Question

If a balloon flying at an altitude of 30 m from an observer at a particular instant is moving horizontally at the rate of 1 m/s away from him, then the rate at which the balloon is moving away directly from the observer at the 40th second is (in m/s)

• A. 1.2
• B. 0.9
• C. 0.6
• D. 0.8

Hint:

In related rates problems, the general procedure is: 1. Identify the given rates and the rate you need to find. 2. Write an equation that relates the variables. 3. Differentiate the equation implicitly with respect to time. 4. Substitute the known values for the variables and their rates at the specific instant to solve for the unknown rate.

Answer 1

The Correct Option is D

Step 1: Model the problem Let the observer be at the origin \( O(0,0) \). Let the altitude of the balloon be \( h = 30 \) m (constant). Let \( x \) be the horizontal distance of the balloon from the observer's vertical line. Let \( s \) be the direct distance (line of sight) from the observer to the balloon. Relation: \( s^2 = x^2 + 30^2 \).
Step 2: Differentiate with respect to time \[ 2s \frac{ds}{dt} = 2x \frac{dx}{dt} \implies \frac{ds}{dt} = \frac{x}{s} \frac{dx}{dt} \] Given horizontal speed \( \frac{dx}{dt} = 1 \) m/s.
Step 3: Find values at t = 40 Assuming the balloon starts horizontally from the observer's position at \( t=0 \) (or \( x=0 \) initially), at \( t=40 \) seconds: \( x = \text{speed} \times \text{time} = 1 \times 40 = 40 \) m. Calculate \( s \): \[ s = \sqrt{40^2 + 30^2} = \sqrt{1600 + 900} = \sqrt{2500} = 50 \text{ m} \]
Step 4: Calculate rate of change \[ \frac{ds}{dt} = \frac{40}{50} \times 1 = 0.8 \text{ m/s} \]