Catch It!

Relative velocity of the cars

$=$ $60-42 = 18kmph$ $=5ms^{-1}$

For simplicity ;

$\Rightarrow$ We can imagine a frame where the car travelling at $42kmph$ is at rest & the car travelling at $60kmph$ overtakes it with a speed of $5ms^{-1}$ .

Now , since each car is $5m$ long , Therefore , in our frame ,

$\Rightarrow$ The overtaking car travels a 'relative' distance of $\color{#3D99F6}{5+5=10m}$ with the relative speed of $5ms^{-1}$ for overtaking.

Hence, $\text {Time taken}$ $=$ $\dfrac {10}{5}$ $=2s$ .

Now , coming back to the original frame ,

This car travels for $2s$ with a speed of $60kmph$ or $\dfrac {50}{3} ms^{-1}$ .

Hence ,

$\text {Total road distance used for overtaking}$

$=$ $\dfrac {50}{3} \times 2$

$=$ $\color{#20A900}{\boxed {33.33 m}}$