ANSWERS: 1
  • Let X = speed of faster train Let Y = speed of slower train Let T = the time in seconds that the faster train covers 4km Let T + 15 = the time in seconds that the slower train covers 4km speed x time = distance The equation for the faster train travling the 4 km is: X(T) = 4km The equation for the slower train travling the 4 km is: Y(T+15) = 4km In 15 minutes the difference between the distance traveled by the faster train and the slower train is 1 km: 15 minutes = 900 seconds X(900) - Y(900) = 1km Let's substitute the equations for X and Y into the above equation: X=4km/T Y=4km/(T+15) 900(4/T) - 900{4/(T+15)}= 1 km Multiply both sides by T(T+15) 900{4(T+15)} - 900{4T}= T(T+15) Multiplying the term on both sides: 3600T+54000-3600T = T^2 + 15T 54000 = T^2 + 15T Subtracting 54000 from both sides: 0 = T^2 + 15T - 54000 Rewriting the equation into standard quadratic form: T^2 + 15T -54000 = 0 Use the quadratic formula T = [-b ± sqrt{b^2-4ac}]/{2(a)} but we will not use the -b-sqrt{b^2-4ac} value, because that would give us a negative value for time. Substituting a = 1 b = 15 c = -54000 into the quadratic formula: T = [-15+sqrt{15^2 - 4(1)(-54000)}]/{2(1)} T = [-15+sqrt{ 225 + 216000}]/2 T = [-15+sqrt{216225}]/2 T = [-15+465]/2 T = 450/2 T = 225 seconds The speed of the faster train is: X = 4 km/225 sec = 0.0177... km/sec = 17.777... meters/sec You did not ask for it but the speed of the slower train is: Y = 4 km/(225 + 15) Y = 4 km/240 sec = 0.0166... km/sec = 16.666... meters/sec

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