There are four traffic lights on a stretch of road and the distance between each traffic light is 300 m. There is a 10-s lag time between successive green lights: The second light turns green 10 s after the first light turns green, the third light turns green 10 s after the second light does (and, thus, 20 s after the first light turns green), and so on. Each light stays green for 15 s. Having watched this sequence for some time, you understand that you can drive through the stretch of road at a range of constant speeds. You drive through the first light right when it turned green. What is the fastest speed that you can drive through the stretch of road without running any red lights

    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

Newtons laws of motion

2.77 x 10^-3 its for plato user