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    A car of mass 650kg is travelling along a straight road, inclined at 5' to the horizontal. At P, the car's speed is 15m/s and at point Q, 400m down the hill, its speed is 35m/s.
    i. Assuming a constant driving force between P and Q and no resistances, calculate the magnitude of the driving force.
    *I did this and got the correct answer which is 257N*

    ii. Assume instead that the resistance to motion between P and Q can be modelled by a constant force of magnitude 900N. Given that the acceleration at Q is 0, show that the power of the car's engine at this instant is approximately 12.1kW.

    Given that the power is the same as when the car is at P, calculate the car's acceleration at P. (the answer is 0.707 m/s/s)

    I don't understand how to do that second bit at all, I played around a bit (well, tried to answer it) and didn't come anywhere near...although I didn't use anything about the acceleration. I have the mark scheme but it isn't making sense to me. Please help if you can xxx
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    lol we were told 2 ignore the specimen as it was waaaaaaaaaaaaaaay to hard! heh heh
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    Its like this I think:

    Take the direction up the slope as positive.

    P = Fv
    P = (-650gsin5 + 900 - ma)35
    P = (-650gsin5 +900)35 (since a is zero at Q)
    P = 12068.6... W
    P = 12.1 kW to 3sf

    next bit..

    12068 = (-ma + 900 -650gsin5)15
    804.5333... = -650a + 900 - 650gsin5

    a = -0.707 ms^-2

    this is 0.707 ms^-2 DOWN the slope.
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    Thank you so much! I understand it now. Endlessly grateful xxxxxxxxx
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    (Original post by undercover-ange)
    Thank you so much! I understand it now. Endlessly grateful xxxxxxxxx
    No probs
 
 
 
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Updated: January 20, 2005

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