Question

Two cars get closer by $9\,mm$ every second while travelling in the opposite directions. They get closer by $1\,m$ every second while travelling in the same direction, what is the speed of the cars$?$
A) $5\,m{s^{ - 1}}$ and $4\,m{s^{ - 1}}$
B) $4\,m{s^{ - 1}}$ and $3\,m{s^{ - 1}}$
C) $6\,m{s^{ - 1}}$ and $3\,m{s^{ - 1}}$
D) $6\,m{s^{ - 1}}$ and $5\,m{s^{ - 1}}$

Answer 1

Here we know the total time of the crossing cars in the opposite direction and the same direction that when two bodies pass in the same direction, the relative velocity is equal to the number of velocities so that we can measure the distance.

Formula used:
The relative speed in opposite direction,
$\left( {x + y} \right)\,m/s$
The relative speed in same direction,
$\left( {x - y} \right)\,m/s$
Where,
$x,y$ are the distance points

Complete step by step solution:
Given by,
Distance in opposite direction $= 9\,m$
Distance in same direction $= 1\,m$
Let the speed of first and second car is $x$ and $y$
Here,
We need to calculate the speed of both cars
According to the relative speed formula,
We using,
The relative speed in opposite direction,
$\Rightarrow$ $x + y = 9$………….$(i)$
The relative speed in same direction,
$\Rightarrow$ $x - y = 1$………….$(ii)$
Combine the both equation $(i)$ and $(ii)$
By solving,
We get,
$\Rightarrow$ $2x = 10$
On simplifying,
Here,
$\Rightarrow$ $x = 5\,m/s$
Now,
We put the value of $x$ in equation $(i)$
$\Rightarrow$ $5 + y = 9$
On solving,
We get,
$\Rightarrow$ $y = 4\,m/s$
Thus, the speed of the cars $5\,m{s^{ - 1}}$ and $4\,m{s^{ - 1}}$.

Hence, option A is the correct answer.

Note: Relative velocity is the scalar quantity, while relative velocity is the quantity of the vector. One body will make a stationary velocity equal to zero and take the other body's velocity with respect to the stationary body, which is the sum of the velocities of the bodies moving in the opposite direction.