1. Which statement most accurately describes kinematics?
ⓐ. It describes changes of motion only after the forces acting on the body have first been determined.
ⓑ. It describes changes of position with time in a chosen reference frame without specifying the causes of motion.
ⓒ. It describes changes of position only for bodies whose speed remains constant with time.
ⓓ. It describes only the path length covered, without using position or a reference frame.
Correct Answer: It describes changes of position with time in a chosen reference frame without specifying the causes of motion.
Explanation: Kinematics is concerned with the description of motion rather than with the physical causes producing that motion. To describe motion, the position of an object must be specified relative to a chosen reference arrangement. A one-dimensional description therefore requires an origin and a convention for the positive direction. Time is also required because motion involves a change of position with time. Quantities such as position, displacement and velocity can then be introduced without first discussing forces. A body may be at rest or moving depending on the reference frame in which its position is examined. Kinematics is therefore not restricted to constant-speed motion and does not require a force analysis before motion can be described.
2. Along an east-west line, east is initially chosen as the positive direction. A particle moves from \(x=+3\,\text{m}\) to \(x=+8\,\text{m}\). The same origin is retained, but west is now chosen as the positive direction. The new initial coordinate, new final coordinate and new displacement are respectively
ⓐ. \(+3\,\text{m},\ +8\,\text{m},\ -5\,\text{m}\)
ⓑ. \(-3\,\text{m},\ -8\,\text{m},\ +5\,\text{m}\)
ⓒ. \(-3\,\text{m},\ -8\,\text{m},\ -5\,\text{m}\)
ⓓ. \(-8\,\text{m},\ -3\,\text{m},\ +5\,\text{m}\)
Correct Answer: \(-3\,\text{m},\ -8\,\text{m},\ -5\,\text{m}\)
Explanation: \( \textbf{Initial description:} \) East is positive, so the coordinates are \(x_1=+3\,\text{m}\) and \(x_2=+8\,\text{m}\).
\( \textbf{Change of convention:} \) The origin remains fixed, but the positive direction is reversed.
Therefore every coordinate changes sign:
\[
x_1'=-3\,\text{m},\qquad x_2'=-8\,\text{m}.
\]
\( \textbf{Displacement in the new system:} \)
\[
\Delta x'=x_2'-x_1'.
\]
Substituting,
\[
\Delta x'=(-8)-(-3)=-5\,\text{m}.
\]
The negative sign does not mean that the physical motion has changed; it reflects the new positive-direction convention. In the original east-positive system the displacement was \(+5\,\text{m}\), whereas in the west-positive system it is \(-5\,\text{m}\). Thus reversing the axis reverses the algebraic sign of the displacement while leaving the actual motion unchanged.
3. Two cyclists travel east along the same straight road with equal constant speeds and maintain a fixed separation from one another. Which statement is correct?
ⓐ. Both cyclists are at rest relative to the road because their separation is constant.
ⓑ. Each cyclist is moving relative to the other because both are moving relative to the road.
ⓒ. Whether they are moving or at rest is independent of the chosen reference frame.
ⓓ. Each cyclist is at rest relative to the other but moving relative to the road.
Correct Answer: Each cyclist is at rest relative to the other but moving relative to the road.
Explanation: Rest or motion is decided by examining whether the position of an object changes with time in a specified reference frame. Relative to the road, the position of each cyclist continually changes, so each cyclist is moving in the road frame. Relative to one cyclist, however, the other remains at a fixed separation because both have equal velocities in the same direction. Hence the position of one cyclist relative to the other does not change with time. Each cyclist is therefore at rest in the frame attached to the other cyclist. There is no contradiction between being at rest in one frame and moving in another. The fixed separation does not make either cyclist stationary relative to the road. The example shows why statements about motion and rest require an identified reference frame.
4. In a coordinate system with positive direction to the right, a particle moves from \(x_1=-4\,\text{m}\) to \(x_2=+6\,\text{m}\). A new origin is then chosen at the old coordinate \(x=+3\,\text{m}\), while the positive direction is kept unchanged. The new coordinates of the initial and final positions and the displacement are
ⓐ. \(-1\,\text{m},\ +9\,\text{m},\ +10\,\text{m}\)
ⓑ. \(-7\,\text{m},\ +3\,\text{m},\ -10\,\text{m}\)
ⓒ. \(-7\,\text{m},\ +3\,\text{m},\ +10\,\text{m}\)
ⓓ. \(+7\,\text{m},\ -3\,\text{m},\ -10\,\text{m}\)
Correct Answer: \(-7\,\text{m},\ +3\,\text{m},\ +10\,\text{m}\)
Explanation: \( \textbf{Given:} \) The new origin is \(3\,\text{m}\) to the right of the old origin, and the axis direction is unchanged.
For any old coordinate \(x\), the new coordinate is
\[
x'=x-3.
\]
For the initial position,
\[
x_1'=-4-3=-7\,\text{m}.
\]
For the final position,
\[
x_2'=+6-3=+3\,\text{m}.
\]
The displacement in the new system is
\[
\Delta x'=x_2'-x_1'=3-(-7)=+10\,\text{m}.
\]
In the old system,
\[
\Delta x=6-(-4)=+10\,\text{m}.
\]
The constant shift of origin is subtracted from both endpoint coordinates and therefore cancels when their difference is taken. Thus position coordinates depend on the selected origin, but displacement is unchanged when only the origin is shifted and the axis direction is retained.
5. Consider the following statements about modelling motion along a straight railway track.
Statement I: For locating a \(180\,\text{m}\) long train during a journey of several hundred kilometres, representing the train by a point may be an adequate approximation.
Statement II: For finding the time from the instant the front of the same train enters a \(200\,\text{m}\) platform until its rear leaves the platform, the train cannot be treated as a point object without losing essential information.
Statement III: Once an origin and a positive direction are chosen, motion along the straight track can be represented by a single coordinate \(x(t)\).
Statement IV: A train having a negative position coordinate must necessarily be moving in the negative direction.
Which set of statements is correct?
ⓐ. Statement I and Statement II only
ⓑ. Statement II and Statement IV only
ⓒ. Statement I, Statement III and Statement IV only
ⓓ. Statement I, Statement II and Statement III only
Correct Answer: Statement I, Statement II and Statement III only
Explanation: Statement I is correct because the \(180\,\text{m}\) length of the train is negligible compared with a journey extending over hundreds of kilometres when only its overall location is required. Statement II is also correct because the train length directly affects the time required for the complete train to clear a platform. Treating it as a point in that problem would discard a distance that is comparable with the platform length. Statement III is correct because rectilinear motion has only two possible directions along one line, so a single signed coordinate \(x(t)\) can specify position. Statement IV is incorrect. A negative coordinate tells us that the train lies on the negative side of the chosen origin at that instant. It does not by itself reveal whether the train is moving toward \(+x\), toward \(-x\), or is momentarily at rest. Position and direction of motion are distinct pieces of information. Thus the validity of a point-object approximation depends on the scale and purpose of the problem, while the sign of position must not be confused with the direction of motion.
6. Use the arrangement described below.
A straight line has positive direction toward the right. Point P is at \(x=-5\,\text{m}\), point Q is at \(x=+4\,\text{m}\), and point R is at \(x=+1\,\text{m}\). A particle moves from P to Q and then from Q to R.
The total distance travelled and the displacement from P to R are respectively
ⓐ. \(12\,\text{m}\) and \(+6\,\text{m}\)
ⓑ. \(12\,\text{m}\) and \(-6\,\text{m}\)
ⓒ. \(6\,\text{m}\) and \(+6\,\text{m}\)
ⓓ. \(6\,\text{m}\) and \(-12\,\text{m}\)
Correct Answer: \(12\,\text{m}\) and \(+6\,\text{m}\)
Explanation: \( \textbf{First segment:} \) The particle moves from \(x=-5\,\text{m}\) to \(x=+4\,\text{m}\).
Its path length on this segment is
\[
|4-(-5)|=9\,\text{m}.
\]
\( \textbf{Second segment:} \) It then moves from \(x=+4\,\text{m}\) to \(x=+1\,\text{m}\).
The path length on this segment is
\[
|1-4|=3\,\text{m}.
\]
Therefore the total distance is
\[
d=9+3=12\,\text{m}.
\]
For displacement, only the initial and final coordinates are required:
\[
\Delta x=x_f-x_i=1-(-5)=+6\,\text{m}.
\]
The return from \(+4\,\text{m}\) to \(+1\,\text{m}\) increases the distance but reduces the net displacement. Distance counts every path segment as a positive length, whereas displacement depends only on the two endpoints.
7. A particle moves continuously along a straight line from \(x=+12\,\text{m}\) to \(x=-4\,\text{m}\) without reversing its direction. Its distance travelled and displacement are
ⓐ. \(-16\,\text{m}\) and \(+16\,\text{m}\)
ⓑ. \(16\,\text{m}\) and \(+16\,\text{m}\)
ⓒ. \(16\,\text{m}\) and \(-16\,\text{m}\)
ⓓ. \(8\,\text{m}\) and \(-16\,\text{m}\)
Correct Answer: \(16\,\text{m}\) and \(-16\,\text{m}\)
Explanation: The initial position is \(x_i=+12\,\text{m}\), while the final position is \(x_f=-4\,\text{m}\). The displacement is therefore
\[
\Delta x=x_f-x_i=-4-12=-16\,\text{m}.
\]
The negative sign indicates that the net change in position is toward the negative direction. Distance, however, is a path length and cannot be negative. Because the particle does not reverse direction, the path length is simply the magnitude of the endpoint separation:
\[
d=|-4-12|=16\,\text{m}.
\]
Thus \(d=|\Delta x|\) even though the displacement itself is negative. Equality between distance and displacement magnitude does not require motion in the positive direction. It requires motion along the line without reversal.
8. The position of a particle undergoing uniform motion is described by
\[
x=5-2t,
\]
where \(x\) is in metres and \(t\) in seconds. Between \(t=0\) and \(t=4\,\text{s}\), its distance travelled and displacement are respectively
ⓐ. \(8\,\text{m}\) and \(+8\,\text{m}\)
ⓑ. \(3\,\text{m}\) and \(-8\,\text{m}\)
ⓒ. \(8\,\text{m}\) and \(-8\,\text{m}\)
ⓓ. \(13\,\text{m}\) and \(-3\,\text{m}\)
Correct Answer: \(8\,\text{m}\) and \(-8\,\text{m}\)
Explanation: \( \textbf{Initial position:} \) At \(t=0\),
\[
x_i=5-2(0)=5\,\text{m}.
\]
\( \textbf{Final position:} \) At \(t=4\,\text{s}\),
\[
x_f=5-2(4)=5-8=-3\,\text{m}.
\]
The displacement is
\[
\Delta x=x_f-x_i=-3-5=-8\,\text{m}.
\]
The coefficient of \(t\) is constant and negative, so the coordinate decreases uniformly with time. The particle therefore moves continuously in the negative direction throughout the interval and does not reverse. Hence the distance is the magnitude of its displacement:
\[
d=|\Delta x|=8\,\text{m}.
\]
Crossing the origin does not constitute a reversal of motion; the particle simply passes from positive to negative position while continuing in the same direction.
9. In an original coordinate system, positive direction is toward the right. A particle moves from \(x=-6\,\text{m}\) to \(x=+8\,\text{m}\) and then back to \(x=+4\,\text{m}\). A second observer chooses the old point \(x=+10\,\text{m}\) as origin and takes the positive direction toward the left. According to the second observer, the net displacement and the total distance travelled are respectively
ⓐ. \(+10\,\text{m}\) and \(18\,\text{m}\)
ⓑ. \(-10\,\text{m}\) and \(10\,\text{m}\)
ⓒ. \(+18\,\text{m}\) and \(10\,\text{m}\)
ⓓ. \(-10\,\text{m}\) and \(18\,\text{m}\)
Correct Answer: \(-10\,\text{m}\) and \(18\,\text{m}\)
Explanation: \( \textbf{Step 1 — Distance in the physical motion:} \)
From \(-6\,\text{m}\) to \(+8\,\text{m}\), the path length is
\[
|8-(-6)|=14\,\text{m}.
\]
From \(+8\,\text{m}\) back to \(+4\,\text{m}\), the path length is
\[
|4-8|=4\,\text{m}.
\]
Therefore
\[
d=14+4=18\,\text{m}.
\]
Changing coordinates cannot alter this physical path length.
\( \textbf{Step 2 — Transform the endpoint coordinates:} \) With the new origin at old \(x=10\,\text{m}\) and positive direction to the left,
\[
x'=10-x.
\]
Thus the initial coordinate becomes \(x_i'=10-(-6)=16\,\text{m}\), while the final coordinate becomes \(x_f'=10-4=6\,\text{m}\).
\( \textbf{Step 3 — New displacement:} \)
\[
\Delta x'=x_f'-x_i'=6-16=-10\,\text{m}.
\]
The reversal of axis direction reverses the sign of the net displacement, whereas the total distance remains \(18\,\text{m}\).
10. Two buses A and B move east on the same straight road with equal constant velocities while B remains \(30\,\text{m}\) ahead of A. Frame R is fixed to the road with east positive. Frame S is attached to bus A, with its origin at A and east positive.
Statement I: Bus B has a constant coordinate \(+30\,\text{m}\) in frame S.
Statement II: Both buses are moving in frame R.
Statement III: If the positive direction of frame R is changed from east to west, the velocity signs of both buses reverse, but B still remains at rest relative to A.
Statement IV: Merely shifting the fixed origin of frame R to another roadside point makes both buses at rest in frame R.
Which set of statements is correct?
ⓐ. Statement I and Statement II only
ⓑ. Statement II, Statement III and Statement IV only
ⓒ. Statement I, Statement II and Statement III only
ⓓ. Statement I and Statement IV only
Correct Answer: Statement I, Statement II and Statement III only
Explanation: Statement I is correct because the separation between B and A remains \(30\,\text{m}\), so B's position relative to A does not change with time. Therefore B is at rest in the frame attached to A. Statement II is correct because both buses change their positions continuously relative to fixed roadside points. Statement III is also correct. Reversing the positive direction changes the algebraic signs assigned to their eastward velocities, but it does not change their physical motion or their constant separation. Their relative-rest condition therefore survives the coordinate reversal. Statement IV is incorrect because shifting a fixed origin only changes the numerical position coordinates by a constant amount. The coordinates of the moving buses will still change with time in that road-fixed frame. A change of coordinate labels cannot make physically moving buses stationary relative to the road. The decisive distinction is between changing a reference frame attached to a moving object and merely relabelling coordinates within a fixed road frame.
11. A particle moves along a straight line according to the observations below. During each listed one-second interval it moves uniformly between the stated positions, and from \(t=2\,\text{s}\) to \(t=3\,\text{s}\) it remains at \(x=4\,\text{m}\).
| \(t\,(\text{s})\) | \(x\,(\text{m})\) |
|---|
| 0 | \(-2\) |
| 1 | \(+1\) |
| 2 | \(+4\) |
| 3 | \(+4\) |
| 4 | \(+1\) |
For the complete \(0\) to \(4\,\text{s}\) interval, which description is correct?
ⓐ. Distance \(=9\,\text{m}\), displacement \(=+3\,\text{m}\), and the motion is non-uniform.
ⓑ. Distance \(=9\,\text{m}\), displacement \(=-3\,\text{m}\), and the motion is uniform.
ⓒ. Distance \(=3\,\text{m}\), displacement \(=+9\,\text{m}\), and the motion is uniform.
ⓓ. Distance \(=6\,\text{m}\), displacement \(=+3\,\text{m}\), and the motion is non-uniform.
Correct Answer: Distance \(=9\,\text{m}\), displacement \(=+3\,\text{m}\), and the motion is non-uniform.
Explanation: \( \textbf{Step 1 — Segment displacements:} \)
From \(t=0\) to \(1\,\text{s}\), the change is \(+1-(-2)=+3\,\text{m}\).
From \(t=1\) to \(2\,\text{s}\), the change is \(+4-(+1)=+3\,\text{m}\).
From \(t=2\) to \(3\,\text{s}\), the change is \(0\,\text{m}\).
From \(t=3\) to \(4\,\text{s}\), the change is \(+1-(+4)=-3\,\text{m}\).
\( \textbf{Step 2 — Total distance:} \)
\[
d=|3|+|3|+|0|+|-3|=9\,\text{m}.
\]
\( \textbf{Step 3 — Net displacement:} \)
\[
\Delta x=x_f-x_i=1-(-2)=+3\,\text{m}.
\]
The equal-time displacements are \(+3,+3,0,-3\,\text{m}\), so they are not all equal. The particle first moves in the positive direction, then remains at rest, and finally moves in the negative direction. Hence its velocity is not constant over the complete interval, making the motion non-uniform.
12. A particle starts at \(x=-3\,\text{m}\), initially moves in the positive direction, reverses direction exactly once, and finally stops at \(x=+5\,\text{m}\). If its total distance travelled is \(20\,\text{m}\), the coordinate of the turning point is
ⓐ. \(+9\,\text{m}\)
ⓑ. \(+11\,\text{m}\)
ⓒ. \(+7\,\text{m}\)
ⓓ. \(+13\,\text{m}\)
Correct Answer: \(+11\,\text{m}\)
Explanation: \( \textbf{Step 1 — Interpret the reversal:} \) The particle starts at \(-3\,\text{m}\) and initially moves toward \(+x\). Because it reverses exactly once and eventually ends at \(+5\,\text{m}\), the turning point must lie to the right of \(+5\,\text{m}\). Let its coordinate be \(x_T\).
\( \textbf{Step 2 — Distance before reversal:} \)
\[
d_1=x_T-(-3)=x_T+3.
\]
\( \textbf{Step 3 — Distance after reversal:} \)
\[
d_2=x_T-5.
\]
The total distance is given as \(20\,\text{m}\), so
\[
(x_T+3)+(x_T-5)=20.
\]
Therefore
\[
2x_T-2=20,
\]
\[
2x_T=22,
\]
\[
x_T=11\,\text{m}.
\]
A useful check is \(14\,\text{m}\) outward from \(-3\) to \(+11\), followed by \(6\,\text{m}\) back to \(+5\), giving \(14+6=20\,\text{m}\). Thus the turning point is at \(+11\,\text{m}\).
13. Two particles P and Q move uniformly along the same straight line with positions
\[
x_P=-4+3t,\qquad x_Q=8+t,
\]
where \(x\) is in metres and \(t\) in seconds. At what time do they meet, and two seconds after meeting what is their separation and which particle is ahead in the positive direction?
ⓐ. They meet at \(6\,\text{s}\); separation \(4\,\text{m}\), with P ahead.
ⓑ. They meet at \(6\,\text{s}\); separation \(2\,\text{m}\), with P ahead.
ⓒ. They meet at \(4\,\text{s}\); separation \(4\,\text{m}\), with P ahead.
ⓓ. They meet at \(4\,\text{s}\); separation \(4\,\text{m}\), with Q ahead.
Correct Answer: They meet at \(6\,\text{s}\); separation \(4\,\text{m}\), with P ahead.
Explanation: \( \textbf{Step 1 — Meeting condition:} \) The particles meet when their position coordinates are equal.
\[
-4+3t=8+t.
\]
Rearranging,
\[
3t-t=8+4,
\]
\[
2t=12,
\]
so
\[
t=6\,\text{s}.
\]
\( \textbf{Step 2 — Move two seconds beyond the meeting:} \) Two seconds later corresponds to \(t=8\,\text{s}\).
For P,
\[
x_P=-4+3(8)=20\,\text{m}.
\]
For Q,
\[
x_Q=8+8=16\,\text{m}.
\]
Hence the separation is
\[
|20-16|=4\,\text{m}.
\]
Since \(20\,\text{m}\gt 16\,\text{m}\), P is ahead in the positive direction. P moves \(2\,\text{m s}^{-1}\) faster than Q, so after their meeting the separation grows by \(2\,\text{m}\) each second; after \(2\,\text{s}\) it must therefore be \(4\,\text{m}\), confirming the result.
14. Read the motion description below.
A cart starts at \(x=0\). During the first \(2\,\text{s}\) it moves uniformly to \(x=+8\,\text{m}\). During the next \(2\,\text{s}\) it remains at \(x=+8\,\text{m}\). During the final \(2\,\text{s}\) it moves uniformly to \(x=+2\,\text{m}\).
For the complete \(6\,\text{s}\) motion, which statement is correct?
ⓐ. Distance \(=2\,\text{m}\), displacement \(=14\,\text{m}\), and the motion is uniform.
ⓑ. Distance \(=14\,\text{m}\), displacement \(=+2\,\text{m}\), and the motion is non-uniform.
ⓒ. Distance \(=14\,\text{m}\), displacement \(=-2\,\text{m}\), and the motion is uniform.
ⓓ. Distance \(=8\,\text{m}\), displacement \(=+2\,\text{m}\), and the motion is uniform.
Correct Answer: Distance \(=14\,\text{m}\), displacement \(=+2\,\text{m}\), and the motion is non-uniform.
Explanation: \( \textbf{Step 1 — First interval:} \) The cart moves from \(0\) to \(+8\,\text{m}\), so it covers \(8\,\text{m}\).
Its constant velocity during this interval is
\[
v_1=\frac{8-0}{2}=+4\,\text{m s}^{-1}.
\]
\( \textbf{Step 2 — Second interval:} \) It remains at \(+8\,\text{m}\), so the distance covered is \(0\) and its velocity is \(0\).
\( \textbf{Step 3 — Final interval:} \) It moves from \(+8\,\text{m}\) to \(+2\,\text{m}\), covering
\[
|2-8|=6\,\text{m}.
\]
Its velocity in this interval is
\[
v_3=\frac{2-8}{2}=-3\,\text{m s}^{-1}.
\]
Thus the total distance is
\[
d=8+0+6=14\,\text{m}.
\]
The displacement depends only on the endpoints:
\[
\Delta x=2-0=+2\,\text{m}.
\]
Because the velocity changes from \(+4\) to \(0\) to \(-3\,\text{m s}^{-1}\), the complete motion is non-uniform even though each moving segment is individually uniform.
15. A particle starts at \(x=+2\,\text{m}\) and finally reaches \(x=-6\,\text{m}\). It is known that at some intermediate instant it reaches \(x=+10\,\text{m}\). The minimum possible distance travelled and the net displacement for the complete motion are respectively
ⓐ. \(8\,\text{m}\) and \(-8\,\text{m}\)
ⓑ. \(16\,\text{m}\) and \(+8\,\text{m}\)
ⓒ. \(24\,\text{m}\) and \(-8\,\text{m}\)
ⓓ. \(24\,\text{m}\) and \(+8\,\text{m}\)
Correct Answer: \(24\,\text{m}\) and \(-8\,\text{m}\)
Explanation: \( \textbf{Step 1 — Required intermediate position:} \) The particle must reach \(x=+10\,\text{m}\) after starting from \(x=+2\,\text{m}\).
The shortest possible path for this part is
\[
d_1=|10-2|=8\,\text{m}.
\]
\( \textbf{Step 2 — Travel from the required intermediate point to the final point:} \)
From \(+10\,\text{m}\) to \(-6\,\text{m}\), the shortest possible path is
\[
d_2=|-6-10|=16\,\text{m}.
\]
Therefore the minimum total distance is
\[
d_{\min}=8+16=24\,\text{m}.
\]
Any additional reversal or detour along the line would only increase this value.
\( \textbf{Step 3 — Net displacement:} \)
\[
\Delta x=x_f-x_i=-6-(+2)=-8\,\text{m}.
\]
The displacement ignores the visit to \(+10\,\text{m}\) because it depends only on the initial and final positions. Thus the intermediate-position condition increases the minimum path length without changing the net displacement.
16. Particle P moves monotonically along a straight line from \(x=-4\,\text{m}\) to \(x=+8\,\text{m}\). Particle Q starts from the same point, moves first to \(x=+10\,\text{m}\), and then returns to \(x=+8\,\text{m}\). A second observer uses the same origin but reverses the positive direction of the coordinate axis. Which statement correctly compares the complete motions as described by the second observer?
ⓐ. Both displacements are \(+12\,\text{m}\), and the distances travelled by P and Q are in the ratio \(3:4\).
ⓑ. Both displacements are \(-12\,\text{m}\), and the distances travelled by P and Q are in the ratio \(3:4\).
ⓒ. The displacements are \(-12\,\text{m}\) and \(-14\,\text{m}\), and the distances are equal.
ⓓ. Both displacements are \(-12\,\text{m}\), and the distances travelled by P and Q are in the ratio \(4:3\).
Correct Answer: Both displacements are \(-12\,\text{m}\), and the distances travelled by P and Q are in the ratio \(3:4\).
Explanation: \( \textbf{Step 1 — Compare the endpoints:} \) Both particles begin at \(-4\,\text{m}\) and finish at \(+8\,\text{m}\).
In the original axis their common displacement is
\[
\Delta x=8-(-4)=+12\,\text{m}.
\]
When the positive direction of the axis is reversed, the physical motion is unchanged but the algebraic sign of every displacement reverses. Therefore
\[
\Delta x_P'=\Delta x_Q'=-12\,\text{m}.
\]
\( \textbf{Step 2 — Distance travelled by P:} \) P moves monotonically from \(-4\) to \(+8\,\text{m}\), so
\[
d_P=|8-(-4)|=12\,\text{m}.
\]
\( \textbf{Step 3 — Distance travelled by Q:} \)
\[
d_Q=|10-(-4)|+|8-10|=14+2=16\,\text{m}.
\]
Thus
\[
d_P:d_Q=12:16=3:4.
\]
The common endpoints guarantee equal displacements, whereas Q's reversal produces the larger path length. Reversing the coordinate axis changes displacement signs but does not change either physical distance.
17. In an original coordinate system, a particle moves uniformly according to
\[
x=4+2t,
\]
where \(x\) is in metres and \(t\) in seconds. A second observer chooses the old point \(x=10\,\text{m}\) as the new origin, takes the positive direction toward the left, and sets the new clock reading \(\tau=0\) at the instant when the old clock reads \(t=3\,\text{s}\). Which equation correctly describes the same motion for the second observer?
ⓐ. \(X=-2\tau\)
ⓑ. \(X=2\tau\)
ⓒ. \(X=6-2\tau\)
ⓓ. \(X=-6+2\tau\)
Correct Answer: \(X=-2\tau\)
Explanation: The question changes both the spatial coordinate convention and the zero of time, but the physical motion remains the same. The old and new time readings are related by
\[
t=\tau+3.
\]
The new spatial origin is at old coordinate \(x=10\,\text{m}\), and the new positive direction is opposite to the old positive direction. Therefore the spatial transformation is
\[
X=-(x-10)=10-x.
\]
Using the original motion equation,
\[
x=4+2(\tau+3).
\]
Hence
\[
x=4+2\tau+6=10+2\tau.
\]
Substituting into the spatial transformation,
\[
X=10-(10+2\tau).
\]
Thus
\[
X=-2\tau.
\]
At \(\tau=0\), the particle is exactly at the new origin, which provides a useful consistency check. The negative coefficient does not mean that the particle physically reverses; it arises because the second observer has chosen the opposite positive direction.
18. In a road-fixed frame, three particles move along the same straight line according to
\[
x_A=2+4t,\qquad x_B=-5+4t,\qquad x_C=16+t,
\]
where \(x\) is in metres and \(t\) in seconds. A new frame S has its origin attached to particle B and has the same positive direction as the road frame. Which statement is correct?
ⓐ. A remains at \(X=+7\,\text{m}\) in S, while C passes through the origin of S at \(t=7\,\text{s}\) moving toward negative \(X\).
ⓑ. A moves in the positive direction in S, while C remains at \(X=21\,\text{m}\).
ⓒ. A remains at \(X=-7\,\text{m}\) in S, while C crosses the origin of S at \(t=5\,\text{s}\).
ⓓ. A remains at \(X=+7\,\text{m}\) in S, while C passes through the origin of S at \(t=5\,\text{s}\) moving toward positive \(X\).
Correct Answer: A remains at \(X=+7\,\text{m}\) in S, while C passes through the origin of S at \(t=7\,\text{s}\) moving toward negative \(X\).
Explanation: In the frame attached to B, the coordinate of any particle is its road-frame position minus the position of B. For A,
\[
X_A=x_A-x_B.
\]
Substituting,
\[
X_A=(2+4t)-(-5+4t)=7\,\text{m}.
\]
Thus A remains at a fixed coordinate \(+7\,\text{m}\) relative to B because A and B have equal velocities. For C,
\[
X_C=x_C-x_B.
\]
Therefore
\[
X_C=(16+t)-(-5+4t)=21-3t.
\]
C reaches the origin of S when \(X_C=0\):
\[
21-3t=0.
\]
Hence
\[
t=7\,\text{s}.
\]
Since the coefficient of \(t\) in \(X_C\) is negative, C moves toward decreasing \(X\) in frame S. The calculation simultaneously shows that mutual rest depends on equality of motion relative to the selected frame, not on whether the objects are moving relative to the road.
19. A particle starts at \(x=+1\,\text{m}\), ends at \(x=+5\,\text{m}\), travels a total distance of \(14\,\text{m}\), and reverses direction exactly once. Its initial direction of motion is not specified. Which conclusion about the coordinate of the turning point is correct?
ⓐ. It must be \(+10\,\text{m}\).
ⓑ. It must be \(-4\,\text{m}\).
ⓒ. It can be either \(-4\,\text{m}\) or \(+10\,\text{m}\).
ⓓ. It can be either \(-10\,\text{m}\) or \(+4\,\text{m}\).
Correct Answer: It can be either \(-4\,\text{m}\) or \(+10\,\text{m}\).
Explanation: Because the particle reverses exactly once, there are two physically distinct possibilities depending on its initial direction. First suppose it initially moves in the positive direction and turns at \(x_T\gt 5\,\text{m}\). The outward distance is
\[
x_T-1,
\]
and the return distance to \(x=5\,\text{m}\) is
\[
x_T-5.
\]
Using the given total distance,
\[
(x_T-1)+(x_T-5)=14.
\]
Thus
\[
2x_T-6=14,
\]
so
\[
x_T=10\,\text{m}.
\]
Now suppose the particle initially moves in the negative direction and turns at \(x_T\lt 1\,\text{m}\). The two path lengths are then
\[
1-x_T
\]
and
\[
5-x_T.
\]
Hence
\[
(1-x_T)+(5-x_T)=14,
\]
which gives
\[
6-2x_T=14,
\]
and therefore
\[
x_T=-4\,\text{m}.
\]
Both paths satisfy the same endpoints, total distance and exactly one reversal. Since the initial direction was not supplied, both turning points are admissible.
20. In an original coordinate system, a particle follows the ordered path
\[
0\rightarrow +5\,\text{m}\rightarrow -2\,\text{m}\rightarrow +3\,\text{m}.
\]
A second observer keeps the same origin but reverses the positive direction of the axis. For the complete motion, what does the second observer obtain for the net displacement, total distance and number of direction reversals?
ⓐ. \(-3\,\text{m},\ 17\,\text{m},\ 2\)
ⓑ. \(+3\,\text{m},\ 17\,\text{m},\ 2\)
ⓒ. \(-3\,\text{m},\ 14\,\text{m},\ 1\)
ⓓ. \(+3\,\text{m},\ 17\,\text{m},\ 1\)
Correct Answer: \(-3\,\text{m},\ 17\,\text{m},\ 2\)
Explanation: First calculate the physical path length, which cannot depend on the chosen sign convention. The first segment has length
\[
|5-0|=5\,\text{m}.
\]
The second segment has length
\[
|-2-5|=7\,\text{m}.
\]
The final segment has length
\[
|3-(-2)|=5\,\text{m}.
\]
Therefore
\[
d=5+7+5=17\,\text{m}.
\]
In the original frame, the net displacement is
\[
\Delta x=3-0=+3\,\text{m}.
\]
Reversing the axis changes every signed coordinate and every signed displacement to its negative, so the second observer obtains
\[
\Delta X=-3\,\text{m}.
\]
The particle first moves in one direction, then reverses at \(+5\,\text{m}\), and reverses again at \(-2\,\text{m}\). Reversing the coordinate convention swaps the algebraic signs assigned to those directions but does not change the fact that two physical reversals occurred. Thus distance and reversal count remain invariant, while the signed displacement changes sign.
