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    Dimensional analysis - distance and speed

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    Step 1: A 240 horsepower car beginning at a stop goes 1/10 of a mile. At optimum performance, what is the maximum speed in miles per hour the car can reach? Feet per minute?
    Step 2: How many feet have been traveled in a 240 horsepower car when the speed reaches 35 MPH at optimum performance? what fraction of a mile is that distance?

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    step 1)
    1 HP = 746 Watts = 746 J/sec
    That is, the car can do a maximum work of 240*746 Joules/sec
    = 179040 Newton meter/second ----(1)
    It travels a distance of 1/10 mile = 1.6*1/10Km = 0.16Km = 160meters say in t seconds
    We know that,
    Power = work done/time
    Also, work done can be written as the change in kinetic energy.
    That is work = (1/2)m(vf^2 - vi^2)
    where, vf = final velocity = v and vi = 0 = initial velocity
    Hence, power = (1/2)m v^2/time ....(1)
    but, time = distance/velocity = d/v
    ==> power = (1/2)m v^3/distance ...(2)
    or, v^3 = 2* power * distance/m
    or v = cube root(2*power*distance ...

    Solution Summary

    This uses optimum performance of a car at 240 horsepower to relate distance and speed.