Question

A particle performs a rectilinear motion in such a way that its initial velocity has a direction opposite to its uniform acceleration. Let $x_A$ and $x_B$ be the magnitude of displacement in the first 10 seconds and the next 10 seconds respectively. Which of the following options is correct?
A) $x_A<x_B$
B) $x_A<x_B$
C) $x_A<x_B$
D) The information is not sufficient to decide the relation between $x_A$ and $x_B$.

Answer 1

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 $s=ut+{\displaystyle \frac{1}{2}}a{t}^2$ 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 $s=ut+{\displaystyle \frac{1}{2}}a{t}^2$
In the first $t=10\text{s}$ of the motion of the particle, the distance covered is
$\Rightarrow s_{10}=10u-{\displaystyle \frac{1}{2}}a\times {10}^2$
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 $s_{20}=20u-{\displaystyle \frac{1}{2}}a\times {20}^2$
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,
$\Rightarrow 3x_A=30u-150a$ ------- (5)
Now subtract equation (4) from (5) to get,
$\Rightarrow 3x_A-x_B=\left(30-10\right)u-\left(150-150\right)a$
$\Rightarrow 3x_A-x_B=20u$
Since the answer is positive we can say $3x_A>x_B$.

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.