1) Two balls are thrown upward from the edge of the cliff.
The first is thrown with a speed of 48 ft/sec
and the second one is thrown a second later with a speed of 24 ft/sec.
Do the balls ever pass each other?
Let the top of the cliff be y = 0.
At time t, the first ball has height: .y .= .48t - 16t²
. . and the second ball has height: .y .= .24(t - 1) - 16(t - 1)²
They will pass when they both have the same height.
. . 48t - 16t² .= .24(t - 1) - 16(t - 1)²
This simplifies to: .8t = 40 . → . t = 5
The second ball overtakes the first ball 5 second after the first ball is thrown,
. . at a height 160 feet below the top of the cliff.
2) A woman at a point A on the shore of a circular lake with radius 2 mi wants to
arrive at the diametrically opposite point C on the other side of the lake in the
shortest possible time. .She can walk at the rate of 4 mph and row a boat at 2 mph.
How should she proceed?
Draw radius OB = 2.
* * *
* \ *
* \ *
* \ *
* θ \ *
A * - - - - * - - - - * C
* 2 O 2 *
* * *
Let /AOB = θ
She will walk along the circular arc AB.
The length of AB is 2θ miles (where θ is in radians).
At 4 mph, it will take her: .2θ/4 = ½θ hours to walk the arc.
She will row across the chord BC.
The length of the chord can be found with the Law of Cosines.
. . BC² .= .2² + 2² - 2(2)(2)cos(2π - θ) .= .8 + 8·cos θ .= .8(1 + cos θ)
. . BC² .= .16[(1 + cos θ)/2] .= .16·cos²(½θ)
. . Hence: .BC .= .4·cos(½θ) miles.
At 2 mph, it will take her: .4·cos(½θ)/2 = 2·cos(½θ) hours to row across.
Her total time is: .T .= .½θ + 2·cos(½θ) hours
And that is the function we must minimize . . .