101. A particle has position
\[
x=2t^2-8t+3,
\]
where SI units are used. Observer S uses this coordinate system. Observer S' keeps the same origin but reverses the positive direction. Compare the average velocity from \(t=1\,\text{s}\) to \(t=5\,\text{s}\) with the instantaneous velocity at \(t=3\,\text{s}\) in both systems.
ⓐ. Reversing the axis changes the instantaneous velocity sign but not the average velocity sign.
ⓑ. In S both are \(-4\,\text{m s}^{-1}\); in S' both are \(+4\,\text{m s}^{-1}\).
ⓒ. In S the average velocity is \(+4\,\text{m s}^{-1}\) but the instantaneous velocity is zero.
ⓓ. In S both are \(+4\,\text{m s}^{-1}\); in S' both are \(-4\,\text{m s}^{-1}\).
Correct Answer: In S both are \(+4\,\text{m s}^{-1}\); in S' both are \(-4\,\text{m s}^{-1}\).
Explanation: \( \textbf{Step 1 — Average velocity in S:} \)
\[
x(1)=2-8+3=-3\,\text{m},
\]
while
\[
x(5)=2(25)-40+3=13\,\text{m}.
\]
Hence
\[
v_{\rm avg}
=\frac{13-(-3)}{5-1}
=\frac{16}{4}
=4\,\text{m s}^{-1}.
\]
\( \textbf{Step 2 — Instantaneous velocity at the midpoint:} \)
\[
v=\frac{dx}{dt}=4t-8.
\]
At \(t=3\,\text{s}\),
\[
v=12-8=4\,\text{m s}^{-1}.
\]
Thus the two quantities happen to be equal in this quadratic motion over the chosen symmetric interval. \( \textbf{Step 3 — Reverse the coordinate axis:} \) Under axis reversal,
\[
X=-x.
\]
Therefore both signed displacement and instantaneous velocity reverse sign. Hence S' measures
\[
V_{\rm avg}=-4\,\text{m s}^{-1}
\]
and
\[
V(3)=-4\,\text{m s}^{-1}.
\]
The equality is preserved even though both algebraic signs change.
102. A particle is claimed to undergo constant acceleration. Its measured positions at \(t=0,1,2,3\,\text{s}\) are \(1,4,9,16\,\text{m}\), respectively. If the claim is correct and the same motion continues, what are its position at \(t=4\,\text{s}\) and its acceleration?
ⓐ. \(20\,\text{m}\) and \(1\,\text{m s}^{-2}\)
ⓑ. \(25\,\text{m}\) and \(2\,\text{m s}^{-2}\)
ⓒ. \(25\,\text{m}\) and \(4\,\text{m s}^{-2}\)
ⓓ. \(24\,\text{m}\) and \(2\,\text{m s}^{-2}\)
Correct Answer: \(25\,\text{m}\) and \(2\,\text{m s}^{-2}\)
Explanation: The successive one-second displacements are
\[
4-1=3\,\text{m},
\]
\[
9-4=5\,\text{m},
\]
and
\[
16-9=7\,\text{m}.
\]
These displacements increase by equal amounts:
\[
5-3=2,\qquad 7-5=2.
\]
For equal time intervals under constant acceleration, successive displacements form a sequence with constant second difference. The next one-second displacement must therefore be
\[
7+2=9\,\text{m}.
\]
Hence
\[
x(4)=16+9=25\,\text{m}.
\]
The position data are recognised as
\[
x=(t+1)^2=t^2+2t+1.
\]
Differentiating,
\[
v=2t+2,
\]
and
\[
a=\frac{dv}{dt}=2\,\text{m s}^{-2}.
\]
Thus both the data pattern and the functional form confirm constant acceleration of \(2\,\text{m s}^{-2}\).
103. The velocity of a particle is measured at equal one-second intervals:
| \(t\,(\text{s})\) | \(v\,(\text{m s}^{-1})\) |
|---|
| 0 | 2 |
| 1 | 6 |
| 2 | 10 |
| 3 | 15 |
Exactly one velocity reading is known to contain an error. If the actual motion has constant acceleration and the first three readings are correct, what should the last reading be, and what is the acceleration?
ⓐ. \(14\,\text{m s}^{-1}\) and \(2\,\text{m s}^{-2}\)
ⓑ. \(14\,\text{m s}^{-1}\) and \(4\,\text{m s}^{-2}\)
ⓒ. \(16\,\text{m s}^{-1}\) and \(4\,\text{m s}^{-2}\)
ⓓ. \(15\,\text{m s}^{-1}\) and \(5\,\text{m s}^{-2}\)
Correct Answer: \(14\,\text{m s}^{-1}\) and \(4\,\text{m s}^{-2}\)
Explanation: With constant acceleration, velocity must change by equal amounts in equal time intervals. From \(t=0\) to \(1\,\text{s}\),
\[
\Delta v=6-2=4\,\text{m s}^{-1}.
\]
From \(t=1\) to \(2\,\text{s}\),
\[
\Delta v=10-6=4\,\text{m s}^{-1}.
\]
The first three readings therefore establish the constant acceleration as
\[
a=\frac{4}{1}=4\,\text{m s}^{-2}.
\]
During the next one-second interval the velocity must again increase by \(4\,\text{m s}^{-1}\). Hence the correct value at \(t=3\,\text{s}\) is
\[
v=10+4=14\,\text{m s}^{-1}.
\]
The listed value \(15\,\text{m s}^{-1}\) is incompatible with the stated constant-acceleration motion. This is a compatibility test: the constant-acceleration condition constrains the complete velocity sequence, not merely its general increasing trend.
104. In coordinate system S, a particle moves according to
\[
x=5-4t+t^2.
\]
A second observer chooses the old point \(x=7\,\text{m}\) as origin, reverses the positive direction, and sets \(\tau=0\) at old time \(t=2\,\text{s}\). Which statement correctly describes the motion in the new coordinate \(X\)?
ⓐ. \(X=6-\tau^2\); at \(\tau=1\,\text{s}\), \(V=-2\,\text{m s}^{-1}\) and \(A=-2\,\text{m s}^{-2}\), so the particle is speeding up.
ⓑ. \(X=6+\tau^2\); at \(\tau=1\,\text{s}\), \(V=+2\,\text{m s}^{-1}\) and \(A=+2\,\text{m s}^{-2}\), so the particle is speeding up.
ⓒ. \(X=\tau^2-6\); at \(\tau=1\,\text{s}\), \(V=+2\,\text{m s}^{-1}\) and \(A=-2\,\text{m s}^{-2}\), so the particle is slowing down.
ⓓ. \(X=6-\tau^2\); at \(\tau=1\,\text{s}\), \(V=-2\,\text{m s}^{-1}\) and \(A=+2\,\text{m s}^{-2}\), so the particle is slowing down.
Correct Answer: \(X=6-\tau^2\); at \(\tau=1\,\text{s}\), \(V=-2\,\text{m s}^{-1}\) and \(A=-2\,\text{m s}^{-2}\), so the particle is speeding up.
Explanation: The time transformation is
\[
t=\tau+2.
\]
Substitute this into the original position:
\[
x=5-4(\tau+2)+(\tau+2)^2.
\]
Expanding,
\[
x=5-4\tau-8+\tau^2+4\tau+4,
\]
so
\[
x=1+\tau^2.
\]
The new origin is old \(x=7\,\text{m}\), and the new positive direction is reversed. Therefore
\[
X=7-x.
\]
Hence
\[
X=7-(1+\tau^2)=6-\tau^2.
\]
The new velocity is
\[
V=\frac{dX}{d\tau}=-2\tau.
\]
At \(\tau=1\,\text{s}\),
\[
V=-2\,\text{m s}^{-1}.
\]
The acceleration is
\[
A=\frac{dV}{d\tau}=-2\,\text{m s}^{-2}.
\]
Velocity and acceleration have the same negative direction, so the speed is increasing. The coordinate transformation changes algebraic signs but preserves the physical classification as constant acceleration.
105. A particle has a quadratic position function
\[
x=A+Bt+Ct^2.
\]
Its instantaneous velocity is \(-4\,\text{m s}^{-1}\) at \(t=1\,\text{s}\) and \(+4\,\text{m s}^{-1}\) at \(t=3\,\text{s}\). Also, \(x=5\,\text{m}\) at \(t=2\,\text{s}\). What are the distance travelled and displacement from \(t=0\) to \(t=4\,\text{s}\)?
ⓐ. Distance \(=16\,\text{m}\), displacement \(=0\)
ⓑ. Distance \(=8\,\text{m}\), displacement \(=0\)
ⓒ. Distance \(=16\,\text{m}\), displacement \(=+8\,\text{m}\)
ⓓ. Distance \(=8\,\text{m}\), displacement \(=-8\,\text{m}\)
Correct Answer: Distance \(=16\,\text{m}\), displacement \(=0\)
Explanation: The velocity is
\[
v=\frac{dx}{dt}=B+2Ct.
\]
At \(t=1\,\text{s}\),
\[
B+2C=-4.
\]
At \(t=3\,\text{s}\),
\[
B+6C=+4.
\]
Subtracting,
\[
4C=8,
\]
so
\[
C=2.
\]
Then
\[
B+4=-4,
\]
giving
\[
B=-8.
\]
Using \(x(2)=5\,\text{m}\),
\[
5=A-16+8,
\]
so
\[
A=13.
\]
Therefore
\[
x=13-8t+2t^2.
\]
The velocity is
\[
v=-8+4t.
\]
Thus \(v=0\) at
\[
t=2\,\text{s},
\]
and the velocity changes from negative to positive there, so a reversal occurs. The relevant positions are
\[
x(0)=13\,\text{m},\qquad x(2)=5\,\text{m},\qquad x(4)=13\,\text{m}.
\]
Hence the distance is
\[
|5-13|+|13-5|=8+8=16\,\text{m}.
\]
The displacement is
\[
13-13=0.
\]
The zero displacement therefore results from a symmetric out-and-back motion under constant acceleration, not from absence of motion.
106. A particle's velocity changes uniformly from \(-12\,\text{m s}^{-1}\) at \(t=0\) to \(+8\,\text{m s}^{-1}\) at \(t=10\,\text{s}\). Which statement is correct?
ⓐ. It slows down for \(4\,\text{s}\), remains at rest for \(2\,\text{s}\), and speeds up for \(4\,\text{s}\).
ⓑ. It speeds up for \(6\,\text{s}\) and slows down for \(4\,\text{s}\).
ⓒ. It slows down for \(6\,\text{s}\), reverses direction instantaneously, and speeds up for the next \(4\,\text{s}\).
ⓓ. Its acceleration changes sign when its velocity becomes zero.
Correct Answer: It slows down for \(6\,\text{s}\), reverses direction instantaneously, and speeds up for the next \(4\,\text{s}\).
Explanation: Uniform change of velocity means constant acceleration:
\[
a=\frac{8-(-12)}{10}
=\frac{20}{10}
=2\,\text{m s}^{-2}.
\]
The velocity is therefore
\[
v=-12+2t.
\]
The particle reaches zero velocity when
\[
0=-12+2t,
\]
so
\[
t=6\,\text{s}.
\]
For the first \(6\,\text{s}\), velocity is negative while acceleration is positive. The two directions oppose each other, so the speed decreases. At \(t=6\,\text{s}\), the velocity is zero only instantaneously. Since the acceleration remains positive, the velocity becomes positive immediately afterward. From \(6\) to \(10\,\text{s}\), velocity and acceleration have the same positive direction, so the speed increases. Thus the particle slows for \(6\,\text{s}\), reverses, and then speeds up for \(4\,\text{s}\).
107. Three particles have the following velocity and constant acceleration at \(t=0\):
| Particle | \(v_0\,(\text{m s}^{-1})\) | \(a\,(\text{m s}^{-2})\) |
|---|
| P | \(-6\) | \(+2\) |
| Q | \(+6\) | \(+2\) |
| R | \(-6\) | \(-2\) |
Which description is correct for the first \(4\,\text{s}\)?
ⓐ. All three have equal speeds at \(t=4\,\text{s}\) because their acceleration magnitudes are equal.
ⓑ. P slows down without reversing; Q slows down; R reverses after \(3\,\text{s}\).
ⓒ. P and R both reverse after \(3\,\text{s}\), while Q remains at constant speed.
ⓓ. P reverses after \(3\,\text{s}\); Q speeds up throughout; R speeds up throughout.
Correct Answer: P reverses after \(3\,\text{s}\); Q speeds up throughout; R speeds up throughout.
Explanation: For P,
\[
v_P=-6+2t.
\]
It reaches zero at
\[
t=3\,\text{s}.
\]
Before that instant, velocity is negative and acceleration positive, so P slows down. Afterward both are positive, so P moves in the positive direction and speeds up. For Q,
\[
v_Q=6+2t,
\]
which remains positive while the acceleration is also positive. Therefore Q speeds up throughout the \(4\,\text{s}\) interval. For R,
\[
v_R=-6-2t.
\]
Both velocity and acceleration are negative at all stated times, so the magnitude of velocity increases continuously. R therefore speeds up in the negative direction and never approaches zero velocity. Equal acceleration magnitudes do not determine identical speed trends; the relative signs of velocity and acceleration are essential.
108. The position-time graph of a particle is described below.
The \(x-t\) graph is a smooth parabola corresponding to constant acceleration. At \(t=1\,\text{s}\), the tangent slope is \(-6\,\text{m s}^{-1}\). At \(t=5\,\text{s}\), the tangent slope is \(+6\,\text{m s}^{-1}\).
Which conclusion follows?
ⓐ. \(a=+3\,\text{m s}^{-2}\), \(v=0\) at \(t=3\,\text{s}\), and the particle reverses there.
ⓑ. The acceleration is \(+6\,\text{m s}^{-2}\), and the particle reverses at \(t=2\,\text{s}\).
ⓒ. The average acceleration is zero because the two velocities have equal magnitudes.
ⓓ. The particle cannot reverse because the graph is a parabola.
Correct Answer: \(a=+3\,\text{m s}^{-2}\), \(v=0\) at \(t=3\,\text{s}\), and the particle reverses there.
Explanation: The tangent slope of an \(x-t\) graph gives instantaneous velocity. Therefore
\[
v_1=-6\,\text{m s}^{-1}
\]
at \(t_1=1\,\text{s}\), and
\[
v_2=+6\,\text{m s}^{-1}
\]
at \(t_2=5\,\text{s}\). Because the acceleration is constant,
\[
a=\frac{v_2-v_1}{t_2-t_1}.
\]
Thus
\[
a=\frac{6-(-6)}{5-1}
=\frac{12}{4}
=3\,\text{m s}^{-2}.
\]
Velocity therefore changes linearly with time. It must reach zero halfway in velocity change between the two stated values, which occurs \(2\,\text{s}\) after \(t=1\,\text{s}\):
\[
t=3\,\text{s}.
\]
Before this instant the velocity is negative, and afterward it is positive. Hence the particle reverses direction at \(t=3\,\text{s}\). Equal speed magnitudes at two times do not imply zero average acceleration when the velocity directions are opposite.
109. Consider the following Assertion and Reason.
Assertion: A particle undergoing constant non-zero acceleration can pass smoothly through an instant of zero velocity and continue moving afterward.
Reason: Zero instantaneous velocity does not require zero acceleration, and if the acceleration remains non-zero the velocity can change sign across that instant.
ⓐ. 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: Constant non-zero acceleration means that velocity continues to change at a fixed rate throughout the interval. A particle may therefore reach
\[
v=0
\]
at one particular instant without its acceleration vanishing. If the acceleration persists, then immediately after that instant the velocity acquires the sign determined by the continuing acceleration. When the velocity sign differs on the two sides of the zero-velocity instant, the particle reverses direction. There is no requirement that the motion terminate at the stopping instant. Treating \(v=0\) as permanent rest would incorrectly ignore the non-zero acceleration. Thus the Assertion is true. The Reason identifies the precise kinematic mechanism and correctly explains the Assertion.
110. A particle initially has negative velocity and a constant positive acceleration large enough that the particle eventually moves with positive velocity. Which sequence correctly describes the motion?
ⓐ. Speed increases \(\rightarrow\) velocity becomes zero \(\rightarrow\) speed decreases \(\rightarrow\) reversal
ⓑ. Speed remains constant \(\rightarrow\) velocity changes sign \(\rightarrow\) speed increases
ⓒ. Velocity becomes zero \(\rightarrow\) speed decreases \(\rightarrow\) reversal \(\rightarrow\) acceleration changes sign
ⓓ. Speed decreases \(\rightarrow\) velocity becomes zero \(\rightarrow\) reversal \(\rightarrow\) speed increases
Correct Answer: Speed decreases \(\rightarrow\) velocity becomes zero \(\rightarrow\) reversal \(\rightarrow\) speed increases
Explanation: Initially,
\[
v\lt 0,\qquad a\gt 0.
\]
Velocity and acceleration are therefore oppositely directed, so the magnitude \(|v|\) decreases. The particle slows down while continuing in the negative direction. Eventually the increasing algebraic velocity reaches
\[
v=0.
\]
This is the instantaneous-rest point. Because the acceleration remains positive, the velocity then becomes positive. The change of velocity sign marks the reversal of direction. After the reversal,
\[
v\gt 0,\qquad a\gt 0,
\]
so velocity and acceleration are in the same direction and the speed increases. The acceleration need not change sign anywhere in this sequence. Thus the natural temporal order is slowing down, momentary zero velocity, reversal, and then speeding up.
111. Two particles P and Q have their positions recorded at equal one-second intervals.
| \(t\,(\text{s})\) | \(x_P\,(\text{m})\) | \(x_Q\,(\text{m})\) |
|---|
| 0 | 0 | 0 |
| 1 | 2 | 3 |
| 2 | 8 | 6 |
| 3 | 18 | 9 |
Assume each motion is described exactly by the simplest polynomial consistent with all the data. Which statement is correct?
ⓐ. P has constant velocity \(6\,\text{m s}^{-1}\), while Q has constant acceleration \(3\,\text{m s}^{-2}\).
ⓑ. Both have constant acceleration, but P's acceleration is twice Q's.
ⓒ. P has constant acceleration \(4\,\text{m s}^{-2}\), while Q has constant velocity \(3\,\text{m s}^{-1}\).
ⓓ. Both have constant velocity because their positions increase at every observation.
Correct Answer: P has constant acceleration \(4\,\text{m s}^{-2}\), while Q has constant velocity \(3\,\text{m s}^{-1}\).
Explanation: For P, the successive one-second displacements are
\[
2,\quad 6,\quad 10\,\text{m}.
\]
The first differences are not constant, so P does not have constant velocity. Their successive increases are
\[
6-2=4,\qquad 10-6=4.
\]
A constant second difference for equal time intervals is the signature of a quadratic \(x(t)\), hence constant acceleration. In fact the data fit
\[
x_P=2t^2,
\]
so
\[
v_P=4t
\]
and
\[
a_P=4\,\text{m s}^{-2}.
\]
For Q, each one-second displacement is
\[
3\,\text{m}.
\]
Thus the velocity is constant:
\[
v_Q=3\,\text{m s}^{-1}.
\]
Its acceleration is therefore zero. Merely observing that both positions increase is insufficient to classify the acceleration; the pattern of change with equal time intervals is decisive.
112. Consider the following statements about uniformly accelerated one-dimensional motion.
Statement I: Constant acceleration requires equal changes in velocity during equal time intervals.
Statement II: For constant non-zero acceleration, velocity is a linear function of time and position is a quadratic function of time.
Statement III: If a particle has zero velocity at one instant during constant acceleration, its acceleration must also be zero at that instant.
Statement IV: Reversing the coordinate axis reverses the algebraic signs of both velocity and acceleration but does not change whether the physical motion has constant acceleration.
Which set of statements is correct?
ⓐ. Statement I, Statement II and Statement IV only
ⓑ. Statement I and Statement II only
ⓒ. Statement II, Statement III and Statement IV only
ⓓ. Statement I, Statement II, Statement III and Statement IV
Correct Answer: Statement I, Statement II and Statement IV only
Explanation: Statement I is correct because constant acceleration means
\[
a=\frac{\Delta v}{\Delta t}
\]
has the same value over equal intervals, so equal time intervals produce equal velocity changes. Statement II is also correct. Integrating the constant rate of velocity change gives a velocity that varies linearly with time, while the corresponding position dependence is quadratic for non-zero constant acceleration. Statement III is incorrect. A particle can be momentarily at rest while acceleration remains non-zero; a turning point under constant acceleration is a standard example. Statement IV is correct because reversing the spatial axis changes every one-dimensional vector component sign. Thus \(v\) and \(a\) both reverse algebraic sign, but an acceleration that was constant remains constant in magnitude and direction relative to the newly defined axis. Coordinate choice changes representation, not the physical classification of the motion. Hence Statements I, II and IV are correct.
113. The position-time graph of a particle contains the point \((t,x)=(3\,\text{s},-5\,\text{m})\). Which conclusion follows from this point alone?
ⓐ. At \(t=3\,\text{s}\), the particle is moving with velocity \(-5\,\text{m s}^{-1}\).
ⓑ. At \(t=3\,\text{s}\), the particle must be moving in the negative direction.
ⓒ. At \(t=3\,\text{s}\), the particle has travelled a distance of \(5\,\text{m}\).
ⓓ. At \(t=3\,\text{s}\), \(x=-5\,\text{m}\); velocity cannot be inferred from that single graph point.
Correct Answer: At \(t=3\,\text{s}\), \(x=-5\,\text{m}\); velocity cannot be inferred from that single graph point.
Explanation: On a position-time graph, the vertical coordinate directly represents the particle's position at the corresponding time. Therefore the plotted point tells us that
\[
x(3\,\text{s})=-5\,\text{m}.
\]
The negative sign only places the particle on the negative side of the chosen origin. Velocity is obtained from the slope of the position-time graph, not from the value of the position coordinate itself. A single isolated point does not provide a slope because the behaviour of the graph around that point is unknown. The particle could be moving toward positive \(x\), moving toward negative \(x\), or even be instantaneously at rest while located at \(-5\,\text{m}\). The distance already travelled also cannot be obtained from one position coordinate unless the previous path is known. Thus the graph ordinate gives position, while motion rate requires slope information.
114. The position-time graph of a particle is described below.
The horizontal axis is time \(t\) in seconds and the vertical axis is position \(x\) in metres. A single straight line joins the points \((1,6)\) and \((5,-2)\).
Between \(t=1\,\text{s}\) and \(t=5\,\text{s}\), the displacement and velocity are respectively
ⓐ. \(-8\,\text{m}\) and \(+2\,\text{m s}^{-1}\)
ⓑ. \(-8\,\text{m}\) and \(-2\,\text{m s}^{-1}\)
ⓒ. \(+8\,\text{m}\) and \(-2\,\text{m s}^{-1}\)
ⓓ. \(-2\,\text{m}\) and \(-8\,\text{m s}^{-1}\)
Correct Answer: \(-8\,\text{m}\) and \(-2\,\text{m s}^{-1}\)
Explanation: The particle's initial position at \(t=1\,\text{s}\) is
\[
x_i=6\,\text{m},
\]
while at \(t=5\,\text{s}\),
\[
x_f=-2\,\text{m}.
\]
Therefore the displacement is
\[
\Delta x=x_f-x_i=-2-6=-8\,\text{m}.
\]
The graph is a straight line, so its slope is constant and equals the particle's constant velocity:
\[
v=\frac{\Delta x}{\Delta t}
=\frac{-8}{5-1}
=-2\,\text{m s}^{-1}.
\]
The negative slope means that the position decreases uniformly with time. Passing from positive to negative position during the interval means that the particle crosses the chosen origin, but this does not imply a direction reversal. Since the slope remains negative throughout, the particle continues moving in the negative direction with constant velocity.
115. The position-time graph of a particle is described below.
From \(t=0\) to \(t=2\,\text{s}\), the graph is horizontal at \(x=4\,\text{m}\). From \(t=2\,\text{s}\) to \(t=5\,\text{s}\), it is a straight line joining \((2,4)\) to \((5,10)\).
Which statement is correct?
ⓐ. The particle is at rest throughout because both graph segments are straight.
ⓑ. The particle moves at \(+2\,\text{m s}^{-1}\) for the first \(2\,\text{s}\), then remains at rest.
ⓒ. Rest from \(0\) to \(2\,\text{s}\); then \(v=+2\,\text{m s}^{-1}\); total displacement \(=+6\,\text{m}\).
ⓓ. The particle reverses direction at \(t=2\,\text{s}\) because the graph changes its slope.
Correct Answer: Rest from \(0\) to \(2\,\text{s}\); then \(v=+2\,\text{m s}^{-1}\); total displacement \(=+6\,\text{m}\).
Explanation: A horizontal segment on a position-time graph has zero slope. Therefore from \(t=0\) to \(2\,\text{s}\),
\[
v=0,
\]
and the particle remains at \(x=4\,\text{m}\). For the second segment, the slope is
\[
v=\frac{10-4}{5-2}
=\frac{6}{3}
=+2\,\text{m s}^{-1}.
\]
Thus the particle begins moving in the positive direction after \(t=2\,\text{s}\). Its complete displacement is obtained from the endpoint positions:
\[
\Delta x=10-4=+6\,\text{m}.
\]
A change from zero slope to positive slope is a transition from rest to motion, not a direction reversal. Direction reversal requires velocity to change from one sign to the opposite sign.
116. The position-time graph of a particle consists of straight segments joining the points \((0,0)\), \((2,6)\), \((5,0)\) and \((7,-8)\), with SI units. Which statement is correct?
ⓐ. The speed order is \(v_3\gt v_1\gt v_2\), and the only direction reversal occurs at \(t=2\,\text{s}\).
ⓑ. The speeds are equal in all three intervals because each graph segment is straight.
ⓒ. The speed order is \(v_3\gt v_1\gt v_2\), and the only direction reversal occurs at \(t=5\,\text{s}\).
ⓓ. The speeds in the three intervals are in the order \(v_1\gt v_2\gt v_3\), and reversals occur at both \(t=2\,\text{s}\) and \(t=5\,\text{s}\).
Correct Answer: The speed order is \(v_3\gt v_1\gt v_2\), and the only direction reversal occurs at \(t=2\,\text{s}\).
Explanation: For the first segment,
\[
v_1=\frac{6-0}{2-0}=+3\,\text{m s}^{-1},
\]
so its speed is \(3\,\text{m s}^{-1}\). For the second segment,
\[
v_2=\frac{0-6}{5-2}=-2\,\text{m s}^{-1},
\]
so its speed is \(2\,\text{m s}^{-1}\). For the third segment,
\[
v_3=\frac{-8-0}{7-5}=-4\,\text{m s}^{-1},
\]
so its speed is \(4\,\text{m s}^{-1}\). Hence
\[
|v_3|\gt |v_1|\gt |v_2|.
\]
At \(t=2\,\text{s}\), the slope changes from positive to negative, so the particle reverses direction. At \(t=5\,\text{s}\), the slope changes from \(-2\) to \(-4\,\text{m s}^{-1}\), but both slopes are negative. The particle therefore changes speed there without reversing direction.
117. Consider the following statements about smooth position-time graphs.
Statement I: If an \(x-t\) graph crosses \(x=0\) with a non-zero slope, the particle crosses the chosen spatial origin without being instantaneously at rest.
Statement II: At a smooth local maximum of an \(x-t\) graph, the tangent is horizontal; if the slope changes from positive before the maximum to negative after it, the particle reverses direction.
ⓐ. Both Statement I and Statement II are correct.
ⓑ. Statement I is correct, but Statement II is incorrect.
ⓒ. Statement I is incorrect, but Statement II is correct.
ⓓ. Both Statement I and Statement II are incorrect.
Correct Answer: Both Statement I and Statement II are correct.
Explanation: Statement I is correct because \(x=0\) identifies a location, namely the chosen origin. Whether the particle is moving or at rest at that location depends on the slope of the \(x-t\) graph. A non-zero slope means a non-zero instantaneous velocity, so the particle simply passes through the origin. Statement II is also correct. At a smooth local maximum, the tangent slope is zero, which gives
\[
v=0
\]
at that instant. If the slope is positive immediately before the maximum, the particle is moving toward increasing \(x\). If the slope is negative immediately afterward, it moves toward decreasing \(x\). The velocity therefore changes sign, which establishes a genuine direction reversal. Zero position and zero velocity are thus completely different graph conditions.
118. The position-time graph of a particle is described below.
The graph is smooth. At \(t=4\,\text{s}\), the tangent to the curve passes through \((3\,\text{s},7\,\text{m})\) and \((5\,\text{s},3\,\text{m})\). The actual curve itself passes through \((2\,\text{s},10\,\text{m})\) and \((6\,\text{s},6\,\text{m})\).
The instantaneous velocity at \(t=4\,\text{s}\) and the average velocity from \(t=2\,\text{s}\) to \(t=6\,\text{s}\) are respectively
ⓐ. \(+2\,\text{m s}^{-1}\) and \(-1\,\text{m s}^{-1}\)
ⓑ. \(-1\,\text{m s}^{-1}\) and \(-2\,\text{m s}^{-1}\)
ⓒ. \(-2\,\text{m s}^{-1}\) and \(-1\,\text{m s}^{-1}\)
ⓓ. \(-2\,\text{m s}^{-1}\) and \(+1\,\text{m s}^{-1}\)
Correct Answer: \(-2\,\text{m s}^{-1}\) and \(-1\,\text{m s}^{-1}\)
Explanation: Instantaneous velocity is obtained from the slope of the tangent to the \(x-t\) curve. Using the two stated points on the tangent,
\[
v(4)=\frac{3-7}{5-3}
=\frac{-4}{2}
=-2\,\text{m s}^{-1}.
\]
Average velocity over a finite interval is instead determined by the secant joining the actual graph points at the interval endpoints. Hence
\[
v_{\rm avg}
=\frac{x(6)-x(2)}{6-2}
=\frac{6-10}{4}
=-1\,\text{m s}^{-1}.
\]
The tangent and secant slopes need not be equal for a curved graph. Both are negative here, so both indicate net motion toward decreasing \(x\), but they describe different aspects of the motion. The tangent gives the instantaneous rate at \(t=4\,\text{s}\), whereas the secant gives the average rate over four seconds.
119. In coordinate system S, a particle has a straight-line position-time graph joining \((0,-4)\) and \((3,5)\), with SI units. A second observer uses the fixed coordinate transformation
\[
X=2-x.
\]
Which statement about the graph in the \(X-t\) system is correct?
ⓐ. It joins \((0,-2)\) and \((3,7)\), has slope \(+3\,\text{m s}^{-1}\), and the displacement is unchanged.
ⓑ. It joins \((0,6)\) and \((3,-3)\), has slope \(-3\,\text{m s}^{-1}\); the displacement changes sign while the physical distance remains \(9\,\text{m}\).
ⓒ. It joins \((0,6)\) and \((3,-3)\), has slope \(+3\,\text{m s}^{-1}\), and the distance travelled becomes \(-9\,\text{m}\).
ⓓ. It joins \((0,-6)\) and \((3,3)\), and the particle becomes non-uniform in the new coordinates.
Correct Answer: It joins \((0,6)\) and \((3,-3)\), has slope \(-3\,\text{m s}^{-1}\); the displacement changes sign while the physical distance remains \(9\,\text{m}\).
Explanation: The new position coordinate is defined by
\[
X=2-x.
\]
At \(t=0\), the old coordinate is \(-4\,\text{m}\), so
\[
X(0)=2-(-4)=6\,\text{m}.
\]
At \(t=3\,\text{s}\),
\[
X(3)=2-5=-3\,\text{m}.
\]
Thus the new graph joins \((0,6)\) and \((3,-3)\). Its slope is
\[
V=\frac{-3-6}{3}=-3\,\text{m s}^{-1}.
\]
The original displacement was
\[
5-(-4)=+9\,\text{m},
\]
whereas the new displacement is \(-9\,\text{m}\). Reversing the spatial orientation therefore reverses the signed displacement and velocity. The physical path length remains \(9\,\text{m}\), and the straight graph remains a constant-slope graph, so the motion remains uniform.
120. The position-time graph of a particle consists of straight segments joining \((0,-2)\), \((2,4)\), \((4,4)\), \((7,-5)\) and \((9,1)\), where SI units are used. Which set of results is correct for the complete \(0\) to \(9\,\text{s}\) interval?
ⓐ. Distance \(=15\,\text{m}\), displacement \(=+3\,\text{m}\), average speed \(=\dfrac{5}{3}\,\text{m s}^{-1}\), one reversal
ⓑ. Distance \(=21\,\text{m}\), displacement \(=-3\,\text{m}\), average velocity \(=-\dfrac{1}{3}\,\text{m s}^{-1}\), two reversals
ⓒ. Distance \(=21\,\text{m}\), displacement \(=+3\,\text{m}\), average speed \(=\dfrac{7}{3}\,\text{m s}^{-1}\), two reversals
ⓓ. Distance \(=18\,\text{m}\), displacement \(=+3\,\text{m}\), average speed \(=2\,\text{m s}^{-1}\), one reversal
Correct Answer: Distance \(=21\,\text{m}\), displacement \(=+3\,\text{m}\), average speed \(=\dfrac{7}{3}\,\text{m s}^{-1}\), two reversals
Explanation: From \(-2\) to \(+4\,\text{m}\), the first segment contributes
\[
d_1=6\,\text{m}.
\]
The horizontal segment from \(t=2\) to \(4\,\text{s}\) contributes zero distance. The next segment from \(+4\) to \(-5\,\text{m}\) contributes
\[
d_3=9\,\text{m},
\]
and the final segment from \(-5\) to \(+1\,\text{m}\) contributes
\[
d_4=6\,\text{m}.
\]
Hence
\[
d=6+0+9+6=21\,\text{m}.
\]
The displacement depends only on the endpoints:
\[
\Delta x=1-(-2)=+3\,\text{m}.
\]
Thus
\[
v_{\rm avg,speed}=\frac{21}{9}=\frac{7}{3}\,\text{m s}^{-1}.
\]
The graph changes from positive slope to rest and then negative slope, giving one reversal of moving direction, and later changes from negative slope to positive slope at \(t=7\,\text{s}\), giving a second reversal. The horizontal interval itself is a period of rest and does not add distance.
