Respuesta :
3 m/s The green car's slowest feasible speed will be such that it coincides with the blue car's speed when the distance between them equals zero. Simple explanation The blue automobile might go slower if the green car is moving faster than it is at the point where the distance between them equals 0. Furthermore, the green automobile will never catch up to the blue car if the blue car moves faster than the green car before they collide.
The formula for the distance of the blue automobile as a function of T is b(T) = 0.5 * 0.3 * T2 = 0.15*T2. The formula for the green car's distance as a function of T is g(T) = (T-5)*V. The blue car's speed as a function of time is given by v(T) = 0.3T. The equation for the position of the green automobile can be rewritten as g(T) = (T-5)*0.3T given the justification for the ideal speed. Set the equations to be equal to one another since we want the distance between the green and blue automobiles to be the same. (T-5) *0.3T = 0.15*T^2 Then, calculate T (T-5)*0.3T = 0.15*T2, where 0.3T2 - 1.5T equals 0.15T2 and 1.5T - 0.15T2 = 0. Our quadratic equation now has the values A = 0.15, B = -1.5, and C = 0.
To determine the roots, which are -4.5 and 10, use the quadratic formula. The interception will happen at T=10 because the number of 10 implies that it will, hence the velocity must be v(T) = 0.3T v(10) = 0.3*10 = 3 m/s. The green car must travel at least 3 m/s in order to catch up to the blue car.
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