An excellent swimmer is confident he can judge his speed in the water exactly. he bets he can swim two lengths of a 50-meter olympic size pool at an average speed of precisely 2 meters per second. on his first lane he is a little bit slow and ends up going at 1.5 meters per second. which average speed does he have to reach on his second lane to win the bet?

First lane:

V = 1.5 m/s
D = 50 m
t = D/V = (50m/1.5m/s)  =

Total time to win the bet

V = 2m/s
D=100m
t = 100m /  2m/s = 50 s

Second lane

t left = 50 s - 50/1.5 s
D = 50 m

V = D/t = 50m/[50 -50/1.5] = 3m/s

  he needs to average 3 m/s on the second lane to wing the bet.

    m₁v₁ + m₂v₂ = (m₁ + m₂)vmomentum is conserved.the left-hand side of the equation represents the momentum in the system before the collision.the right-hand side represents the momentum in the system after the collisionm₁ and v₁ represents the mass and the velocity of the physics blockm₂ and v₂ represents the mass and velocity of the ball of puttyv represents the final velocity of the combined mass.the question states the two masses stick together. they combined mass has its own velocity with the body mass the sum of the two objects' massestheir combined momentum is (m₁ + m₂)vto find the magitude of (m₁ + m₂)v, we need to add the vectors on the left-hand side of the equation: m₁v₁ + m₂v₂.note the diagram. i have positioned the vectors in accordance with their directions.the magnitude of the resultant can be calculated using the pythagorean theorem since this is a right triangle situation (north and west are perpendicular)|(m₁ + m₂)v| =  √[ (m₁v₁)² + (m₂v₂)² ]|(m₁ + m₂)v| =  √[ (1.5kg·8.0m/s)² + (0.40kg·20m/s)² ]|(m₁ + m₂)v| =  14.4 kg m/sfor two significant figures (assuming that 20m/s is two), this rounds to 14 kg m/s