The plank leaning against the wall forms a right triangle with one side 12 meters, and the hypotenuse 13 meters. (We make the reasonable assumption that the wall is perpendicular to the floor or earth.) The problem is made easier because I recognize that it's a 5 - 12 - 13 right triangle. This saves us from having to use trig functions to determine the third side.
The point where the plank touches against the wall is 5 meters up (That's the third side of the triangle. If this isn't clear, draw it.).
(a) Moving the bottom at 1/4 meter/second for 28 seconds moves the bottom inward 7 meters, so now it's 5 meters from the wall.
(b) How lucky. Since the bottom of the triangle is 5 meters from the wall and the hypotenuse is still 13 meters, we still have a 5 - 12 - 13 right triangle. Since the top of the plank is now 12 meters up, and it started from 5 meters up, it traveled 7 meters in 28 seconds, its rate was 7/28 meters/second, which we reduce the fraction to 1/4 meter/second.
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(b) How fast is the top of the plank moving up the wall 28 sconds after we start pushing?
Since you push the plank for 28 seconds, then at the end of those 28 seconds the plank is no longer moving.
Looking at it that way, the answer is zero m/sec. Do you think the question is really asking for the average rate?
I having hard time understanding it still! Maybe if you can drow it be easier, if you can’t thats fine. Thank you
+128460 After 28 sec. the plank will be 12 - (1/4)(28) = 12 - 7 = 5 meters from the wall
For the second part...we have that
x^2 + y^2 = r^2 take the derivative with respect to time
2x (dx/dt) + 2y (dy/dt) = 2r(dr/dt)
r = 13 m
dr/dt = 0 [the plank length is not changing ]
dx/dt = -1/4 m/s
At 28 sec, x = 5 meters and y = sqrt (13^2 - 5^2) = sqrt (169 - 25) = sqrt (144) = 12 meters
dy/dt = what we are looking for
So we have
2(5m)(-1/4 m/s) + 2(12m)(dy/dt) = 0
(10m) (-1/4 m/s) = - (24m)(dy/dt)
-2.5 (m^2/s) = (-24m) (dy/dt)
2.5 m^2/s = (24m) (dy/dt) divide both sides by 24 m
(2.5/24)(m/s) = dy/dt
(5/48)m/s = dy/dt
