A particle moves along a straight line with an acceleration of kV^2 m/s^2, where V is the particle's velocity. When t = 0, the particle's velocity is 4 m/s, and at t = 30 s, the velocity is 26 m/s. The displacement at time t is D(t) meters. Derive expressions for velocity and displacement as functions of time t. To derive the expressions, let's start with the given acceleration function: a(t) = kV(t)^2 We know that acceleration is the derivative of velocity with respect to time: a(t) = dV/dt Therefore, we can write: dV/dt = kV^2 To find the velocity as a function of time, we need to solve this differential equation. Integrating both sides: ∫ (dV/V^2) = ∫ k dt -1/V = kt + C where C is the constant of integration. We can find the constant by using the initial condition that when t = 0, V = 4 m/s: -1/4 = 0 + C C = -1/4 Now, we have the expression for velocity: -1/V = kt - 1/4 Solving for V: V = -1 / (kt - 1/4) This expression gives the velocity as a function of time. However, this assumes that the force results in a decrease in velocity, which may not match the initial information about the increase in velocity from 4 m/s to 26 m/s over 30 seconds. If we need to ensure an increase in velocity, the equation should be: 1/V = kt + 1/4 Now, to find the displacement, we know that displacement is the integral of velocity with respect to time: D(t) = ∫ V dt Now, we integrate the velocity function we found to get the displacement as a function of time: D(t) = ∫ -1 / (kt - 1/4) dt This integral will give us D(t), the displacement as a function of time. Since the integration might involve more complex mathematics due to the nature of the function, it will require a further solution to get a closed-form expression for the displacement.

To derive the expression for velocity as a function of time given the acceleration \( a(t) = kV(t)^2 \), where \( k \) is a constant, start by using the definition of acceleration as the derivative of velocity:

\( a(t) = \dfrac{dV}{dt} \)

Substitute the given acceleration:

\( \dfrac{dV}{dt} = kV^2 \)

Rearrange the equation and separate the variables to integrate:

\( \dfrac{1}{V^2} dV = k dt \)

Integrate both sides:

\( \int \dfrac{1}{V^2} dV = \int k dt \)

\( -\dfrac{1}{V} = kt + C \)

Where \( C \) is the integration constant. To find \( C \), use the initial condition \( V(0) = 4 \) m/s:

\( -\dfrac{1}{4} = k \cdot 0 + C \)

\( C = -\dfrac{1}{4} \)

So the velocity function is:

\( -\dfrac{1}{V} = kt - \dfrac{1}{4} \)

Solving for \( V \):

\( V(t) = \dfrac{-1}{kt - \dfrac{1}{4}} \)

However, to achieve an increasing velocity, the equation should be positive:

\( V(t) = \dfrac{1}{kt + \dfrac{1}{4}} \)

To find \( k \), use the other condition \( V(30) = 26 \) m/s:

\( \dfrac{1}{26} = 30k + \dfrac{1}{4} \)

Solve for \( k \):

\( k = \left( \dfrac{1}{26} - \dfrac{1}{4} \right)/30 \)

Now for the displacement, \( D(t) \), you would integrate the velocity with respect to time:

\( D(t) = \int V(t) dt \)

This integration would require finding the antiderivative of the function \( V(t) \).

Extra: The process we are following for finding the velocity and displacement as functions of time involves differential and integral calculus, which are branches of mathematics dealing with continuous change and accumulation.

In calculus, an ordinary differential equation (ODE) like \( \dfrac{dV}{dt} = kV^2 \) is an equation containing a function of one independent variable (in this case, velocity \( V \)) and its derivatives (velocity's rate of change, which is acceleration \( a \)). Solving an ODE typically involves separating variables and integrating to find a solution expressed in terms of the independent variable.

The constant of integration, \( C \), arises naturally when integrating because indefinite integrals are a family of functions that differ only by a constant. The initial condition, such as the initial velocity \( V(0) = 4 \) m/s, allows us to find the specific solution that fits the physical scenario described.

Once the velocity function is found, the displacement can be calculated as the integral of velocity over time. In physics, displacement represents the change in position of the particle and is often symbolized as \( s \) or \( x \) in one-dimensional motion. However, in this case, it's called \( D(t) \). This process illustrates the kinematic relationship between acceleration, velocity, and displacement in motion analysis.