Analyzing Particle Motion and Velocity Equation

The problem describes the motion of a particle starting from the origin. We are given its velocity as a function of its position \(x\). The velocity is defined by the differential equation: \(\rm \frac{dx}{dt}=x+1\).

The particle starts at the origin, which means at time \(t=0\), its position \(x=0\). We need to find the time it takes for the particle to travel a distance of 24 m. Since it starts at \(x=0\), traversing a distance of 24 m means reaching the position \(x=24\).

Solving the Velocity Differential Equation

To find the relationship between position \(x\) and time \(t\), we need to solve the given differential equation \(\rm \frac{dx}{dt}=x+1\). This is a first-order separable differential equation.

We can separate the variables \(x\) and \(t\) as follows:

\(\rm \frac{dx}{x+1} = dt\)

Now, we integrate both sides of the equation:

\(\rm \int \frac{dx}{x+1} = \int dt\)

Integrating the left side gives \(\ln|x+1|\) and integrating the right side gives \(t\). We must also include a constant of integration, let's call it \(C\):

\(\rm \ln|x+1| = t + C\)

Applying Initial Conditions to Find Constant

We know that the particle starts at the origin, so at \(t=0\), \(x=0\). We can use these initial conditions to find the value of the constant \(C\).

Substitute \(t=0\) and \(x=0\) into the integrated equation:

\(\rm \ln|0+1| = 0 + C\)

\(\rm \ln|1| = C\)

Since \(\ln(1)=0\), we get:

\(\rm 0 = C\)

So, the specific relationship between \(x\) and \(t\) for this particle's motion is:

\(\rm \ln|x+1| = t\)

Since the particle starts at \(x=0\) and moves to \(x=24\), \(x+1\) will always be positive, so we can write:

\(\rm \ln(x+1) = t\)

Calculating Time to Traverse 24 m

We need to find the time \(t\) when the particle's position is \(x=24\) m. We use the equation we derived: \(\rm t = \ln(x+1)\).

Substitute \(x=24\) into the equation:

\(\rm t = \ln(24+1)\)

\(\rm t = \ln(25)\)

We can simplify \(\ln(25)\) using the logarithm property \(\ln(a^b) = b \ln(a)\). Since \(25 = 5^2\), we have:

\(\rm t = \ln(5^2)\)

\(\rm t = 2 \ln(5)\)

Therefore, the time taken by the particle to traverse a distance of 24 m is \(2 \ln(5)\) seconds.

The time taken is \(2 \ln(5)\) seconds.

Revision Table: Particle Kinematics

• Velocity: The rate of change of position with respect to time, \(\rm v = \frac{dx}{dt}\).
• Differential Equation: An equation involving an unknown function and its derivatives. Solving it helps find the function.
• Separable Differential Equation: A type of differential equation that can be written in the form \(\rm f(x)dx = g(t)dt\).
• Integration: The process of finding the antiderivative; used here to find \(x(t)\) from \(v(x)\).
• Initial Conditions: Specific values of variables (like \(x=0\) at \(t=0\)) used to find the constant of integration.
• Logarithm Properties: Rules like \(\ln(a^b) = b \ln(a)\) used for simplification.

Additional Information: Exponential Growth

The differential equation \(\rm \frac{dx}{dt}=x+1\) shows that the velocity of the particle is linearly dependent on its position. Specifically, the velocity increases as the position \(x\) increases. The term '+1' means even at \(x=0\), the velocity is 1 m/s. The solution \(\rm t = \ln(x+1)\) can be rewritten as \(\rm x+1 = e^t\), or \(\rm x = e^t - 1\). This equation describes the position \(x\) as a function of time \(t\).

Let's check this solution:

• At \(t=0\), \(x = e^0 - 1 = 1 - 1 = 0\), which matches the initial condition.
• The velocity is \(\rm \frac{dx}{dt} = \frac{d}{dt}(e^t - 1) = e^t\).
• From the equation \(\rm x = e^t - 1\), we have \(\rm e^t = x+1\).
• So, the velocity \(\rm \frac{dx}{dt} = e^t = x+1\), which matches the given differential equation.

The position function \(x(t) = e^t - 1\) shows exponential growth. As time increases, the position grows exponentially, leading to an exponentially increasing velocity. This explains why the particle reaches 24m relatively quickly.