Asked by Anonymous
Hillary and Charlene both drove from City A to City B. At 10a, Hilary left City A and drove at an average speed of 120km/h. Charlene drove at an average speed of 144 km/h and took 50 mins. She arrived at City B at the same time as Hillary. Find the time Charlene left City A.
Answers
Answered by
Reiny
"Charlene drove at an average speed of 144 km/h and took 50 mins"
---> the distance she drove is 144(50/60) = 120 km
since Hillary went the same distance at 120 km/h, her time was
120/120 = 1 hour
Can you take it from there?
---> the distance she drove is 144(50/60) = 120 km
since Hillary went the same distance at 120 km/h, her time was
120/120 = 1 hour
Can you take it from there?
Answered by
henry2,
d1 = d2.
V1*T1 = V2*T2.
120 * T1= 144 * (50/60),
T1 = 1 h. = Hillary's driving time.
10 AM + 1h = 11:00 AM = their arrival time.
11:00 - 55 = 10:60 - 55 = 10:05 AM. = The time Charlene left city A.
V1*T1 = V2*T2.
120 * T1= 144 * (50/60),
T1 = 1 h. = Hillary's driving time.
10 AM + 1h = 11:00 AM = their arrival time.
11:00 - 55 = 10:60 - 55 = 10:05 AM. = The time Charlene left city A.
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