Answer
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Hint:
Displacement is a vector quantity that has both direction and magnitude. It refers to the difference between the initial position and the final position of the particle for a given time interval. The magnitude of displacement is always positive and can be determined if the velocity of the particle in a given time interval is known.
Formula used:
Newton’s first equation of motion gives the distance covered by a body as $s = ut + \dfrac{1}{2}a{t^2}$ where $u$ is the initial speed of the body, $a$ is the acceleration of the body and $t$ is the time taken to cover the distance.
Complete step by step answer:
Step 1: List the key points provided in the question.
The initial velocity of the particle undergoing a rectilinear motion is directed opposite to its uniform acceleration.
Let $u$ be the initial velocity of the particle and $a$ be the uniform acceleration of the particle.
It is given that at points A and B, the magnitudes of the displacement of the particle are ${x_A}$ and ${x_B}$ respectively.
We have to find the relation between ${x_A}$ and ${x_B}$.
Step 2: Using Newton’s first equation of motion, express the displacements of the particle in terms of its initial velocity and acceleration.
Newton’s first equation of motion gives the distance covered by the particle as $s = ut + \dfrac{1}{2}a{t^2}$
In the first $t = 10{\text{s}}$ of the motion of the particle, the distance covered is
$\Rightarrow {s_{10}} = 10u - \dfrac{1}{2}a \times {10^2}$
On simplifying this becomes, ${s_{10}} = 10u - 50a$.
Then the magnitude of displacement of the particle will be
$\Rightarrow {x_A} = \left| {{s_{10}} - {s_0}} \right|$ -------- (1).
Substituting for ${s_{10}} = 10u - 50a$ in equation (1) we get,
$\Rightarrow {x_A} = \left| {10u - 50a - 0} \right| = 10u - 50a$
Thus the magnitude of displacement for the first 10 seconds is ${x_A} = 10u - 50a$ ---------- (2).
Now the distance covered in $t = 20{\text{s}}$ is ${s_{20}} = 20u - \dfrac{1}{2}a \times {20^2}$
On simplifying this becomes, ${s_{20}} = 20u - 200a$.
Then the magnitude of displacement of the particle will be
$\Rightarrow {x_B} = \left| {{s_{20}} - {s_{10}}} \right|$ ----------- (3).
Substituting for ${s_{10}} = 10u - 50a$ and ${s_{20}} = 20u - 200a$ in equation (3) we get,
$\Rightarrow {x_B} = \left| {\left( {20u - 200a} \right) - \left( {10u - 50a} \right)} \right| = 10u - 150a$
Thus the magnitude of displacement for the next 10 seconds is
$\Rightarrow {x_B} = 10u - 150a$ ---------- (4).
Step 3: Using equations (2) and (4) obtain a relation between ${x_A}$ and ${x_B}$.
Equation (2) gives ${x_A} = 10u - 50a$ and equation (4) gives ${x_B} = 10u - 150a$.
Multiply equation (2) by 3 to get,
$\Rightarrow 3{x_A} = 30u - 150a$ ------- (5)
Now subtract equation (4) from (5) to get,
$\Rightarrow 3{x_A} - {x_B} = \left( {30 - 10} \right)u - \left( {150 - 150} \right)a$
$ \Rightarrow 3{x_A} - {x_B} = 20u$
Since the answer is positive we can say $3{x_A} > {x_B}$.
But unless the value of the initial velocity $u$ is known, we cannot determine which displacement is greater. So the correct option is D.
Note:
Alternate method:
We consider two cases.
Case 1: The initial velocity of the particle is $u = 5a$.
Then substituting for $u = 5a$ in equations (2) and (4) we get, ${x_A} = 0$ and ${x_B} > 0$
$ \Rightarrow {x_A} < {x_B}$
Case 2: The initial velocity of the particle is $u = 15a$.
Then substituting for $u = 15a$ in equations (2) and (4) we get, ${x_A} > 0$ and ${x_B} = 0$
$ \Rightarrow {x_A} > {x_B}$
But the relations obtained in the two cases contradict each other. So the correct relation between the magnitude of the displacements cannot be determined unless $u$ and $a$ are known.
Displacement is a vector quantity that has both direction and magnitude. It refers to the difference between the initial position and the final position of the particle for a given time interval. The magnitude of displacement is always positive and can be determined if the velocity of the particle in a given time interval is known.
Formula used:
Newton’s first equation of motion gives the distance covered by a body as $s = ut + \dfrac{1}{2}a{t^2}$ where $u$ is the initial speed of the body, $a$ is the acceleration of the body and $t$ is the time taken to cover the distance.
Complete step by step answer:
Step 1: List the key points provided in the question.
The initial velocity of the particle undergoing a rectilinear motion is directed opposite to its uniform acceleration.
Let $u$ be the initial velocity of the particle and $a$ be the uniform acceleration of the particle.
It is given that at points A and B, the magnitudes of the displacement of the particle are ${x_A}$ and ${x_B}$ respectively.
We have to find the relation between ${x_A}$ and ${x_B}$.
Step 2: Using Newton’s first equation of motion, express the displacements of the particle in terms of its initial velocity and acceleration.
Newton’s first equation of motion gives the distance covered by the particle as $s = ut + \dfrac{1}{2}a{t^2}$
In the first $t = 10{\text{s}}$ of the motion of the particle, the distance covered is
$\Rightarrow {s_{10}} = 10u - \dfrac{1}{2}a \times {10^2}$
On simplifying this becomes, ${s_{10}} = 10u - 50a$.
Then the magnitude of displacement of the particle will be
$\Rightarrow {x_A} = \left| {{s_{10}} - {s_0}} \right|$ -------- (1).
Substituting for ${s_{10}} = 10u - 50a$ in equation (1) we get,
$\Rightarrow {x_A} = \left| {10u - 50a - 0} \right| = 10u - 50a$
Thus the magnitude of displacement for the first 10 seconds is ${x_A} = 10u - 50a$ ---------- (2).
Now the distance covered in $t = 20{\text{s}}$ is ${s_{20}} = 20u - \dfrac{1}{2}a \times {20^2}$
On simplifying this becomes, ${s_{20}} = 20u - 200a$.
Then the magnitude of displacement of the particle will be
$\Rightarrow {x_B} = \left| {{s_{20}} - {s_{10}}} \right|$ ----------- (3).
Substituting for ${s_{10}} = 10u - 50a$ and ${s_{20}} = 20u - 200a$ in equation (3) we get,
$\Rightarrow {x_B} = \left| {\left( {20u - 200a} \right) - \left( {10u - 50a} \right)} \right| = 10u - 150a$
Thus the magnitude of displacement for the next 10 seconds is
$\Rightarrow {x_B} = 10u - 150a$ ---------- (4).
Step 3: Using equations (2) and (4) obtain a relation between ${x_A}$ and ${x_B}$.
Equation (2) gives ${x_A} = 10u - 50a$ and equation (4) gives ${x_B} = 10u - 150a$.
Multiply equation (2) by 3 to get,
$\Rightarrow 3{x_A} = 30u - 150a$ ------- (5)
Now subtract equation (4) from (5) to get,
$\Rightarrow 3{x_A} - {x_B} = \left( {30 - 10} \right)u - \left( {150 - 150} \right)a$
$ \Rightarrow 3{x_A} - {x_B} = 20u$
Since the answer is positive we can say $3{x_A} > {x_B}$.
But unless the value of the initial velocity $u$ is known, we cannot determine which displacement is greater. So the correct option is D.
Note:
Alternate method:
We consider two cases.
Case 1: The initial velocity of the particle is $u = 5a$.
Then substituting for $u = 5a$ in equations (2) and (4) we get, ${x_A} = 0$ and ${x_B} > 0$
$ \Rightarrow {x_A} < {x_B}$
Case 2: The initial velocity of the particle is $u = 15a$.
Then substituting for $u = 15a$ in equations (2) and (4) we get, ${x_A} > 0$ and ${x_B} = 0$
$ \Rightarrow {x_A} > {x_B}$
But the relations obtained in the two cases contradict each other. So the correct relation between the magnitude of the displacements cannot be determined unless $u$ and $a$ are known.
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