A speeding car passes a stop at 10:15 am. At 10:15 am it is 450 m East of the stop light and at 10:30 am it is located 900 m East of the stop light. What is the average velocity of the car from the time between 10:30 am?
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Step-by-step explanation:
3(x)+9(3-x)
=15-6x=12
⇒x=2
So the time he runs is 3-2-1 hour
Distance covered by running is 1x9=9 km
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