Asked by Claire
A and B are 525 km apart. Michael left A at 8am and travelled to B at a constant speed of 75 km/h. Chad left A some time later on his way to B. Chad passed Michael at 10:30 am and then travelled at an average speed of 90 km/h for the rest of his journey. How far would Michael still need to travel when Chad arrives in B?
Answers
Answered by
Henry
T1 = 10:30 - 8:00 = 2.5 b.
d1 = V1*T1 = 75 * 2.5 = 187.5 km.
d2 = V*T, (525-187.5) = 80*T, T = 4.22 h. For Chad to reach B.
D1 = (525-187.5) - 75*4.22.
D1 = 337.5 - 316.5 = 21 = Dist. Michael still need to travel.
d1 = V1*T1 = 75 * 2.5 = 187.5 km.
d2 = V*T, (525-187.5) = 80*T, T = 4.22 h. For Chad to reach B.
D1 = (525-187.5) - 75*4.22.
D1 = 337.5 - 316.5 = 21 = Dist. Michael still need to travel.
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