COMEDK 2024 Shift 1 Paper

Let the speeds of the two objects be v1 and v2.
When the two objects are moving in the same direction, the relative speed is given by:
|v1−v2|
According to the problem, this relative speed results in an increase in the distance between them by 6 m in one second. Therefore, we have:
|v1−v2|=6
When the two objects are moving towards each other, the relative speed is given by:
v1+v2
In this case, the distance decreases by 8 m in one second, so we have:
v1+v2=8
We now have two equations:
‌|v1−v2|=6
‌v1+v2=8
We can solve these equations by considering the absolute value condition in two cases:
Case 1: v1≥v2
v1−v2=6‌‌(. from |v−1−v−2|=6)
v1+v2=8
Adding these two equations:
(v1−v2)+(v1+v2)=6+8

‌2v1=14
‌v1=7
Substituting v1=7 into v1+v2=8 :
‌7+v2=8
‌v2=1

‌‌ Case 2: ‌v1<v2
‌v2−v1=6‌‌(‌ from ‌|v−1−v−2|=6)
‌v1+v2=8
Adding these two equations:
‌(v2−v1)+(v1+v2)=6+8
‌2v2=14
‌v2=7
Substituting v2=7 into v1+v2=8 :

‌v1+7=8
‌v1=1
From both cases, we find that the speeds of the objects are 7ms−1 and 1ms−1.
Therefore, the correct answer is:
Option B: 7ms−1 and 1ms−1