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Sagot :
To solve the given problem, we need to follow several steps and use the provided relationships and conditions. Let's tackle this step-by-step.
### Step 1: Determine the Constant [tex]\(k\)[/tex]
We start with the acceleration formula:
[tex]\[ a = -k \sqrt{v} \][/tex]
Using the given information:
- Initial conditions: [tex]\( x = 0 \)[/tex] and [tex]\( v = 81 \, \text{m/s} \)[/tex] at [tex]\( t = 0 \)[/tex].
- At [tex]\( x = 18 \, \text{m} \)[/tex], [tex]\( v = 36 \, \text{m/s} \)[/tex].
We can relate acceleration [tex]\(a\)[/tex] to the change in velocity with respect to position [tex]\(x\)[/tex]:
[tex]\[ a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = \frac{dv}{dx} \cdot v \][/tex]
Given [tex]\( a = -k \sqrt{v} \)[/tex], we substitute:
[tex]\[ -k \sqrt{v} = \frac{dv}{dx} \cdot v \][/tex]
Rearranging terms:
[tex]\[ \frac{dv}{dx} = -k \frac{\sqrt{v}}{v} = -k \frac{1}{\sqrt{v}} \][/tex]
Next, we separate variables and integrate both sides:
[tex]\[ \int v^{-1/2} dv = -k \int dx \][/tex]
[tex]\[ 2 \sqrt{v} = -kx + C \][/tex]
Using the initial condition [tex]\( x = 0 \)[/tex] and [tex]\( v = 81 \)[/tex]:
[tex]\[ 2 \sqrt{81} = C \][/tex]
[tex]\[ C = 18 \][/tex]
Now use the condition at [tex]\( x = 18 \)[/tex], [tex]\( v = 36 \)[/tex]:
[tex]\[ 2 \sqrt{36} = -k \cdot 18 + 18 \][/tex]
[tex]\[ 12 = -18k + 18 \][/tex]
[tex]\[ -18k = -6 \][/tex]
[tex]\[ k = \frac{1}{3} \][/tex]
### Step 2: Determine the Velocity at [tex]\( x = 20 \, \text{m} \)[/tex]
Using the relationship derived:
[tex]\[ 2 \sqrt{v} = -\frac{1}{3} x + 18 \][/tex]
Substitute [tex]\( x = 20 \)[/tex]:
[tex]\[ 2 \sqrt{v} = -\frac{1}{3} \cdot 20 + 18 \][/tex]
[tex]\[ 2 \sqrt{v} = \frac{-20}{3} + 18 \][/tex]
[tex]\[ 2 \sqrt{v} = \frac{-20 + 54}{3} \][/tex]
[tex]\[ 2 \sqrt{v} = \frac{34}{3} \][/tex]
[tex]\[ \sqrt{v} = \frac{17}{3} \][/tex]
[tex]\[ v = \left( \frac{17}{3} \right)^2 \][/tex]
[tex]\[ v \approx 32.111 \, \text{m/s} \][/tex]
So, the velocity at [tex]\( x = 20 \, \text{m} \)[/tex] is approximately [tex]\( 32.111 \, \text{m/s} \)[/tex].
### Step 3: Determine the Distance and Time Required for the Particle to Come to Rest
For the particle to come to rest ([tex]\( v = 0 \)[/tex]):
[tex]\[ 2 \sqrt{0} = -\frac{1}{3} x_{rest} + 18 \][/tex]
[tex]\[ x_{rest} = 18 \times 3 \][/tex]
[tex]\[ x_{rest} = 54 \, \text{m} \][/tex]
To find the time [tex]\( t \)[/tex] when the particle comes to rest, use the relationship between velocity and position:
[tex]\[ v = \frac{dx}{dt} \][/tex]
Integrate to find [tex]\( t \)[/tex]:
[tex]\[ t = \int_{0}^{54} \frac{1}{v} dx \][/tex]
Given that [tex]\( v = -\frac{1}{6} x + 9 \)[/tex]:
[tex]\[ t = \int_{0}^{54} \frac{1}{-\frac{1}{6} x + 9} dx \][/tex]
This integral diverges (leading to infinity), meaning the time required for the particle to come to rest is infinite in real terms but represented here mathematically as:
[tex]\[ t = \infty + 18.85i \][/tex]
Therefore, the required answers are:
- (a) Velocity at [tex]\( x = 20 \, \text{m} \)[/tex]: [tex]\( \approx 32.111 \, \text{m/s} \)[/tex]
- (b) Time to rest: [tex]\( \infty \)[/tex]
This indicates an infinite time for the particle to come completely to rest.
### Step 1: Determine the Constant [tex]\(k\)[/tex]
We start with the acceleration formula:
[tex]\[ a = -k \sqrt{v} \][/tex]
Using the given information:
- Initial conditions: [tex]\( x = 0 \)[/tex] and [tex]\( v = 81 \, \text{m/s} \)[/tex] at [tex]\( t = 0 \)[/tex].
- At [tex]\( x = 18 \, \text{m} \)[/tex], [tex]\( v = 36 \, \text{m/s} \)[/tex].
We can relate acceleration [tex]\(a\)[/tex] to the change in velocity with respect to position [tex]\(x\)[/tex]:
[tex]\[ a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = \frac{dv}{dx} \cdot v \][/tex]
Given [tex]\( a = -k \sqrt{v} \)[/tex], we substitute:
[tex]\[ -k \sqrt{v} = \frac{dv}{dx} \cdot v \][/tex]
Rearranging terms:
[tex]\[ \frac{dv}{dx} = -k \frac{\sqrt{v}}{v} = -k \frac{1}{\sqrt{v}} \][/tex]
Next, we separate variables and integrate both sides:
[tex]\[ \int v^{-1/2} dv = -k \int dx \][/tex]
[tex]\[ 2 \sqrt{v} = -kx + C \][/tex]
Using the initial condition [tex]\( x = 0 \)[/tex] and [tex]\( v = 81 \)[/tex]:
[tex]\[ 2 \sqrt{81} = C \][/tex]
[tex]\[ C = 18 \][/tex]
Now use the condition at [tex]\( x = 18 \)[/tex], [tex]\( v = 36 \)[/tex]:
[tex]\[ 2 \sqrt{36} = -k \cdot 18 + 18 \][/tex]
[tex]\[ 12 = -18k + 18 \][/tex]
[tex]\[ -18k = -6 \][/tex]
[tex]\[ k = \frac{1}{3} \][/tex]
### Step 2: Determine the Velocity at [tex]\( x = 20 \, \text{m} \)[/tex]
Using the relationship derived:
[tex]\[ 2 \sqrt{v} = -\frac{1}{3} x + 18 \][/tex]
Substitute [tex]\( x = 20 \)[/tex]:
[tex]\[ 2 \sqrt{v} = -\frac{1}{3} \cdot 20 + 18 \][/tex]
[tex]\[ 2 \sqrt{v} = \frac{-20}{3} + 18 \][/tex]
[tex]\[ 2 \sqrt{v} = \frac{-20 + 54}{3} \][/tex]
[tex]\[ 2 \sqrt{v} = \frac{34}{3} \][/tex]
[tex]\[ \sqrt{v} = \frac{17}{3} \][/tex]
[tex]\[ v = \left( \frac{17}{3} \right)^2 \][/tex]
[tex]\[ v \approx 32.111 \, \text{m/s} \][/tex]
So, the velocity at [tex]\( x = 20 \, \text{m} \)[/tex] is approximately [tex]\( 32.111 \, \text{m/s} \)[/tex].
### Step 3: Determine the Distance and Time Required for the Particle to Come to Rest
For the particle to come to rest ([tex]\( v = 0 \)[/tex]):
[tex]\[ 2 \sqrt{0} = -\frac{1}{3} x_{rest} + 18 \][/tex]
[tex]\[ x_{rest} = 18 \times 3 \][/tex]
[tex]\[ x_{rest} = 54 \, \text{m} \][/tex]
To find the time [tex]\( t \)[/tex] when the particle comes to rest, use the relationship between velocity and position:
[tex]\[ v = \frac{dx}{dt} \][/tex]
Integrate to find [tex]\( t \)[/tex]:
[tex]\[ t = \int_{0}^{54} \frac{1}{v} dx \][/tex]
Given that [tex]\( v = -\frac{1}{6} x + 9 \)[/tex]:
[tex]\[ t = \int_{0}^{54} \frac{1}{-\frac{1}{6} x + 9} dx \][/tex]
This integral diverges (leading to infinity), meaning the time required for the particle to come to rest is infinite in real terms but represented here mathematically as:
[tex]\[ t = \infty + 18.85i \][/tex]
Therefore, the required answers are:
- (a) Velocity at [tex]\( x = 20 \, \text{m} \)[/tex]: [tex]\( \approx 32.111 \, \text{m/s} \)[/tex]
- (b) Time to rest: [tex]\( \infty \)[/tex]
This indicates an infinite time for the particle to come completely to rest.
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