Question

A car travelling at a constant speed of 20m/s overtakes another car which is moving at a constant acceleration of 2 m/s^2 and it's initially at rest. Assume the length of each car to be 5 meters. What is the total road distance used in overtaking?

Answer 1

the length of distance covered in overtaking is 309.74 m.

A car travelling at a constant speed of 20m/s overtakes another car which is moving at a constant acceleration of 2 m/s² and it's initially at rest. The length of each car is 5 m.

• We have to find the total road distance used in overtaking.

• This can be solved using kinematic equations of relative motion.

Time taken to travel d distance by first car = (d/20) sec

So distance traveled by 2nd car = ut + 1/2 at² = 0 × t + 1/2 × 2 × (d/20)² = (d/20)²

To overtake the 2nd car,

distance traveled by first car + 2 × length of car = distance traveled by 2nd car

⇒ d + 10 = (d/20)²

⇒400d + 4000 = d²

⇒d² - 400d - 4000 = 0

⇒d = {400 ± √(160000 + 16000)}/2

= (400 ± √(176000)}/2

≈ 309.74 m

Therefore the length of distance covered in overtaking is 309.74 m.