In the drawing above two rugby players start 41 meters apart at rest. Player 1 accelerates at 2.2 m/s^2 and player 2 accelerates at 1.7 m/s^2. How much time elapses before the players collide?

Respuesta :

Answer:

The time that elapses before the players collide is 4.59 secs

Step-by-step explanation:

The distance between the two players is 41 meters. Hence, they would have total distance of 41 meters by the time they collide.

We can write that

S₁ + S₂ = 41 m

Where S₁ is the distance covered by Player 1 before collision

and S₂ is the distance covered by Player 2 before collision

From one of the equations of kinematics for linear motion,

[tex]S = ut[/tex] + [tex]\frac{1}{2} at^{2}[/tex]

Where [tex]S[/tex] is distance

[tex]u[/tex] is the initial velocity

[tex]a[/tex] is acceleration

and [tex]t[/tex] is time

Since the players collide at the same time, then time spent by player 1 before collision equals time spent by player 2 before collision

That is, t₁ = t₂ = t

Where t₁ is the time spent by player 1

and t₂ is the time spent by player 2

For player 1

[tex]S[/tex] = S₁

[tex]u[/tex] = 0 m/s ( The player starts at rest)

[tex]a[/tex] = 2.2 m/s²

Then,

S₁ = [tex]0(t)[/tex] + [tex]\frac{1}{2} (2.2)t^{2}[/tex]

S₁ = [tex]1.1t^{2}[/tex]

[tex]t^{2} = \frac{S_{1} }{1.1}[/tex]

For player 2

[tex]S[/tex] = S₂

[tex]u[/tex] = 0 m/s ( The player starts at rest)

[tex]a[/tex] = 1.7 m/s²

Then,

S₂ = [tex]0(t)[/tex] + [tex]\frac{1}{2} (1.7)t^{2}[/tex]

S₂ = [tex]0.85t^{2}[/tex]

[tex]t^{2} = \frac{S_{2} }{0.85}[/tex]

Since the time spent by both players is equal, We can  write that

[tex]t^{2}[/tex] = [tex]t^{2}[/tex]

Then,

[tex]\frac{S_{1} }{1.1} = \frac{S_{2} }{0.85}[/tex]

[tex]0.85S_{1} = 1.1S_{2}[/tex]

From, S₁ + S₂ = 41 m

S₂ = 41  - S₁

Then,

[tex]0.85S_{1} = 1.1 ( 41 -S_{1} )[/tex]

[tex]0.85S_{1} = 45.1 - 1.1S_{1}[/tex]

[tex]0.85S_{1} + 1.1S_{1} = 45.1 \\1.95S_{1} = 45.1\\S_{1} = \frac{45.1}{1.95} \\S_{1} = 23.13 m[/tex]

This is the distance covered by the first player. We can then put this value into [tex]t^{2} = \frac{S_{1} }{1.1}[/tex] to determine how much time elapses before the players collide.

[tex]t^{2} = \frac{S_{1} }{1.1}[/tex]

[tex]t^{2} = \frac{23.13 }{1.1}[/tex]

[tex]t^{2} = 21.03\\t = \sqrt{21.03} \\t = 4.59 secs[/tex]

Hence, the time that elapses before the players collide is 4.59 secs