q, velocity remains positive but decreases to zero. Thus position continues increasing while its slope becomes progressively smaller, finally becoming horizontal, so Q matches 2. For R, velocity is negative but increases toward zero. The particle therefore continues toward decreasing \(x\), while the negative slope of the \(x-t\) graph becomes less steep and finally horizontal, giving R-3. For S, velocity changes sign from negative to positive. The \(x-t\) slope therefore changes from negative to zero to positive, producing a local minimum, so S matches 4. The translation between the two graph types depends on remembering that velocity itself is the slope of the position-time graph.
206. Which of the following graph pairings is physically inconsistent for the same one-dimensional motion over the stated interval?
ⓐ. A straight \(x-t\) line with positive slope paired with a horizontal \(v-t\) line above the time axis
ⓑ. A downward-opening parabolic \(x-t\) curve with a local maximum paired with a straight \(v-t\) line of negative slope crossing from positive to negative velocity
ⓒ. An upward-opening parabolic \(x-t\) curve with a local minimum paired with a straight \(v-t\) line of positive slope crossing from negative to positive velocity
ⓓ. A horizontal \(x-t\) segment paired over the same interval with a horizontal \(v-t\) line at a non-zero velocity
Correct Answer: A horizontal \(x-t\) segment paired over the same interval with a horizontal \(v-t\) line at a non-zero velocity
Explanation: In option A, a straight \(x-t\) line with positive slope represents constant positive velocity, exactly matching a horizontal positive \(v-t\) line. Option B is also consistent: constant negative acceleration produces a downward-opening quadratic \(x-t\) graph and a linearly decreasing \(v-t\) graph. If velocity crosses zero from positive to negative, the \(x-t\) curve has a local maximum. Option C is the corresponding positive-acceleration case, producing a local minimum when velocity changes from negative to positive. In option D, however, a horizontal \(x-t\) segment means position is constant. Its slope, and therefore velocity, must be zero throughout that interval. A horizontal \(v-t\) line at a non-zero value instead represents constant non-zero velocity. These two descriptions cannot represent the same motion over the same interval.
207. A particle has a straight-line velocity-time graph from \(t=0\) to \(6\,\text{s}\). Its initial velocity is \(+u\), and its final velocity is \(-\dfrac{u}{2}\). The displacement during the \(6\,\text{s}\) interval is \(+9\,\text{m}\). Which set of values is correct?
ⓐ. \(u=3\,\text{m s}^{-1}\), \(a=-0.75\,\text{m s}^{-2}\), reversal at \(t=4\,\text{s}\), distance \(=7.5\,\text{m}\)
ⓑ. \(u=6\,\text{m s}^{-1}\), \(a=-1.5\,\text{m s}^{-2}\), reversal at \(t=4\,\text{s}\), distance \(=15\,\text{m}\)
ⓒ. \(u=6\,\text{m s}^{-1}\), \(a=-1\,\text{m s}^{-2}\), reversal at \(t=6\,\text{s}\), distance \(=9\,\text{m}\)
ⓓ. \(u=9\,\text{m s}^{-1}\), \(a=-2\,\text{m s}^{-2}\), reversal at \(t=3\,\text{s}\), distance \(=18\,\text{m}\)
Correct Answer: \(u=6\,\text{m s}^{-1}\), \(a=-1.5\,\text{m s}^{-2}\), reversal at \(t=4\,\text{s}\), distance \(=15\,\text{m}\)
Explanation: The displacement equals the signed area under the straight \(v-t\) graph:
\[
9=\frac{u+(-u/2)}{2}(6).
\]
Therefore
\[
9=\frac{u}{4}(6)=\frac{3u}{2},
\]
which gives
\[
u=6\,\text{m s}^{-1}.
\]
The final velocity is therefore
\[
-\frac u2=-3\,\text{m s}^{-1}.
\]
The constant acceleration is
\[
a=\frac{-3-6}{6}=-1.5\,\text{m s}^{-2}.
\]
The velocity reaches zero when
\[
0=6-1.5t,
\]
so
\[
t=4\,\text{s}.
\]
The positive triangular area before reversal is
\[
\frac12(4)(6)=12\,\text{m}.
\]
The negative triangular area after reversal has magnitude
\[
\frac12(2)(3)=3\,\text{m}.
\]
Hence the total distance is
\[
d=12+3=15\,\text{m}.
\]
208. A particle moves with constant acceleration and its position-time graph crosses \(x=0\) at \(t=1\,\text{s}\) and again at \(t=5\,\text{s}\). At \(t=0\), its position is \(+5\,\text{m}\). Which statement correctly describes the motion from \(t=0\) to \(t=6\,\text{s}\)?
ⓐ. \(x=t^2-6t+5\); \(v\) changes from \(-6\) to \(+6\,\text{m s}^{-1}\), reversal occurs at \(t=3\,\text{s}\), and distance \(=18\,\text{m}\).
ⓑ. \(x=-t^2+6t+5\); velocity changes from \(+6\) to \(-6\,\text{m s}^{-1}\), and the total distance is \(10\,\text{m}\)
ⓒ. \(x=t^2-5t+5\); the particle reverses at \(t=2.5\,\text{s}\) and finishes at the origin
ⓓ. \(x=5(t-1)(t-5)\); the particle has acceleration \(10\,\text{m s}^{-2}\) and never reverses
Correct Answer: \(x=t^2-6t+5\); \(v\) changes from \(-6\) to \(+6\,\text{m s}^{-1}\), reversal occurs at \(t=3\,\text{s}\), and distance \(=18\,\text{m}\).
Explanation: Since the position is zero at \(t=1\,\text{s}\) and \(t=5\,\text{s}\), the quadratic position function can be written as
\[
x=k(t-1)(t-5).
\]
At \(t=0\),
\[
5=k(-1)(-5)=5k,
\]
so
\[
k=1.
\]
Therefore
\[
x=(t-1)(t-5)=t^2-6t+5.
\]
The velocity is
\[
v=\frac{dx}{dt}=2t-6,
\]
and the acceleration is
\[
a=2\,\text{m s}^{-2}.
\]
At \(t=0\), \(v=-6\,\text{m s}^{-1}\), while at \(t=6\,\text{s}\), \(v=+6\,\text{m s}^{-1}\). Velocity becomes zero at
\[
t=3\,\text{s},
\]
so the particle reverses there. The turning position is
\[
x(3)=9-18+5=-4\,\text{m}.
\]
Also,
\[
x(6)=36-36+5=5\,\text{m},
\]
so the particle returns to its initial position by \(t=6\,\text{s}\). It travels \(9\,\text{m}\) from \(+5\) to \(-4\,\text{m}\) and another \(9\,\text{m}\) back, giving
\[
d=18\,\text{m}.
\]
209. Consider the following statements about scalar and vector quantities.
Statement I: Speed is a scalar, whereas velocity is a vector.
Statement II: Any physical quantity having a negative numerical value must be a vector.
Statement III: Two vectors having equal magnitudes need not be equal vectors.
Statement IV: The zero vector has zero magnitude, and a unique direction need not be assigned to it.
Which set of statements is correct?
ⓐ. Statement I and Statement II only
ⓑ. Statement I, Statement III and Statement IV only
ⓒ. Statement II, Statement III and Statement IV only
ⓓ. Statement I, Statement II, Statement III and Statement IV
Correct Answer: Statement I, Statement III and Statement IV only
Explanation: A scalar is completely specified by its magnitude together with the appropriate unit and does not require a spatial direction. Speed is therefore a scalar, whereas velocity requires both magnitude and direction and is a vector, so Statement I is correct. A negative algebraic value does not by itself make a quantity a vector. For example, a scalar temperature may be represented by a negative numerical value on a chosen temperature scale, so Statement II is incorrect. Two vectors are equal only when they have both the same magnitude and the same direction. Equal magnitude alone is therefore insufficient, making Statement III correct. A zero vector has magnitude zero. Since it has no non-zero length that singles out a spatial direction, its direction need not be specified uniquely. Thus Statement IV is also correct.
210. Which of the following groups contains only scalar quantities?
ⓐ. Distance, speed, mass, time
ⓑ. Displacement, speed, mass, time
ⓒ. Distance, velocity, acceleration, time
ⓓ. Displacement, velocity, speed, acceleration
Correct Answer: Distance, speed, mass, time
Explanation: Distance specifies the total path length and has magnitude but no spatial direction, so it is a scalar. Speed is also a scalar because it is the magnitude of velocity. Mass and time likewise require no spatial direction and are scalars. Displacement is a vector because it specifies both magnitude and direction from the initial position to the final position. Velocity is a vector because its direction is part of its physical specification. Acceleration is also a vector because it describes the directional rate of change of velocity. Therefore only option A contains scalar quantities exclusively.
211. Three non-zero vectors are described as follows.
Vector \(\vec A\) has magnitude \(5\) units and points due north from point P. Vector \(\vec B\) also has magnitude \(5\) units and points due north from a different point Q. Vector \(\vec C\) has magnitude \(5\) units and points due south.
Which statement is correct?
ⓐ. \(\vec A\ne\vec B\) because their tails are at different points, and \(\vec C=\vec A\)
ⓑ. \(\vec A=-\vec B\), and \(\vec C=\vec B\)
ⓒ. \(\vec A=\vec B\), and \(\vec C=-\vec A\)
ⓓ. All three vectors are equal because their magnitudes are equal
Correct Answer: \(\vec A=\vec B\), and \(\vec C=-\vec A\)
Explanation: Equality of free vectors depends on magnitude and direction, not on the location at which the arrows are drawn. Vectors \(\vec A\) and \(\vec B\) both have magnitude \(5\) units and point due north. Therefore
\[
\vec A=\vec B.
\]
Vector \(\vec C\) has the same magnitude but points in the opposite direction. The negative of a vector has equal magnitude and exactly opposite direction. Hence
\[
\vec C=-\vec A.
\]
The different drawing positions of \(\vec A\) and \(\vec B\) do not alter their equality. Equal magnitude alone, however, is not enough to make \(\vec C\) equal to them.
212. Let \(\vec A\ne\vec 0\). Two vectors are defined by
\[
\vec B=-3\vec A,\qquad \vec C=-\frac12\vec B.
\]
Which statement is correct?
ⓐ. \(|\vec B|=-3A\), while \(\vec C\) is opposite to \(\vec A\)
ⓑ. \(|\vec B|=3A\) opposite to \(\vec A\); \(|\vec C|=\dfrac{3A}{2}\) along \(\vec A\)
ⓒ. \(\vec C=\dfrac12\vec A\)
ⓓ. \(\vec C\) is the zero vector because two negative scalars are involved
Correct Answer: \(|\vec B|=3A\) opposite to \(\vec A\); \(|\vec C|=\dfrac{3A}{2}\) along \(\vec A\)
Explanation: Multiplication of a vector by a real number \(\lambda\) changes its magnitude by the factor \(|\lambda|\). A negative multiplier also reverses the direction. Therefore
\[
\vec B=-3\vec A
\]
has magnitude
\[
|\vec B|=3A
\]
and points opposite to \(\vec A\). Now
\[
\vec C=-\frac12(-3\vec A)=\frac32\vec A.
\]
The two negative factors produce a positive multiplier. Thus \(\vec C\) points in the same direction as \(\vec A\) and has magnitude \(3A/2\). A vector magnitude itself is non-negative, so writing \(|\vec B|=-3A\) would be incorrect.
213. A non-zero vector \(\vec P\) is multiplied by a real number \(\lambda\) to obtain \(\vec Q\). It is observed that \(\vec Q\) has twice the magnitude of \(\vec P\) and points exactly opposite to it. A third vector is defined by \(\vec R=-\dfrac12\vec Q\). Which result is correct?
ⓐ. \(\lambda=+2\) and \(\vec R=-\vec P\)
ⓑ. \(\lambda=-\dfrac12\) and \(\vec R=2\vec P\)
ⓒ. \(\lambda=-2\) and \(\vec R=\vec P\)
ⓓ. \(\lambda=+2\) and \(\vec R=\vec P\)
Correct Answer: \(\lambda=-2\) and \(\vec R=\vec P\)
Explanation: Since the magnitude of \(\vec Q\) is twice that of \(\vec P\),
\[
|\lambda|=2.
\]
The direction of \(\vec Q\) is opposite to \(\vec P\), so the multiplier must be negative. Therefore
\[
\lambda=-2
\]
and
\[
\vec Q=-2\vec P.
\]
Now apply the second scaling:
\[
\vec R=-\frac12\vec Q.
\]
Substituting \(\vec Q=-2\vec P\),
\[
\vec R=-\frac12(-2\vec P)=\vec P.
\]
Thus the two successive direction reversals restore the original direction and magnitude.
214. Two displacement vectors are added head-to-tail. The first has magnitude \(3\,\text{m}\) toward east and the second has magnitude \(4\,\text{m}\) toward north.
Starting from O, draw a \(3\,\text{m}\) arrow due east to P. From P, draw a \(4\,\text{m}\) arrow due north to Q. The resultant is the vector from O directly to Q.
What are the magnitude and direction of the resultant?
ⓐ. \(7\,\text{m}\), \(45^\circ\) north of east
ⓑ. \(5\,\text{m}\), with \(\tan\theta=\dfrac{3}{4}\) measured north of east
ⓒ. \(1\,\text{m}\), due north
ⓓ. \(5\,\text{m}\), with \(\tan\theta=\dfrac{4}{3}\) measured north of east
Correct Answer: \(5\,\text{m}\), with \(\tan\theta=\dfrac{4}{3}\) measured north of east
Explanation: The eastward and northward vectors are perpendicular. Their resultant therefore forms the hypotenuse of a right triangle with sides \(3\,\text{m}\) and \(4\,\text{m}\). Hence
\[
R=\sqrt{3^2+4^2}=5\,\text{m}.
\]
If \(\theta\) is measured from east toward north, the eastward component is adjacent to \(\theta\) and the northward component is opposite. Therefore
\[
\tan\theta=\frac{4}{3}.
\]
The magnitudes \(3\) and \(4\) cannot simply be added to give \(7\,\text{m}\) because their directions differ. The head-to-tail construction correctly gives the vector from the initial point O to the final point Q.
215. Two vectors have magnitudes \(7\) units and \(11\) units. Which set contains only possible magnitudes of their resultant?
ⓐ. \(2,\ 10,\ 20\)
ⓑ. \(4,\ 10,\ 18\)
ⓒ. \(3,\ 18,\ 19\)
ⓓ. \(0,\ 7,\ 18\)
Correct Answer: \(4,\ 10,\ 18\)
Explanation: For two vectors of fixed magnitudes \(A\) and \(B\), the resultant magnitude must satisfy
\[
|A-B|\le R\le A+B.
\]
Here
\[
|11-7|=4
\]
and
\[
11+7=18.
\]
Therefore every possible resultant must lie between \(4\) and \(18\) units inclusive. The lower limit \(4\) occurs when the vectors are antiparallel. The upper limit \(18\) occurs when they are parallel in the same direction. Intermediate values such as \(10\) can occur for suitable included angles. Values below \(4\) or above \(18\) are impossible for these two vector magnitudes.
216. Two vectors \(\vec A\) and \(\vec B\) have equal magnitude \(A\). The magnitude of their resultant \(\vec A+\vec B\) is also \(A\). What is the angle between \(\vec A\) and \(\vec B\)?
ⓐ. \(60^\circ\)
ⓑ. \(90^\circ\)
ⓒ. \(120^\circ\)
ⓓ. \(180^\circ\)
Correct Answer: \(120^\circ\)
Explanation: For two vectors of equal magnitude \(A\) with included angle \(\theta\),
\[
R^2=A^2+A^2+2A^2\cos\theta.
\]
The resultant magnitude is given as
\[
R=A.
\]
Therefore
\[
A^2=2A^2(1+\cos\theta).
\]
Dividing by \(A^2\),
\[
1=2+2\cos\theta.
\]
Hence
\[
\cos\theta=-\frac12.
\]
Therefore
\[
\theta=120^\circ.
\]
Because the vectors have equal magnitudes, their resultant lies along the internal angle bisector, making an angle \(60^\circ\) with each vector.
217. Two vectors have magnitudes \(6\) units and \(8\) units. Their resultant initially has magnitude \(10\) units. The \(6\)-unit vector is kept fixed, while the \(8\)-unit vector is rotated so that the angle between them becomes \(60^\circ\). What is the new resultant magnitude?
ⓐ. \(10\) units
ⓑ. \(14\) units
ⓒ. \(2\sqrt{37}\) units
ⓓ. \(2\sqrt{13}\) units
Correct Answer: \(2\sqrt{37}\) units
Explanation: The initial resultant satisfies
\[
10^2=6^2+8^2+2(6)(8)\cos\theta.
\]
Since
\[
100=36+64,
\]
the cosine term must initially be zero. Thus the original vectors are perpendicular. After rotation, the included angle is \(60^\circ\). The new resultant \(R'\) satisfies
\[
R'^2=6^2+8^2+2(6)(8)\cos60^\circ.
\]
Substituting,
\[
R'^2=36+64+96\left(\frac12\right)=148.
\]
Hence
\[
R'=\sqrt{148}=2\sqrt{37}.
\]
Reducing the included angle from \(90^\circ\) to \(60^\circ\) increases the resultant because the two vectors become more nearly aligned.
218. Two vectors each have magnitude \(10\) units. Vector \(\vec A\) points along the positive \(x\)-direction, while vector \(\vec B\) makes an angle \(120^\circ\) counterclockwise from \(\vec A\). Which statement about \(\vec A+\vec B\) is correct?
ⓐ. Magnitude \(20\) units directed at \(60^\circ\) from \(+x\)
ⓑ. Magnitude \(10\) units directed at \(60^\circ\) from \(+x\)
ⓒ. Magnitude \(10\) units directed at \(120^\circ\) from \(+x\)
ⓓ. Magnitude \(5\sqrt3\) units directed along \(+y\)
Correct Answer: Magnitude \(10\) units directed at \(60^\circ\) from \(+x\)
Explanation: Resolve the two vectors into components. Vector \(\vec A\) has components
\[
(10,0).
\]
For \(\vec B\),
\[
B_x=10\cos120^\circ=-5
\]
and
\[
B_y=10\sin120^\circ=5\sqrt3.
\]
Thus the resultant components are
\[
R_x=10-5=5,
\]
\[
R_y=5\sqrt3.
\]
Its magnitude is
\[
R=\sqrt{5^2+(5\sqrt3)^2}
=\sqrt{25+75}=10.
\]
Its direction satisfies
\[
\tan\alpha=\frac{5\sqrt3}{5}=\sqrt3,
\]
so \(\alpha=60^\circ\). The result also agrees with the fact that two equal vectors have their resultant along the angle bisector.
219. Three vectors in a plane satisfy
\[
\vec A+\vec B+\vec C=\vec0.
\]
Their Cartesian components are
\[
\vec A=(4,3),\qquad \vec B=(-2,5).
\]
Which result for \(\vec C\) is correct?
ⓐ. \(\vec C=(2,8)\) with magnitude \(2\sqrt{17}\)
ⓑ. \(\vec C=(-6,2)\) with magnitude \(2\sqrt{10}\)
ⓒ. \(\vec C=(-2,-8)\) with magnitude \(2\sqrt{17}\)
ⓓ. \(\vec C=(2,-8)\) with magnitude \(2\sqrt{17}\)
Correct Answer: \(\vec C=(-2,-8)\) with magnitude \(2\sqrt{17}\)
Explanation: The condition
\[
\vec A+\vec B+\vec C=\vec0
\]
means that
\[
\vec C=-(\vec A+\vec B).
\]
First add the \(x\)-components:
\[
A_x+B_x=4+(-2)=2.
\]
Then add the \(y\)-components:
\[
A_y+B_y=3+5=8.
\]
Therefore
\[
\vec A+\vec B=(2,8),
\]
so
\[
\vec C=(-2,-8).
\]
Its magnitude is
\[
|\vec C|=\sqrt{(-2)^2+(-8)^2}
=\sqrt{68}=2\sqrt{17}.
\]
The negative signs indicate that \(\vec C\) must point into the quadrant opposite to the resultant of \(\vec A\) and \(\vec B\) so that the vector sum closes to zero.
220. A vector of magnitude \(10\) units makes an angle \(37^\circ\) above the positive \(x\)-axis, where \(\cos37^\circ=\dfrac45\) and \(\sin37^\circ=\dfrac35\). A second vector of magnitude \(5\) units points along the negative \(x\)-axis. What is the magnitude of their resultant?
ⓐ. \(5\sqrt5\) units
ⓑ. \(3\sqrt5\) units
ⓒ. \(11\) units
ⓓ. \(\sqrt{61}\) units
Correct Answer: \(3\sqrt5\) units
Explanation: The first vector has components
\[
A_x=10\cos37^\circ=8
\]
and
\[
A_y=10\sin37^\circ=6.
\]
The second vector points along negative \(x\), so its components are
\[
B_x=-5,\qquad B_y=0.
\]
Therefore the resultant components are
\[
R_x=8-5=3,
\]
\[
R_y=6.
\]
Its magnitude is
\[
R=\sqrt{3^2+6^2}
=\sqrt{45}
=3\sqrt5.
\]
The resultant is not obtained by subtracting the two vector magnitudes because only their \(x\)-components oppose each other; the \(y\)-component of the first vector remains unchanged.
