Manuel a physics student leaves his dorm at a speed of 1.2 m/s heading for the p

Question

Manuel a physics student leaves his dorm at a speed of 1.2 m/s heading for the physics building 95 m away. Just as he leaves his dorm, Amy another physics student, leaves her physics building and heads towards Manuel at a steady 1.6 m/s. Manuel immediately spots her and begins speeding up at a constant rate of .075 m/s^2. Set the origin of the coordinates (x=0) at the physics building with the positive x direction pointing towards Manuel's dorm. Set time t=0 when Manuel and Amy leave their respective buildings. At what values of x and t do Manuel and Amy meet?

Explanation / Answer

The speed of Manuel V1 = 1.2m/s the speed of Amy V2 = 1.6m/s the acceleration a = 0.075m/s^2 the distance between two buildingsd = 95m When the students are met at a distance x from physics building x = 1.6 t and for Manuel 95 - x = 1.2m/s t + (1/2)(0.075) (t^2) 95 - 1.6t = 1.2t + 0.0375t^2 0.0375t^2 +2.8t -95 = 0 By solving the above equation                t = 25.3s Then the distance x = 1.6(25.3s)                              = 40.5m

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