The balloon leaves the ground 80 ft. from an observer and rises vertically upward at 5 ft/s.
(a) How fast is the balloon receding from the observer after 12 seconds?
(b) find the corresponding acceleration?
How do I sketch it?
and what equation should I use?
Please show the solution/s
Thank You~!
Thank You!!
To sketch the scenario, you can draw a coordinate system with an observer at the origin (0,0) and the balloon rising vertically upwards. Mark a point at (80, 0) to represent the initial position of the balloon.
To find the speed at which the balloon is receding from the observer after 12 seconds, we can use the Pythagorean Theorem. The distance between the observer and the balloon at any given time is the hypotenuse of a right triangle, with the horizontal distance representing 80 ft and the vertical distance representing the height of the balloon.
(a) To find the rate of change of horizontal distance (dx/dt), we differentiate the horizontal distance equation with respect to time:
dx/dt = 0 ft/s (the balloon is not moving horizontally)
To find the rate of change of vertical distance (dy/dt), we differentiate the vertical distance equation with respect to time:
dy/dt = 5 ft/s
To find the rate of change of the hypotenuse (dr/dt), we use the Pythagorean theorem:
r^2 = x^2 + y^2
Differentiating with respect to time:
2r(dr/dt) = 2x(dx/dt) + 2y(dy/dt)
Simplifying:
dr/dt = (x(dx/dt) + y(dy/dt)) / r
Plugging in the values:
x = 80 ft
dx/dt = 0 ft/s
y = 5 ft
dy/dt = 5 ft/s
r = sqrt(x^2 + y^2) = sqrt(80^2 + 5^2) = sqrt(6565) ft
dr/dt = (80 * 0 + 5 * 5) / sqrt(6565) = 1.924 ft/s
Therefore, the balloon is receding from the observer at a speed of approximately 1.924 ft/s after 12 seconds.
(b) To find the corresponding acceleration, we differentiate the vertical velocity equation with respect to time:
d^2y/dt^2 = 0 (the balloon is not accelerating vertically)
Again using the Pythagorean theorem:
(d/dr)(dr/dt) = (dx/dt * dx/dt + dy/dt * dy/dt) / r
Plugging in the given values:
dx/dt = 0 ft/s
dy/dt = 5 ft/s
r = sqrt(6565) ft
(d/dr)(dr/dt) = (0 * 0 + 5 * 5) / sqrt(6565) = 0.076 ft/s^2
Therefore, the corresponding acceleration of the balloon is approximately 0.076 ft/s^2.
To sketch the scenario, you can start by drawing a horizontal line to represent the ground and mark a point 80 ft away from it to represent the observer. From this point, draw a vertical line to represent the balloon rising upward. Make sure to label the upward direction as positive.
To solve the problem, we need to apply some basic principles of calculus. We can use derivatives to find the velocity and acceleration of the balloon at any given time.
Let's denote the height of the balloon above the ground at time t as h(t) and its distance from the observer at time t as d(t). We're given that d(t) = 80 ft (constant) and h'(t) = 5 ft/s (rate of change of height).
(a) To find how fast the balloon is receding from the observer after 12 seconds, we need to find d'(t) (rate of change of distance). Since the balloon is rising vertically, d'(t) can also be interpreted as the velocity of the balloon moving away from the observer in the horizontal direction.
To find d'(t), we can differentiate d(t) = 80 ft with respect to time t. Since d(t) is constant, its derivative will be 0. Therefore, we conclude that d'(t) = 0 ft/s. This means that after 12 seconds, the balloon is not receding from the observer; it maintains a constant distance of 80 ft.
(b) To find the acceleration of the balloon at any given time, we need to differentiate the velocity h'(t) = 5 ft/s with respect to time t. The derivative of h'(t) gives us the acceleration a(t).
So, a(t) = h''(t). To find h''(t), we need to determine the second derivative of h(t).
Since the balloon is rising vertically at a constant rate of 5 ft/s, its acceleration a(t) is zero. This means that the balloon's velocity remains constant; it neither speeds up nor slows down.
In summary:
(a) The balloon is not receding from the observer after 12 seconds.
(b) The acceleration of the balloon is zero.
I hope this helps! Let me know if you have any further questions.
draw a right-angled triangle (side view)
mark the base as 80, and the height of the balloon as h
a) you want to know how fast the balloon is receding, so we are talking about how fast the hypotenuse is changing.
let the hypo be x
x^2 = 80^2 + h^2
2x dx/dt = 0 + 2h dh/dt, but we know dh/dt = 5
when t = 12, h = 12(5) or 60 ft
x^2 = 80^2 + 60^2
x = 100
so in 2x dx/dt = 2h dh/dt
2(100) dx/dt = 2(60)(5)
dx/dt = 3
a) 3 ft/s