Ablue car pulls away from a red stop-light just after it has turned green with a constant acceleration of 0.3 m/s2. a green car arrives at the position of the stop-light 5 s after the light had turned green. what is the slowest constant speed which the green car can maintain and still catch up to the blue car?

3 m/s The slowest possible velocity for the green car will be such that it's velocity matches the velocity of the blue car at the instant the distance between them reaches 0. Reason is simple. If the green car is going faster than the blue car at the moment they reach a distance of 0, then it would be possible for the blue car to go slower. And if the blue car reaches a velocity greater than the green car prior to their touching, then the green car will never reach the blue car. The blue car's distance as a function of T can be expressed as b(T) = 0.5 * 0.3 * T^2 = 0.15*T^2 The distance the green car as a function of T can be expressed as g(T) = (T-5)*V The velocity of the blue car as a function of T v(T) = 0.3T Given the reasoning for the optimal speed, the equation for the green car's location can be rewritten as g(T) = (T-5)*0.3T And since we want the distance for both the green and blue cars to match, set the equations equal to each other. (T-5)*0.3T = 0.15*T^2 Now solve for T (T-5)*0.3T = 0.15*T^2 0.3T^2 - 1.5T = 0.15T^2 0.15T^2 - 1.5T = 0 We now have a quadratic equation with A = 0.15, B = -1.5, and C=0. Use the quadratic formula to find the roots which are -4.5 and 10. The value of 10 indicates that the interception will occur at T=10, so the velocity has to be v(T) = 0.3T v(10) = 0.3*10 = 3 m/s The minimum speed the green car can have and still reach the blue car is 3 m/s.