Question

What are the speeds of two objects if, when they move uniformly towards each other, they get $4\text{m}$ closer in each second and when they move uniformly in the same direction with the original speeds, they get $4.0\text{m}$ closer in each $10s$?

• A. $2.8\text{m/s}$ and $1.2\text{m/s}$
• B. $5.2\text{m/s}$ and $4.6\text{m/s}$
• C. $3.2\text{m/s}$ and $2.1\text{m/s}$
• D. $2.2\text{m/s}$ and $1.8\text{m/s}$

Answer 1

The correct option is D $2.2\text{m/s}$ and $1.8\text{m/s}$

Let $\overrightarrow{V_A}$ and $\overrightarrow{V_B}$ be the velocities of two objects moving uniformly.
Case 1: When objects are moving opposite to each other:
In this case the relative speed of object $A$ w.r.t object $B$ is given by $V_{AB}=V_A-(-V_B)$ (∵$\overrightarrow{V_A}$ is in the opposite direction to that of object $\overrightarrow{V_B}$)
⟹$V_{AB}=V_A+V_B$
According to question :$(V_A+V_B)\times 1s=4\text{m}$
⟹$V_A+V_B=4\text{m/s}$ (i)
Case 2: When objects are moving in same direction.
In this case the relative speed of object $A$ w.r.t object $B$ is given by $V_{AB}=V_A-V_B$
According to question:
$(V_A-V_B)\times 10s=4\text{m}$
⟹$(V_A-V_B)=0.4\text{m/s}$ (ii)
Using (i) and (ii) we get $V_A=2.2\text{m/s}$ and $V_B=1.8\text{m/s}$