last Q'z before st8

[B3ndythumbs]

A car accerlates from rest, drives a distance at 60 mph, then decerlates to rest. Accerlation and decelaration have equal magnitude but opposite sign. Total distance is 18 miles, total time is 20 minutes. What is the positive accerlation? ANS - .733 ft/s/s

A pane of glass is 24in x 36 in x .125 in thick. It is shattered into square pieces .25 in on a side. What is percent increase in total SA? ANS - 98.3%

What is the largest number of 8.5x11 in rectangular pieces that may be cut from a 2ft by 3ft sheet of poster board? ANS - 8 int

[AuSmith]

1) First, it takes just as long to accelerate as decelerate. Call this time [unparseable or potentially dangerous latex formula] and the time going [unparseable or potentially dangerous latex formula] as [unparseable or potentially dangerous latex formula]. Then, translate. Note, [unparseable or potentially dangerous latex formula] is [unparseable or potentially dangerous latex formula] mile per minute. If the acceleration is [unparseable or potentially dangerous latex formula],

[unparseable or potentially dangerous latex formula]

[unparseable or potentially dangerous latex formula]

What more is there? There's continuity. The car constantly accelerated from rest to [unparseable or potentially dangerous latex formula] mile per minute in [unparseable or potentially dangerous latex formula] minutes. That means [unparseable or potentially dangerous latex formula]. Now we have

[unparseable or potentially dangerous latex formula].

Subtract,

[unparseable or potentially dangerous latex formula].

Finally, convert.

[unparseable or potentially dangerous latex formula]

2) Say we align the piece of glass so that the [unparseable or potentially dangerous latex formula] inch side is vertical and the [unparseable or potentially dangerous latex formula] inch side is horizontal. We cut the glass with [unparseable or potentially dangerous latex formula] horizontal cuts and [unparseable or potentially dangerous latex formula] vertical cuts. You subtract one because each (vertical) cut corresponds to two (columns), and every column but the two outer columns correspond to two cuts.

We want the total surface area gained from the cuts. First, let's do horizontal cuts. We have [unparseable or potentially dangerous latex formula] that are [unparseable or potentially dangerous latex formula] in. long by [unparseable or potentially dangerous latex formula] in. deep. As we noted, there are two sides to a cut. So, the extra surface area from horizontal cuts is [unparseable or potentially dangerous latex formula]. Similarly, vertical cuts gain [unparseable or potentially dangerous latex formula] sq. inches.

[unparseable or potentially dangerous latex formula]

We started with [unparseable or potentially dangerous latex formula]. The percent increase is [unparseable or potentially dangerous latex formula].

3) Think of the [unparseable or potentially dangerous latex formula] ft. side. You're never going to fit more than two sheets of paper in [unparseable or potentially dangerous latex formula] ft., even if you used the shorter side. You might as well put two [unparseable or potentially dangerous latex formula] inch sides together to get [unparseable or potentially dangerous latex formula] inches. Then, [unparseable or potentially dangerous latex formula]. And, of course [unparseable or potentially dangerous latex formula]. Besides you only have [unparseable or potentially dangerous latex formula] sq. ft. to make an extra sheet of paper with, which is [unparseable or potentially dangerous latex formula] sq ft.

I'd like to see a real proof. This might be like the brass monkey problem... how do you most efficiently stack cannon balls? Pyramids! We've been doing it for centuries! But, some mathematician had to write a huge proof of why pyramids were the most efficient.