A 550 kg dragster accelerates from rest to a final speed of 110 m/s in 400 m (about a quarter of a mile) and encounters an average frictional force of 1200 N. What is its average power output in watts and horsepower if this takes 7.30 s?
I keep gettting 455821.918. Not quite sure where I'm going wrong here. Not sure why we're doing more physics in Calculus either.
On
Let $P$ be the average power output. The total amount of work done is $P \cdot (7.3 \,\mbox{s})$. But the total amount of work done is also the sum of these two quantities:
Part 1 is just force times distance, or if the force is not constant, you can use force averaged over distance. You don't say how the force was averaged, so I'll assume it's averaged over distance, since the answer to the problem is indeterminate otherwise. Then the first part of the work done is
$$(1200 \,\mbox{N}) \cdot (400 \,\mbox{m}) = 4.8 \times 10^5 \mbox{J}.$$
Part 2 is given by the kinetic energy formula,
$$\frac{1}{2} (550 \,\mbox{kg}) (110 \,\mbox{m/s})^2 = 3.3275 \times 10^6 \mbox{J}$$
Taken together, that's $3.8075 \times 10^6 \mbox{J}$. Divide by the time $(7.3 \,\mbox{s})$ and you have average power output of about $521575.3 \, \mbox{W}$.
Although I liked solving physics problems when I took first-year calculus, I share your confusion about what this problem is doing in a calculus course. The solution is just algebra, and there isn't enough information to compare the results with what you would get by integrating functions.
We cannot use the equations of motion $v=u+at$ and its relatives, since the acceleration here is non-constant.
Average Power = $Energy/Time = Work/Time = Force\times Distance/ Time$
$Force = Mass\times Acceleration_{average} = \frac{550\times 110}{7.3} \approx 8287.67$
But friction was acting, so actual force exerted by dragster $= 8287.67+1200=9487.67$
$\therefore P = \frac{9487.67\times 400}{7.3}\approx 519872.39 $
This is in watts. Convert to horsepower.