Answer:
The velocity of the upper end P when Q is 4 m from the wall is -0.8 m/s
Explanation:
Let's denote the distance of the lower end Q from the wall as x and the distance of the upper end P from the ground as y. We are given that the ladder has a length of 5 m, and thus by the Pythagorean theorem, we have:
x^2 + y^2 = 5^2 = 25
Now, we're given that the lower end Q is sliding away from the wall at a constant rate of 0.6 m/s. This means that the rate of change of x with respect to time (dx/dt) is 0.6 m/s. We need to find the rate of change of y with respect to time (dy/dt) when x = 4 m.
Differentiate both sides of the Pythagorean equation with respect to time (t):
d(x^2)/dt + d(y^2)/dt = d(25)/dt
2x(dx/dt) + 2y(dy/dt) = 0
We can plug in the known values and solve for dy/dt when x = 4 m:
2(4)(0.6) + 2y(dy/dt) = 0
To find the value of y when x = 4 m, we can use the Pythagorean equation:
4^2 + y^2 = 25
16 + y^2 = 25
y^2 = 9
y = 3 (since the height must be positive)
Now, we can substitute y = 3 into the equation we derived earlier:
2(4)(0.6) + 2(3)(dy/dt) = 0
4.8 + 6(dy/dt) = 0
Now, solve for dy/dt:
6(dy/dt) = -4.8
(dy/dt) = -4.8 / 6
(dy/dt) = -0.8 m/s
Thus, the velocity of the upper end P when Q is 4 m from the wall is -0.8 m/s. The negative sign indicates that the upper end P is moving downward.