A person has a height 1.8 m. Beginning directly under a streetlight that is 6 m above theground, they walk away from their starting position at 7 m/s. Find the speed at which thetop of the person’s shadow moves.a. 6 m/s. b. 8 m/s c. 10 m/s d. 12 m/s e. 14 m/s

This problem can be solved using similar triangles.

1. First, we set up the similar triangles. The first triangle is formed by the person and their shadow, and the second triangle is formed by the streetlight and the person's shadow.

2. The height of the person (1.8 m) is to the length of the person's shadow as the height of the streetlight (6 m) is to the total length of the shadow (the length of the person's shadow plus the distance the person has walked).

3. We can set up the following equation using the properties of similar triangles: 1.8 / x = 6 / (x + d), where x is the length of the person's shadow and d is the distance the person has walked.

4. We can solve this equation for x to find the length of the person's shadow in terms of d: x = 1.8d / (6 - 1.8) = 0.42857d.

5. The speed at which the top of the person's shadow moves is the derivative of x with respect to time (dx/dt).

6. Since d is increasing at a rate of 7 m/s, we can find dx/dt by differentiating x with respect to time: dx/dt = 0.42857 * 7 = 3 m/s.

7. However, the person is also moving at 7 m/s, so the top of the person's shadow is moving at the speed of the person plus the speed of the shadow: 7 m/s + 3 m/s = 10 m/s.

So, the correct answer is c. 10 m/s.