Question

Two cars $A$ and $B$ move on a straight road with constant velocity. When they move in same direction, they come close by $8\text{m}$ in $10\text{s}$ and when they move in opposite direction towards each other, they come close by $20\text{m in 1 s}$. The speed of the cars A and B (in $m/s$) respectively are

• A. $9.6\text{and}10.4$
• B. $10.4\text{and}9.6$
• C. $20.8\text{and}10.4$
• D. $10.4\text{and}20.8$

Answer 1

The correct option is B $10.4\text{and}9.6$


In first case, when $A$ and $B$ are moving in same direction:
using relative velocity we can say,
$V_A-V_B=\frac{8}{10}.....\left(1\right)$

In second case, when $A$ and $B$ are moving in opposite direction:
using relative velocity we can say,
$V_A+V_B=\mathrm{20........}\left(2\right)$

Adding equations $\left(1\right)$ and $\left(2\right)$, we get:
$2V_A=20.8$
$\Rightarrow V_A=10.4m/s$
$\Rightarrow V_B=9.6m/s$