Questions: A 6.50-ft-tall man walks at 6.00 ft / s toward a street light that is 16.0 ft above the ground. At what rate is the end of the man's shadow moving when he is 7.0 ft from the base of the light? Use the direction in which the distance from the street light increases as the positive direction. The end of the man's shadow is moving at a rate of ft / s. (Round to two decimal places as needed.)

Solution Steps

Let \( x \) be the distance from the man to the street light, and \( s \) be the length of the man's shadow. The height of the street light is \( 16 \) ft, and the height of the man is \( 6.5 \) ft. Using similar triangles, we establish the relationship: \[ \frac{16}{x+s} = \frac{6.5}{s} \]

Rearranging the equation gives us: \[ s = \frac{6.5}{16}(x+s) \] Solving for \( s \) in terms of \( x \) yields: \[ s = \frac{0.6842}{1} x \]

Differentiating both sides with respect to time \( t \) gives: \[ \frac{ds}{dt} = 0.6842 \frac{dx}{dt} \] Given that the man walks towards the light at a rate of \( \frac{dx}{dt} = -6 \) ft/s, we substitute this value: \[ \frac{ds}{dt} = 0.6842 \times (-6) = -4.1053 \text{ ft/s} \]

Final Answer