After robbing a bank in Dodge City, a robber gallops off at 14 mi/h. 20 minutes later, the marshall leaves to pursue the robber at 16 mi/h?How long (in hours) does it take the marshall to catch up to the robber? |
After robbing a bank in Dodge City, a robber gallops off at 14 mi/h. 20 minutes later, the marshall leaves to pursue the robber at 16 mi/h?How long (in hours) does it take the marshall to catch up to the robber? |
Robber has run 20 minutes in advance. Therefore he hasheref already covered the distance
20/60*14=14/3 miles. The marshal starts half an hour later. He can cover a distance of (16-14) m in an hour, Therefore to cover the distance 14/3 miles he needs,
14/3*1/2 hours = 7/3 hours = 2 hours and twenty minutes.
Let t = time for marshall to catch the robber after the bank is robbed
Let VR = the robber's speed
Let VM = the marshall's speed
d = dR = ( VR ) ( t )
d = dM = ( VM ) ( t - 1/3 )
d = dR = dM = ( VR ) ( t ) = ( VM ) ( t - 1/3 )
( 14 ) ( t ) = ( 16 ) ( t - 1/3 ) = 16t - 16/3
Multiply both sides by 3 and get :
( 48 ) ( t ) = ( 48 ) ( t ) - 16
6t = 16
t = 16/6 = 8/3 h = 2 2/3 h <-------------------
Check the answer :
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dR = ( 14 mi/h ) ( 8/3 h ) = 112/3 mi = 37 1/3 mi<----------------
dM = ( 16 mi/h ) ( ( 8/3 - 1/3 ) = ( 16 mi/h ) ( 7/3 h ) = 112/3 mi = 37 1/3 mi <----
d = dR = dM = 37 1/3 mi -----------> [ OK ]