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# Answer to Question #94315 in Mechanics | Relativity for ivwananji

Question #94315
on your wedding day you leave for the church 30.0 minutes before the ceremony is to begin which should be plenty of time since the church is only 10 miles away.on the way however you make an unanticipated for construction work on the road as a result your average speed for the first 15 minutes is only 5.0 mi h what average speed do you need for the rest of the trip to get to the church on time

@$T = 30\text{ min} = 0.5\text{ h}; D = 10\text{ mi}; \\ t_1 = 15\text{ min} = 0.25\text{ h}; v_1 = 5\frac{\text{mi}}{\text{h}}; \\ v_2 = ? \\ D = d_1 + d_2 = v_1t_1 + v_2t_2 = v_1t_1 + v_2(T-t_1); \\ v_2 = \frac{D-v_1t_1}{T-t_1} = \frac{10\text{ mi}-5\frac{\text{mi}}{\text{h}}0.25\text{ h}}{(0.5-0.25)\text{ h}} = 35\frac{\text{mi}}{\text{h}}. @$

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