Answer to Question #138530 in Classical Mechanics for Paul Kachimba

Question

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 10.0 miles away. On the way, however, you have to make an unanticipated stop for construction work on the road. As a result, your average speed for the first 15 minutes is only 5.0 miles/h. What average speed do you need for the rest of the trip to get you to the church on time?

Answer 1

Time= 30 mins which is 0.5 hours

Distance =10 miles

$t_1=15min =0.25h v_1=5mi/h$

$v_2=?$

$D=d_1+d_2=v_1+v_2t_2=v_1t_2+v_2(T-t_1)$

$V_2=\frac{D-v_1t_1}{T-t_1}=\frac{10mi-5\frac{mi}{h}0.25h}{(0.5-0.25)h}=35mi/h$