Question 765549: a car starts from rest and travels towards a jeep. the jeep starts at the same time with the car and travels towards the car. the car and jeep are 10 km apart if the acceleration of the jeep and the car are 7 meters per second squared and 5 meters per secind squared respectively, when and where will they meet?
Answer by Lola77479(27)   (Show Source):
You can put this solution on YOUR website! a car starts from rest and travels towards a jeep. the jeep starts at the same time with the car and travels towards the car. the car and jeep are 10 km apart if the acceleration of the jeep and the car are 7 meters per second squared and 5 meters per secind squared respectively, when and where will they meet?
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The accelerations in opposite directions are added, same as speeds.
--> 12 m/sec/sec
then find the time travel 10000 meters
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s = at^2/2 = 10000
t^2 = 20000/12
t =~ 40.825 seconds to meet
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In that time, the car goes
s = at^2/2 = 5*40.825^2/2
s =~ 4166.7 meters
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For the jeep:
s = 7*40.825^2/2
s =~ 5833.4 meters
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