WBJEE 2024 Math Paper

The particle will come to rest when its velocity is zero. To find the velocity, we differentiate the position function with respect to time:
v=‌

dx

dt

=a√λ‌cos(√λt+b)
Setting the velocity to zero, we get:
a√λ‌cos(√λt+b)=0
This equation is satisfied when:
cos(√λt+b)=0
The cosine function is zero at odd multiples of ‌

π

2

. Therefore, we have:
√λt+b=(2n+1)‌

π

2

, where n is an integer.
Solving for t, we get:
t=‌

(2n+1) π 2 −b

√λ

This gives us an infinite number of times when the particle comes to rest. Let's consider the first two instances ( n=0 and n=1 ):
‌t1=‌

‌ π 2 −b

√λ

‌t2=‌

‌ 3π 2 −b

√λ

Now, we can find the positions of the particle at these times by plugging them back into the original position function:
‌x1=asin‌(√λt1+b)=asin‌(‌

π

2

)=a
‌x2=asin‌(√λt2+b)=asin‌(‌

3π

2

)=−a
The distance between these two points is:
|x2−x1|=|−a−a|=2a
Therefore, the correct answer is Option C: 2a.