Concept Card: Frame of Reference, Distance & Displacement in 1D Motion
1. Frame of Reference:
A frame of reference is a coordinate system (Cartesian axes $x, y, z$) attached to an observer along with a calibrated clock to record the time coordinate $t$.
- Inertial Frame of Reference: A frame in which Newton’s First Law of Motion holds valid (i.e., a body remains at rest or moves with uniform velocity unless acted upon by an external net force). An inertial frame is non-accelerating ($\vec{a}_{\text{frame}} = 0$).
Examples: A stationary laboratory floor (approximated), a train moving at constant velocity on straight tracks. - Non-Inertial Frame of Reference: An accelerated or rotating frame of reference ($\vec{a}_{\text{frame}} \neq 0$). To apply Newton’s second law in such frames, an apparent fictitious force (pseudo force) must be included:
$\vec{F}_{\text{pseudo}} = -m \vec{a}_{\text{frame}}$
2. Position Vector & Path in One Dimension (1D):
For rectilinear motion along the $x$-axis, the position of a particle at any instant $t$ is completely specified by its coordinate $x(t)$ relative to an origin $O$.
$\vec{r}(t) = x(t)\hat{i}$
3. Distance vs. Displacement:
| Characteristic | Distance ($s$) | Displacement ($\Delta \vec{x}$) |
|---|---|---|
| Physical Nature | Scalar quantity ($s \ge 0$) | Vector quantity ($\Delta \vec{x} = \vec{x}_f – \vec{x}_i$) |
| Definition | Total length of actual path traversed | Shortest straight-line distance from initial to final position |
| Path Dependence | Depends on the actual trajectory followed | Independent of path; function of initial & final points only |
| Time Behavior | Cannot decrease with time for moving body ($\frac{ds}{dt} \ge 0$) | Can increase, decrease, become zero, or be negative |
| Fundamental Relation | $\text{Distance} \ge |\text{Displacement}| \implies \frac{\text{Distance}}{|\text{Displacement}|} \ge 1$ (Equality holds if and only if the particle moves in a straight line without reversing direction) | |
4. Calculating Distance in 1D Motion with Turning Points:
If a particle moving along the $x$-axis changes its direction of motion, its velocity becomes instantaneously zero ($v(t) = \frac{dx}{dt} = 0$).
To find the total distance traveled between $t = 0$ and $t = T$ with a turning point at $t = t_0$ ($0 < t_0 < T$):
$\text{Displacement } \Delta x = x(T) – x(0) = \int_{0}^{T} v(t)\,dt$
$\text{Total Distance } s = \int_{0}^{T} |v(t)|\,dt = |x(t_0) – x(0)| + |x(T) – x(t_0)|$
Solved Examples
Example 1 (Distance vs. Displacement with Direction Reversal):
A particle moves along the $x$-axis such that its position as a function of time is given by:
$x(t) = t^2 – 4t + 3\text{ m}$
Find:
(a) The displacement of the particle between $t = 0$ and $t = 5\text{ s}$.
(b) The total distance traveled by the particle between $t = 0$ and $t = 5\text{ s}$.
Solution:
(a) Position at $t = 0$: $x(0) = 0^2 – 4(0) + 3 = +3\text{ m}$.
Position at $t = 5\text{ s}$: $x(5) = 5^2 – 4(5) + 3 = 25 – 20 + 3 = +8\text{ m}$.
$\text{Displacement } \Delta x = x(5) – x(0) = 8 – 3 = +5\text{ m}$.
(b) To calculate distance, check for turning points where velocity $v(t) = 0$:
$v(t) = \frac{dx}{dt} = 2t – 4 = 0 \implies t = 2\text{ s}$.
Since $t = 2\text{ s}$ lies within the interval $[0, 5\text{ s}]$, the particle reverses its direction at $t = 2\text{ s}$.
Position at turning point: $x(2) = 2^2 – 4(2) + 3 = 4 – 8 + 3 = -1\text{ m}$.
Calculate segment distances:
– From $t = 0$ to $t = 2\text{ s}$: $s_1 = |x(2) – x(0)| = |-1 – 3| = |-4| = 4\text{ m}$.
– From $t = 2$ to $t = 5\text{ s}$: $s_2 = |x(5) – x(2)| = |8 – (-1)| = |9| = 9\text{ m}$.
$\text{Total Distance } s = s_1 + s_2 = 4\text{ m} + 9\text{ m} = 13\text{ m}$.
Example 2 (Two-Dimensional Circular & Radial Path):
A particle moves along the perimeter of a semicircular track of radius $R = 7\text{ m}$ from point $A(-R, 0)$ to point $B(+R, 0)$, and then travels along a straight diameter back to the origin $O(0, 0)$.
(a) Determine the total distance covered.
(b) Find the net displacement vector and its magnitude.
Solution:
(a) The path consists of two parts:
1. Semicircular arc from $A$ to $B$: length $s_1 = \pi R = \frac{22}{7} \times 7 = 22\text{ m}$.
2. Straight line from $B(+R, 0)$ to $O(0, 0)$: length $s_2 = R = 7\text{ m}$.
$\text{Total Distance } s = s_1 + s_2 = 22\text{ m} + 7\text{ m} = 29\text{ m}$.
(b) Initial position: $\vec{r}_i = -R\hat{i} = -7\hat{i}\text{ m}$.
Final position: $\vec{r}_f = 0\hat{i} = 0\text{ m}$.
$\Delta \vec{r} = \vec{r}_f – \vec{r}_i = 0 – (-7\hat{i}) = +7\hat{i}\text{ m}$.
Magnitude of displacement: $|\Delta \vec{r}| = 7\text{ m}$, directed along the $+x$-axis.
Example 3 (Multiple Turning Points with Cubic Kinematics):
A particle moves along a straight line such that its velocity is given by $v(t) = 3t^2 – 12t + 9\text{ m/s}$. Find the total distance traveled by the particle in the time interval from $t = 0$ to $t = 4\text{ s}$.
Solution:
1. Find turning points by solving $v(t) = 0$:
$3t^2 – 12t + 9 = 0 \implies t^2 – 4t + 3 = 0 \implies (t – 1)(t – 3) = 0$.
The particle reverses its direction at two points within the interval: $t_1 = 1\text{ s}$ and $t_2 = 3\text{ s}$.
2. Determine the position function $x(t) = \int v(t)\,dt$ assuming $x(0) = 0$:
$x(t) = \int (3t^2 – 12t + 9)\,dt = t^3 – 6t^2 + 9t$.
3. Evaluate position at critical timestamps:
$x(0) = 0\text{ m}$
$x(1) = 1^3 – 6(1)^2 + 9(1) = 1 – 6 + 9 = +4\text{ m}$
$x(3) = 3^3 – 6(3)^2 + 9(3) = 27 – 54 + 27 = 0\text{ m}$
$x(4) = 4^3 – 6(4)^2 + 9(4) = 64 – 96 + 36 = +4\text{ m}$
4. Sum individual segment distances:
– Segment 1 ($0 \le t \le 1\text{ s}$): $s_1 = |x(1) – x(0)| = |4 – 0| = 4\text{ m}$
– Segment 2 ($1 \le t \le 3\text{ s}$): $s_2 = |x(3) – x(1)| = |0 – 4| = 4\text{ m}$
– Segment 3 ($3 \le t \le 4\text{ s}$): $s_3 = |x(4) – x(3)| = |4 – 0| = 4\text{ m}$
$\text{Total Distance } s = 4 + 4 + 4 = 12\text{ m}$.
(Note: Net displacement is $\Delta x = x(4) – x(0) = 4\text{ m}$).
Example 4 (Relative Motion & Frame of Reference):
Two trains $A$ and $B$, each of length $120\text{ m}$, are traveling on parallel tracks with uniform speeds $v_A = 20\text{ m/s}$ and $v_B = 10\text{ m/s}$ in opposite directions. Compute:
(a) The relative speed of train $A$ with respect to train $B$.
(b) The time taken for the trains to completely pass each other.
(c) The distance traveled by train $A$ in the ground frame during the crossing.
Solution:
(a) Choosing the direction of train $A$ as positive $+x$:
$\vec{v}_A = +20\hat{i}\text{ m/s}, \quad \vec{v}_B = -10\hat{i}\text{ m/s}$
Relative velocity of $A$ with respect to $B$:
$\vec{v}_{AB} = \vec{v}_A – \vec{v}_B = 20 – (-10) = 30\hat{i}\text{ m/s}$.
Relative speed $= 30\text{ m/s}$.
(b) For complete crossing, the relative distance to be covered is the sum of their lengths:
$s_{\text{rel}} = L_A + L_B = 120\text{ m} + 120\text{ m} = 240\text{ m}$.
$\text{Time } t = \frac{s_{\text{rel}}}{v_{\text{rel}}} = \frac{240\text{ m}}{30\text{ m/s}} = 8\text{ s}$.
(c) In the ground frame, train $A$ moves at $20\text{ m/s}$ for $8\text{ s}$:
$d_A = v_A \times t = 20\text{ m/s} \times 8\text{ s} = 160\text{ m}$.
Worksheet: 10 Practice Problems
Problem 1 (JEE Main – Single Correct):
A particle moves along a circular path of radius $R$. What is the magnitude of displacement of the particle after it completes three-fourths ($3/4$) of a revolution?
(A) $\frac{3}{2}\pi R$
(B) $\sqrt{2} R$
(C) $2 R$
(D) $\frac{\sqrt{3}}{2} R$
Problem 2 (JEE Main – Single Correct):
The position of an object moving along the $x$-axis is given by $x(t) = 6t – t^2\text{ m}$, where $t$ is in seconds. The total distance traveled by the object between $t = 0$ and $t = 5\text{ s}$ is:
(A) $5\text{ m}$
(B) $9\text{ m}$
(C) $13\text{ m}$
(D) $17\text{ m}$
Problem 3 (JEE Main – Single Correct):
An athlete runs along a semicircular track of radius $r$. The ratio of the distance traveled to the magnitude of displacement is:
(A) $\pi$
(B) $\frac{\pi}{2}$
(C) $2\pi$
(D) $\frac{2}{\pi}$
Problem 4 (JEE Main – Single Correct):
An airplane travels $400\text{ m}$ North, $300\text{ m}$ West, and then flies vertically upwards by $1200\text{ m}$. The magnitude of net displacement of the airplane from its initial point is:
(A) $1900\text{ m}$
(B) $1300\text{ m}$
(C) $1500\text{ m}$
(D) $1700\text{ m}$
Problem 5 (JEE Main – Single Correct):
A wheel of radius $R = 1\text{ m}$ rolls forward without slipping on a horizontal ground through half a revolution. The magnitude of the displacement of the point of the wheel initially in contact with the ground is:
(A) $2\text{ m}$
(B) $\pi\text{ m}$
(C) $\sqrt{\pi^2 + 4}\text{ m}$
(D) $\sqrt{\pi^2 + 2}\text{ m}$
Problem 6 (JEE Advanced – One or More Correct):
Which of the following statements is/are TRUE regarding distance and displacement?
(A) The magnitude of displacement between two points is uniquely defined, but the distance between them can have infinite values.
(B) If the distance covered by a particle over a time interval is zero, its displacement must necessarily be zero.
(C) If the displacement of a particle over a time interval is zero, its distance traveled must necessarily be zero.
(D) For motion along a straight line, distance is equal to the magnitude of displacement if and only if velocity does not change its sign.
Problem 7 (JEE Advanced – One or More Correct):
A particle moves along the $x$-axis such that its coordinate varies as $x(t) = 2t^3 – 9t^2 + 12t + 1\text{ m}$. Which of the following statements is/are correct?
(A) The particle momentarily comes to rest at $t = 1\text{ s}$ and $t = 2\text{ s}$.
(B) The net displacement of the particle between $t = 0$ and $t = 3\text{ s}$ is $9\text{ m}$.
(C) The total distance traveled by the particle between $t = 0$ and $t = 3\text{ s}$ is $11\text{ m}$.
(D) The acceleration of the particle becomes zero at $t = 1.5\text{ s}$.
Problem 8 (JEE Advanced – One or More Correct):
Consider two frames of reference: Frame $S_1$ (fixed to the ground) and Frame $S_2$ (attached to a train accelerating with $\vec{a}_0$ relative to the ground). Which of the following statements is/are correct?
(A) A particle at rest in $S_1$ will appear accelerated with $-\vec{a}_0$ to an observer in $S_2$.
(B) The displacement of a particle measured over a time interval can have different magnitudes in $S_1$ and $S_2$.
(C) Distance covered by a particle between two events is invariant across all frames of reference.
(D) Newton’s second law $\vec{F}_{\text{net}} = m\vec{a}$ holds in $S_2$ only when a pseudo force $\vec{F}_p = -m\vec{a}_0$ is added to real forces.
Problem 9 (JEE Main / Advanced – Numerical Value Type):
A body moving along the $x$-axis has its displacement governed by $x(t) = 8t – 2t^2\text{ m}$, where $t$ is in seconds. The total distance covered by the body from $t = 0$ to $t = 3\text{ s}$ is $d\text{ m}$. Find the value of $d$.
Problem 10 (JEE Main / Advanced – Numerical Value Type):
A drunkard takes $5$ steps forward and $3$ steps backward, followed again by $5$ steps forward and $3$ steps backward, and so on. Each step is $1\text{ m}$ long and takes $1\text{ second}$. A deep pit is located at a distance of $13\text{ m}$ from the starting point on the forward path. The drunkard falls into the pit after a total time of $T\text{ seconds}$. Determine the integer value of $T$.
Solutions & Explanations
Answer Key Summary:
1. (B) | 2. (C) | 3. (B) | 4. (B) | 5. (C) | 6. (A, B, D) | 7. (A, B, C, D) | 8. (A, B, D) | 9. 10 | 10. 37
Solution 1:
Let the circle be centered at $(0, 0)$. Initial position: $\vec{r}_i = R\hat{i}$.
After $\frac{3}{4}$ revolution ($270^\circ$), the particle reaches $(0, -R)$ so $\vec{r}_f = -R\hat{j}$.
Displacement vector: $\Delta \vec{r} = \vec{r}_f – \vec{r}_i = -R\hat{i} – R\hat{j}$.
Magnitude: $|\Delta \vec{r}| = \sqrt{(-R)^2 + (-R)^2} = \sqrt{2} R$.
Correct Answer: (B)
Solution 2:
$x(t) = 6t – t^2 \implies v(t) = \frac{dx}{dt} = 6 – 2t$.
Setting $v(t) = 0 \implies t = 3\text{ s}$ (turning point).
– At $t = 0$: $x(0) = 0\text{ m}$.
– At $t = 3\text{ s}$: $x(3) = 6(3) – 3^2 = 18 – 9 = 9\text{ m}$.
– At $t = 5\text{ s}$: $x(5) = 6(5) – 5^2 = 30 – 25 = 5\text{ m}$.
Total distance $s = |x(3) – x(0)| + |x(5) – x(3)| = |9 – 0| + |5 – 9| = 9 + 4 = 13\text{ m}$.
Correct Answer: (C)
Solution 3:
For a semicircle of radius $r$:
Distance traveled = half circumference = $\pi r$.
Displacement = straight line diameter = $2 r$.
Ratio $= \frac{\text{Distance}}{|\text{Displacement}|} = \frac{\pi r}{2 r} = \frac{\pi}{2}$.
Correct Answer: (B)
Solution 4:
Choosing axes: North along $+y$, East along $+x$, Vertically upwards along $+z$.
Displacement vector: $\Delta \vec{r} = -300\hat{i} + 400\hat{j} + 1200\hat{k}\text{ m}$.
Magnitude: $|\Delta \vec{r}| = \sqrt{(-300)^2 + (400)^2 + (1200)^2} = \sqrt{90000 + 160000 + 1440000} = \sqrt{1690000} = 1300\text{ m}$.
Correct Answer: (B)
Solution 5:
In pure rolling through half a revolution:
The horizontal forward displacement of the wheel’s axis and contact point is half the circumference: $\Delta x = \pi R$.
The vertical displacement of the point originally at bottom (now at the top) is the diameter: $\Delta y = 2 R$.
Net displacement: $|\Delta \vec{r}| = \sqrt{(\Delta x)^2 + (\Delta y)^2} = \sqrt{(\pi R)^2 + (2 R)^2} = R\sqrt{\pi^2 + 4}$.
Since $R = 1\text{ m}$, $|\Delta \vec{r}| = \sqrt{\pi^2 + 4}\text{ m}$.
Correct Answer: (C)
Solution 6:
– (A) True: Infinite trajectories connect two points, but only one direct displacement vector exists.
– (B) True: Distance is strictly non-negative; distance $= 0$ implies particle did not move at all, hence displacement $= 0$.
– (C) False: A closed loop has zero displacement but non-zero distance.
– (D) True: Without direction reversal, path length equals $|\Delta x|$.
Correct Answer: (A, B, D)
Solution 7:
$x(t) = 2t^3 – 9t^2 + 12t + 1$.
$v(t) = 6t^2 – 18t + 12 = 6(t^2 – 3t + 2) = 6(t – 1)(t – 2)$.
$v = 0$ at $t = 1\text{ s}$ and $t = 2\text{ s}$ (Statement A is True).
Evaluate positions:
$x(0) = 1\text{ m}$
$x(1) = 2 – 9 + 12 + 1 = 6\text{ m}$
$x(2) = 2(8) – 9(4) + 12(2) + 1 = 16 – 36 + 24 + 1 = 5\text{ m}$
$x(3) = 2(27) – 9(9) + 12(3) + 1 = 54 – 81 + 36 + 1 = 10\text{ m}$.
– Displacement between $t = 0$ and $t = 3\text{ s}$: $\Delta x = x(3) – x(0) = 10 – 1 = 9\text{ m}$ (Statement B is True).
– Distance: $s = |x(1) – x(0)| + |x(2) – x(1)| + |x(3) – x(2)| = |6 – 1| + |5 – 6| + |10 – 5| = 5 + 1 + 5 = 11\text{ m}$ (Statement C is True).
– Acceleration: $a(t) = \frac{dv}{dt} = 12t – 18 = 0 \implies t = 1.5\text{ s}$ (Statement D is True).
Correct Answer: (A, B, C, D)
Solution 8:
– (A) True: $\vec{a}_{P/S_2} = \vec{a}_{P/S_1} – \vec{a}_{S_2/S_1} = 0 – \vec{a}_0 = -\vec{a}_0$.
– (B) True: Displacement is frame-dependent (e.g., ball dropped in moving train has straight vertical displacement in train frame, parabolic in ground frame).
– (C) False: Distance depends on the chosen frame of reference.
– (D) True: Standard definition of non-inertial reference frames.
Correct Answer: (A, B, D)
Solution 9:
$x(t) = 8t – 2t^2 \implies v(t) = 8 – 4t = 0 \implies t = 2\text{ s}$ (turning point).
Positions:
$x(0) = 0\text{ m}$
$x(2) = 8(2) – 2(2^2) = 16 – 8 = 8\text{ m}$
$x(3) = 8(3) – 2(3^2) = 24 – 18 = 6\text{ m}$.
Total distance $d = |x(2) – x(0)| + |x(3) – x(2)| = |8 – 0| + |6 – 8| = 8 + 2 = 10\text{ m}$.
Correct Answer: 10
Solution 10:
In each cycle of $8\text{ steps}$ ($5\text{ forward} + 3\text{ backward}$):
– Net forward displacement $= 5 – 3 = 2\text{ m}$.
– Time taken $= 5 + 3 = 8\text{ seconds}$.
After $4\text{ full cycles}$ ($4 \times 8 = 32\text{ seconds}$):
– Net forward displacement $= 4 \times 2 = 8\text{ m}$.
In the next cycle, the drunkard steps forward by $5\text{ m}$:
– Position reached $= 8\text{ m} + 5\text{ m} = 13\text{ m}$.
At this exact moment, he reaches the pit and falls in!
Time for these last $5\text{ steps} = 5\text{ seconds}$.
Total time $T = 32\text{ s} + 5\text{ s} = 37\text{ seconds}$.
Correct Answer: 37