Question

1-43. a man who is 6 feet tall is walking at a rate of 4 feet/second, directly away from a light that is located 10 feet above the ground. a. how long is the man’s shadow when he is 20 feet away from the light? b. how long is the man’s shadow seconds after he starts walking? the answer will be in terms of t

Explanation:

Step1: Set up similar - triangle proportion

Let $x$ be the length of the man's shadow. The height of the light is $h_1 = 10$ feet, the height of the man is $h_2=6$ feet. The distance of the man from the light is $d$. By similar - triangles, we have $\frac{10}{6}=\frac{d + x}{x}$. Cross - multiplying gives $10x=6(d + x)$.

Step2: Simplify the equation

Expand the right - hand side: $10x=6d+6x$. Subtract $6x$ from both sides: $10x−6x = 6d$, so $4x = 6d$, and $x=\frac{3}{2}d$.

Step3: Solve part (a)

When $d = 20$ feet, substitute $d$ into the formula $x=\frac{3}{2}d$. Then $x=\frac{3}{2}\times20=30$ feet.

Step4: Solve part (b)

The man's speed is $v = 4$ feet/second. The distance $d$ of the man from the light $t$ seconds after he starts walking is $d = 4t$. Substitute $d = 4t$ into the formula $x=\frac{3}{2}d$. We get $x=\frac{3}{2}\times4t=6t$ feet.

Answer:

a. 30 feet
b. $6t$ feet