A road perpendicular to a highway leads to a farmhouse located a = 1.2 km away. An automobile travels past the farmhouse at a speed of v= 84 km/h. How fast is the distance between the automobile and the farmhouse increasing when the automobile is 3.6 km past the intersection of the highway and the road? Let / denote the distance between the automobile and the farmhouse, and let s denote the distance past the intersection of the highway and the road. v km/h Automobile (Use decimal notation. Give your answer to three decimal places.)
A road perpendicular to a highway leads to a farmhouse located a = 1.2 km away. An automobile travels past the farmhouse at a speed of v= 84 km/h. How fast is the distance between the automobile and the farmhouse increasing when the automobile is 3.6 km past the intersection of the highway and the road? Let / denote the distance between the automobile and the farmhouse, and let s denote the distance past the intersection of the highway and the road. v km/h Automobile (Use decimal notation. Give your answer to three decimal places.)
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Transcribed Image Text:## Problem Statement
A road perpendicular to a highway leads to a farmhouse located \( a = 1.2 \) km away.
An automobile travels past the farmhouse at a speed of \( v = 84 \) km/h. How fast is the distance between the automobile and the farmhouse increasing when the automobile is 3.6 km past the intersection of the highway and the road?
Let \( l \) denote the distance between the automobile and the farmhouse, and let \( s \) denote the distance past the intersection of the highway and the road.
(Use decimal notation. Give your answer to three decimal places.)
## Diagram Explanation
The diagram illustrates a right triangle formed by the perpendicular road, the highway, and the straight-line distance from the farmhouse to the automobile.
- The vertical side represents the distance \( a = 1.2 \) km from the farmhouse to the highway.
- The horizontal side represents \( s \), the distance traveled by the automobile past the intersection of the highway and the road.
- The hypotenuse represents \( l \), the distance between the automobile and the farmhouse.
- The automobile is shown traveling horizontally away from the intersection of the highway and road, at a speed of \( v = 84 \) km/h.
![### Problem Description
A 5-meter ladder leans against a wall. Assume the bottom slides away from the wall at a rate of 0.5 m/s.
### Diagram Explanation
In the diagram, there are three positions of the ladder shown at different times (t = 0, t = 1, t = 2). The ladder's position becomes more horizontal over time as the bottom slides away from the wall.
- The x-axis is the horizontal distance from the wall to the bottom of the ladder.
- The y-axis is the height of the top of the ladder on the wall.
Beneath this, there is a right triangle diagram illustrating the relationship:
- The ladder is the hypotenuse of the triangle, measuring 5 meters.
- `h` is the height (vertical leg of the triangle).
- `x` is the horizontal distance from the wall (horizontal leg of the triangle).
### Variables and Problem Requirements
Let:
- `h` be the height of the ladder’s top at time `t`.
- `x` be the distance from the wall to the ladder’s bottom.
You need to find the velocity of the top of the ladder, denoted as \( \frac{dh}{dt} \), at \( t = 2 \) seconds given:
- The initial distance \( x = 1.5 \) meters at \( t = 0 \).
### Solution Process
Using the Pythagorean theorem:
\[ h^2 + x^2 = 5^2 \]
Differentiate both sides with respect to time \( t \):
\[ 2h \frac{dh}{dt} + 2x \frac{dx}{dt} = 0 \]
Given \( \frac{dx}{dt} = 0.5 \) m/s, find \( \frac{dh}{dt} \) at \( t = 2 \).
Use decimal notation and provide your answer to three decimal places.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4d2e368b-4254-4228-bfaa-fb4bb1de38c6%2Faf3798d4-d3de-4c10-b1e9-f5213101c77a%2F5l82gng_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem Description
A 5-meter ladder leans against a wall. Assume the bottom slides away from the wall at a rate of 0.5 m/s.
### Diagram Explanation
In the diagram, there are three positions of the ladder shown at different times (t = 0, t = 1, t = 2). The ladder's position becomes more horizontal over time as the bottom slides away from the wall.
- The x-axis is the horizontal distance from the wall to the bottom of the ladder.
- The y-axis is the height of the top of the ladder on the wall.
Beneath this, there is a right triangle diagram illustrating the relationship:
- The ladder is the hypotenuse of the triangle, measuring 5 meters.
- `h` is the height (vertical leg of the triangle).
- `x` is the horizontal distance from the wall (horizontal leg of the triangle).
### Variables and Problem Requirements
Let:
- `h` be the height of the ladder’s top at time `t`.
- `x` be the distance from the wall to the ladder’s bottom.
You need to find the velocity of the top of the ladder, denoted as \( \frac{dh}{dt} \), at \( t = 2 \) seconds given:
- The initial distance \( x = 1.5 \) meters at \( t = 0 \).
### Solution Process
Using the Pythagorean theorem:
\[ h^2 + x^2 = 5^2 \]
Differentiate both sides with respect to time \( t \):
\[ 2h \frac{dh}{dt} + 2x \frac{dx}{dt} = 0 \]
Given \( \frac{dx}{dt} = 0.5 \) m/s, find \( \frac{dh}{dt} \) at \( t = 2 \).
Use decimal notation and provide your answer to three decimal places.
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