SOLVE THE QUESTION BELOW.
A speeding car passes a stop light at 10:00am. At 10:15am it is 450m East of the stop light and at 10:30am it is located 900m East of the stop light.
What is the average velocity of the car from the time between 10:15am and 10:30am?
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Answered by
4
Answer:
Let the time he walks be x hours
From the given condition
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 1×9=9 km
Answered by
4
Answer:
kya hua umesh
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