Question

In the drawing above two rugby players start 51 meters apart at rest. Player 1 accelerates at 1.9 m/s^2 and player 2 accelerates at 1.2 m/s^2. How much time elapses before the players collide

Answer 1

12.1 s

Step-by-step explanation:

Since the two rugby players start at rest, we use the equation of motion

x - x₀ = ut + 1/2at² where x₀ = initial position, x = final position, u = initial velocity, a = acceleration and t = time

Now for Player 1, x₀ = 0 , u = 0 and a = 1.9 m/s²

So, x₁ - 0 = 0t + 1/2(1.9 m/s²)t² = 0.95t²

x₁ = 0.95t²

Now for Player 2, x₀ = 51 m , u = 0 m/s and a = 1.2 m/s²

So, x₂ - 51 = 0t + 1/2(1.2 m/s²)t² = 0.6t²

x₂ - 51 = 0.6t²

x₂ = 51 + 0.6t²

We find the time when x₁ = x₂, when the players collide

So, equating both expressions, we have

0.95t² = 51 + 0.6t²

collecting like terms, we have

0.95t² - 0.6t² = 51

0.35t² = 51

t² = 51/0.35

t² = 145.714

t = ±√145.714

t = ± 12.07 s

t ≅ ±12.1 s

Since t cannot be negative, we take t = + 12.1 s.

So, 12.1 s elapses before the players collide.