Question

A balloon is rising vertically above a​ level, straight road at a constant rate of 5 ft divided by sec. Just when the balloon is 106 ft above the​ ground, a bicycle moving at a constant rate of 17 ft divided by sec passes under it. How fast is the distance s (t )between the bicycle and balloon increasing 6 seconds​ later?

Answer 1

db = 14.2 ft/sec

Step-by-step explanation:

from the question we are given the following values:

height of the balloon (h) = 106 ft

rate of increase in height (speed) = 5ft/sec

rate of movement of the bicycle (speed) = 17 ft/sec

tine (t) = 6 sec

• The height of the balloon, the horizontal distance of the bicycle and the distance of the balloon from the bicycle after three seconds all form three sides of a triangle
• The distance between the balloon and the bicycle is the hypotenuse (b)
• The horizontal distance is the adjacent side (a)
• The height of the balloon is the opposite side (h)

Therefore

b^{2} = a^{2} + h^{2} .....equation 1

the height (h) after 6 seconds becomes = 106 + (5 x 6) = 136 ft

the horizontal distance (a) after 6 seconds = 17 x 6 = 102 ft

therefore

b^{2} = 102^{2} + 136^{2}

b = 170 ft

differentiating equation 1 in terms of time it becomes

2b\frac{db}[dt} = 2a\frac{da}[dt} + 2h\frac{dh}[dt}

where da = rate of change of horizontal distance

db = rate of change of vertical height

2 x 170 x \frac{db}[dt} = (2 x 102 x \frac{17}[dt}) + (2 x 136 x \frac{5}[dt})

2 x 170 x db = (2 x 102 x 17) + (2 x 136 x 5)

db = 14.2 ft/sec