Question

. A balloon is rising at a constant speed of . A boy is cycling along a straight road at a speed of . When he passes under the balloon, it is ft above him. How fast is the distance between the boy and the balloon increasing s later?

Answer 1

The distance between the balloon and the boy is increasing at a rate of 15.8 ft/s

Explanation:

In this problem we have to consider both the motion of the boy and the motion of the balloon.

Taking the position of the boy when he is just below the balloon as origin (x=0, y=0), we have:

- The motion of the boy is a uniform motion along the x-direction, so its x-position at time t is given by

$x(t) = v_b t$

where $v_b=15 ft/s$ is the velocity of the boy and t is the time.

So he is at 45 ft away from the origin, along the x-axis.

- The motion of the balloon is also a uniform motion but along the y-direction, so its y-position at time t is given by

$y(t) =y_0 +v't$

where

$y_0 = 45 ft$ is the initial vertical position of the balloon

v' =5 ft/s is the velocity of the balloon

Substituting t = 3 s,

$y(3)=45+(5)(3)=60 ft$

So the balloon is 60 ft above the origin, along the y-axis.

Since the two motions are perpendicular to each other, we can find the distance between the boy and the balloon at time t = 3 s by applying Pythagorean's theorem:

$d=\sqrt{x^2+y^2}=\sqrt{45^2+60^2}=75 ft$

However, here we are interested in how fast is this distance is increasing: which means, we are interested in finding the relative speed between the boy and the balloon.

As we said, the two objects move in perpendicular directions with speeds:

$v_x = 15 ft/s\\v_y=5 ft/s$

So, we can apply Pythagorean's theorem again to find their relative speed:

$v=\sqrt{v_x^2+v_y^2}=\sqrt{15^2+5^2}=15.8 ft/s$

So, their distance is increasing at a rate of 15.8 ft/s.

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