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So here is my problem, I have ten minutes to push a1814.37 kgs automobile, that is...

So here is my problem, I have ten minutes to push a1814.37 kgs automobile, that is not operational, 92 m down the street to my friends house. The streets are level; however , I am experiencing a 85.5 km/h wind blowing in the opposite direction of travel. The question is can I get there in 10 minutes, and if I could, what needs to be my average speed.

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Answer #1

Net average speed required to travel 92 m, 2et 92 m 10×60 s = 0.15 ms-1 This is the vector sum of the velocities of the wind

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