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?

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,

+

Where is distance

is the initial velocity

is acceleration

and 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

= S₁

= 0 m/s ( The player starts at rest)

= 2.2 m/s²

Then,

S₁ = +

S₁ =

For player 2

= S₂

= 0 m/s ( The player starts at rest)

= 1.7 m/s²

Then,

S₂ = +

S₂ =

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

=

Then,

From, S₁ + S₂ = 41 m

S₂ = 41  - S₁

Then,

This is the distance covered by the first player. We can then put this value into to determine how much time elapses before the players collide.

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

an

y = 6x -3y -12

y +3y +12 = 6x

=> . x = { 2y / 3 +2 } this is equation of line in terms of x

=> . if line meets the x-axis the co-ordinates of meeting point would be ( x , 0 )

=> . x = 2 ( 0 ) / 3 + 2

=> . x = 2

so, the point is ( 2 , 0 )