Question
Question: A particle performs a rectilinear motion in such a way that its initial velocity has a direction opp...
A particle performs a rectilinear motion in such a way that its initial velocity has a direction opposite to its uniform acceleration. Let xA and xB be the magnitude of displacement in the first 10 seconds and the next 10 seconds respectively. Which of the following options is correct?
A) xA<xB
B) xA<xB
C) xA<xB
D) The information is not sufficient to decide the relation between xA and xB.
Solution
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+21at2 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 xA and xB respectively.
We have to find the relation between xA and xB.
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+21at2
In the first t=10s of the motion of the particle, the distance covered is
⇒s10=10u−21a×102
On simplifying this becomes, s10=10u−50a.
Then the magnitude of displacement of the particle will be
⇒xA=∣s10−s0∣ -------- (1).
Substituting for s10=10u−50a in equation (1) we get,
⇒xA=∣10u−50a−0∣=10u−50a
Thus the magnitude of displacement for the first 10 seconds is xA=10u−50a ---------- (2).
Now the distance covered in t=20s is s20=20u−21a×202
On simplifying this becomes, s20=20u−200a.
Then the magnitude of displacement of the particle will be
⇒xB=∣s20−s10∣ ----------- (3).
Substituting for s10=10u−50a and s20=20u−200a in equation (3) we get,
⇒xB=∣(20u−200a)−(10u−50a)∣=10u−150a
Thus the magnitude of displacement for the next 10 seconds is
⇒xB=10u−150a ---------- (4).
Step 3: Using equations (2) and (4) obtain a relation between xA and xB.
Equation (2) gives xA=10u−50a and equation (4) gives xB=10u−150a.
Multiply equation (2) by 3 to get,
⇒3xA=30u−150a ------- (5)
Now subtract equation (4) from (5) to get,
⇒3xA−xB=(30−10)u−(150−150)a
⇒3xA−xB=20u
Since the answer is positive we can say 3xA>xB.
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, xA=0 and xB>0
⇒xA<xB
Case 2: The initial velocity of the particle is u=15a.
Then substituting for u=15a in equations (2) and (4) we get, xA>0 and xB=0
⇒xA>xB
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.