Question

Two cars $\mathrm{A}$ and $\mathrm{B}$ are moving with the same speed of $45\mathrm{km}/\mathrm{h}$ along the same direction. If a third car $\mathrm{C}$ comes from the opposite direction with a speed of $36\mathrm{km}/\mathrm{h}$ meets two cars in an interval of $5\min$, the distance of separation of two cars $\mathrm{A}$ and $\mathrm{B}$ should be (in km).

• A. $6.75km$
• B. $7.25km$
• C. $5.55km$
• D. $8.35km$
• E. $4.75km$

Answer 1

The correct option is A

$6.75km$

Step 1. Given data

Two cars $\mathrm{A}$ and $\mathrm{B}$ are moving with same speed = $45\mathrm{km}/\mathrm{h}$ (along same direction)

Third car $\mathrm{C}$ coming from the opposite direction with a speed = $36\mathrm{km}/\mathrm{h}$

Car C meets the other two cars in an interval of $5\min$.

We have to find distance between two cars.

Step 2. Formula to be used

Relative velocity is given as,

Velocity of approach $=\frac{\mathrm{Distance}\mathrm{between}\mathrm{two}\mathrm{cars}}{\mathrm{Time}\mathrm{difference}\mathrm{in}\mathrm{meeting}\mathrm{of}\mathrm{two}\mathrm{cars}}$

Step 3. Calculating the relative velocity

${\mathrm{V}}_{\mathrm{A}}={\mathrm{V}}_{\mathrm{B}}$

$=45\mathrm{km}/\mathrm{hr}$

Let us assume that, $\mathrm{x}$ is the distance between $\mathrm{A}$ and $\mathrm{B}$.

So, the velocity of approach between two cars is,

${\mathrm{V}}_{\mathrm{AC}}={\mathrm{V}}_{\mathrm{A}}+{\mathrm{V}}_{\mathrm{B}}$

$=45+36$

$=81\mathrm{km}/\mathrm{hr}$

Now,

${\mathrm{V}}_{\mathrm{BC}}={\mathrm{V}}_{\mathrm{B}}+{\mathrm{V}}_{\mathrm{C}}$

$=45+36$

$=81\mathrm{km}/\mathrm{hr}$

Step 4. Calculating the distance between two cars

From the given,

$∆\mathrm{t}=5\min$

Put all the values in the formula of the velocity of approach we get,

$81=\frac{\mathrm{x}\times 60}{5}$

$\mathrm{x}=\frac{5\times 81}{60}$

$=6.75\mathrm{km}$

Hence, option (A) is the correct answer.