A lamp post 3m high is 6m from a wall. A 2m man tall is walking directly from the post toward at 2.5m/s. How fast is his 1.5 from the wall
Question makes no sense.
It appears whole words are missing.
A lamp post 3m high is 6m from a wall. A 2m man tall is walking directly from the post toward at 2.5m/s. How fast is his 1.5 from the wall
The corrections are= "How fast is his shadow moving up the wall when he is 1.5m from the wall". I encountered the question, but still don't know the answer.
To determine how fast the man's shadow is moving on the wall, we can use similar triangles. Let's first identify the similar triangles created by the lamp post, the man, and their shadows.
In the larger triangle, the lamp post serves as the vertical side, and the wall serves as the horizontal side. The smaller triangle formed by the man and his shadow has the same shape as the larger triangle.
We know the height of the lamp post is 3m and the distance between the lamp post and the wall is 6m. Therefore, the ratio of the height of the smaller triangle to the height of the larger triangle is equal to the ratio of their corresponding bases.
Height of the smaller triangle (man's height) / Height of the larger triangle (lamp post height) = Distance of the smaller triangle (man's distance from the post) / Distance of the larger triangle (distance from the post to the wall)
Let's plug in the values we have:
2m (man's height) / 3m (lamp post height) = x (man's distance from the wall) / 6m (distance from the post to the wall)
Now, we can solve for x, which represents the distance of the man from the wall:
Cross-multiplying, we get:
2m * 6m = 3m * x
12m² = 3m * x
x = 12m² / 3m
x = 4m
Therefore, the man is 4m from the wall.
Now, to determine how fast the man's distance from the wall is changing, we need to differentiate the equation with respect to time.
Differentiating the equation x = 4m, we get:
dx/dt = d/dt(4m)
Since the man is walking directly from the post toward the wall, the change in his distance from the wall (dx) is the same as his speed, which is 2.5 m/s:
dx/dt = 2.5 m/s
Therefore, the man's distance from the wall is changing at a rate of 2.5 m/s.