# 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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##### Solution Summary

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

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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 ...

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