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Note that the solutions are written in the format you would actually submit as a solution, namely as a Maple file. Maple commands appear red. Maple responses are black with cursive type. Text comments appear black with typewriter text.
b.) Acceleration is represented by the slope on a v vs.
t graph, so the greatest positive acceleration occurs
where the graph has its greatest positive slope. Visual
inspection shows that this occurs during the time interval
from 4 to 8 seconds. The slope here is
A = (vf - v0)/(tf - t0) = (2 - (-2))/(8 - 4) = 1 m/s^2.
c.) The initial and final velocities are the same for 2 seconds and 4 seconds, hence the average acceleration is 0.
Since there are two unknowns (the height and v0) we can't solve this problem with just one equation. We decided to solve for v0 first since we note that we do know the total distance for a complete trip of the ball. We assumethat the ball is hit and then caught at the same height, i.e. the batter's bat and the catcher's glove are about the same height above the ground. We were not told this by the wording of the problem, but we take it as a reasonable assumption. In this case, t=12 seconds and x-x0 = 0, so we use the equation x = x0 + v0t - (1/2)gt^2 (- on the acceleration term since gravity accelerates in the negative x direction in our convention).
For the deceleration portion, we have
Note that we have put the signs into the equation
rather than on the values of the acceleration and
deceleration so all quantities in the equations are
to be considered as magnitudes, i.e. positive
values.
Now, after a little practice, we learn that we can
just put in the appropropriate values of acceleration
and deceleration and let Maple's solve
command do the work while we lounge. As usual, we
don't include below Maple's response except where
its especially relevant (e.g. when it gives us the
answer)
The last two values are the answers desired. These are the times for deceleration and acceleration in terms of seconds. You can check them by using Maple's subs command.
This problem is difficult to solve unless we use relative quantities, i.e. instead of relating displacement (and hence velocity and acceleration) to a fixed origin, we choose our coordinate system origin to be one of the moving vehicles. Therefore, our displacement, velocity, and acceleration values will be relative to either the car or truck.
Let's say that we choose the front of the truck to be our origin. This then becomes our way of determining the position of the truck as well as the position of objects relative to the truck. For the car, let's choose the rear to mark its position (any point on the car would do, we simply have to choose it and stick with it). Therefore, our initial quantities are:
The final relative quantities after the pass are:
Now let's invoke Maple (remember to restart Maple at this point if you haven't done so)
whereas for the car, we have
In Maple, we plot the position for both car and truck on the same graph and graphically find the point where the rear of the car just passes the front of the truck by clicking on the intersection of the two curves.
leastsquares := proc(xx,yy)
local xb,yb,i,n;
if nops(xx) = nops(yy) then
n := nops(xx);
xb := sum(xx[i]/n, i=1..n);
yb := sum(yy[i]/n, i=1..n);
RETURN(simplify(evalf(yb+(x-xb)*
sum((xx[i]-xb)*(yy[i]-yb),i=1..n)/
sum((xx[i]-xb)^2,i=1..n))))
else RETURN(`Error: lengths of lists do not match`)
fi
end;
merge := proc(xx,yy)
local i,n,zz;
if nops(xx) = nops(yy) then
zz := NULL;
n := nops(xx);
for i to n do zz := zz, [xx[i],yy[i]] od;
RETURN([zz])
else RETURN(`Error: lengths of lists do not match`)
fi
end;
residuals := (y,h,line)->merge(y,map(a->a[2]-subs(x=a[1],line),merge(y,h)));
Now with the programs loaded, we can enter the data from the problem.
year := [1896,1900,1904,1908,1912,1920,1924,1928,1932,1936,1948,
1952,1956,1960,1964,1968,1972,1976,1980,1984,1988,1992];
height := [130.,130.,137.75,146,155.5,161,155.5,165.25,169.75,
171.25,169.25,179,179.5,185,200.75,212.5,216.5,216.5,
227.5,226.25,237.25,228.25];
line:=leastsquares(year,height);
Now plot the line and the data
with(plots,display):
datplot := plot(merge(year,height)):
linplot := plot(line,x=1896..2000):
display({datplot,linplot});
Since we want to be sure we are looking at a meaningful fit, we also need to plot the residuals plot(residuals(year,height,line));
We can see an obvious trend to the residuals. The height cycles above and below the predicted height about every 25 years. It appears as though 1996 is just at the point where the actual height will lie BELOW the prediction. Hence, instead of estimating 236 inches (which we get by clicking on the graph of height vs. year for year=1996), we should estimate about 10-15 inches lower, say 225 inches. Now, to get the estimate for 2096, let's plot the line for a much longer period. linplot := plot(line,x=1896..2000): display({datplot,linplot});
Again, if we assume a 25-year cycle, we should again find the actual winning height somewhat below the predicted height. With only 3 cycles to guide us however, that's not a very certain prediction. If the actual trend were to turn out to be 20 years for example, then the actual winning height would lie ABOVE our prediction. Hence, the best estimate we can make is probably the prediction from our line. At 2096, Maple tells us that the winning height should be 343 inches. We can get this by clicking on the graph at x = 2096 or with the Maple command subs(x=2096,line). To get the predicted year for which the height would be 1250 feet, use solve solve(line=1250*12, x); Maple returns the year, 15750! Yowsir!
lenth := [6.5,11.0,13.2,15.0,18.1,23.0,24.4,26.5,30.6,34.3,37.5,41.5];
period := [0.51,0.67,0.73,0.79,0.89,0.98,1.01,1.08,1.13,1.25,1.28,1.33];
line:= leastsquares(lenth,period);
datplot := plot(merge(lenth,period), style=POINT, symbol=cross):
linplot := plot(line, x=0..50):
display({datplot,linplot});
That looks fine; let's plot the residuals plot(residuals(lenth,period,line));
Now we notice that the residuals indicate serious problems with the
# #Enter the data as an ordered list of ordered pairs as told # airlist := [[10, 141], [20,145],[25,148],[30,150],[35,153],[50,161],[80,175]]; # #Note that I called the ordered list of ordered pairs "airlist". #Now make the plot of the data using circles for each point # Before we do that, just load the plots functions to be sure # we have everything we need for later. # with(plots): plot(airlist, x=0..100, style=point,symbol=circle);
# #Now load Maple's statistics package to do the fit to # the data # with(stats): # #To fit, we need to arrange the data into a list of x points # and a list of y points that correspond to them. Here, we # call these "xlst" and "ylst" # xlst := [10,20,25,30,35,50,80]; ylst := [141,145,148,150,153,161,175]; # #Now, we do the least squares fit to the data with the Maple # fit function. The solution will be the function (i.e. y(x)) # which best fits the data. If we do not give Maple's fit # a function of our own, it assumes we think the data fits a # line. We will name our best fit function "bestlin". So ... # fit[leastsquare[[x,bestlin]]]([xlst,ylst]);
#
#Now assign the solution to the name "bestlin" and produce
# a plot.
assign(");
#
#Since the equation is obviously that of a line,
# we won't bother plotting the line itself. Instead, we
# save the plot as a Maple plot structure with the name
# "bstlinplot" and plot it AND the data plot (which we
# will chall "dataplot", duhh!) simultaneously.
#
bstlinplot := plot(bestlin, x=0..100):
dataplot := plot(airlist, x=0..100, style=point,symbol=circle):
#
#Note that we used colon (:) instead of semicolon (;)
# above to keep us from having to see Maple's plot structure!
#
display({bstlinplot,dataplot});
#
#Since V = at + b, at V = 0, we have t = -b/a. Hence,
# we can get the zero-volume temperature from our
# "bestlin" fit to the data.
#
bestlin;
evalf(");
# #Using Maple as a calculator, we find -b/a # -135.6186253/.4946784922;
# #So, the lowest achievable temperature is -274 degrees # Celsius. #Now we are out of Maple territory. In order to create a new temperature scale which has zero volume at T = 0 while maintaining the same slope for V vs. T as for V vs. t, we need to consider the plot just made. The zero for volume is the x-intercept on our graph, namely the value of x (t in this case) where the value of y (V for our plot) is zero. To maintain a line of the same slope, we need to scoot it along the x-axis (maintaining it as parallel to the original V vs. t) until it goes through the (0,0) point. This is just the same as adding a constant to the x-intercept for the V vs. t line. The new x-intercept should be zero, so, the constant we want to add is +b/a or +274 degrees Celsius. Why, well ...
V = at + b = aT ===> T = t + b/a
Next: Solutions to Week 2
Up: Table of Contents