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When studying kinematics, a core concept is the velocity function, which describes how an object's speed and direction change over time. In our exercise, we start with the acceleration function given by \(a_x(t) = At - Bt^2\). To find the velocity function \(v_x(t)\), we need to integrate the acceleration with respect to time. The general idea of integration in this context is to determine how velocity accumulates over time as acceleration acts. Performing this integration, we get:

• \( v_x(t) = \int (At - Bt^2) \, dt = \frac{A}{2} t^2 - \frac{B}{3} t^3 + C \)

As a rule of thumb, whenever you integrate, a constant of integration \(C\) will appear. To find this constant, we apply the condition that the motorcycle starts from rest, meaning \(v_x(0) = 0\). Substituting into the equation gives \(C = 0\), resulting in:

• \( v_x(t) = \frac{A}{2} t^2 - \frac{B}{3} t^3 \)

The position function \(x(t)\) helps us understand the location of an object over time in kinematics. To derive the position function from the velocity function, we integrate \(v_x(t)\) with respect to time. The velocity function derived earlier is \(v_x(t) = \frac{A}{2} t^2 - \frac{B}{3} t^3\). By integrating this function, you compute the position:

• \( x(t) = \int \left( \frac{A}{2} t^2 - \frac{B}{3} t^3 \right) \, dt = \frac{A}{6} t^3 - \frac{B}{12} t^4 + D \)

To determine the constant of integration \(D\), we use the initial condition stating that the motorcycle is at the origin when \(t = 0\), hence \(x(0) = 0\). This yields \(D = 0\), so:

• \( x(t) = \frac{A}{6} t^3 - \frac{B}{12} t^4 \)

This equation provides the path of the motorcycle over time, allowing us to calculate its position at any given moment.

Acceleration is a vector quantity in kinematics that represents the rate of change of velocity over time. It's crucial for determining how an object's speed increases or decreases. In our scenario, the acceleration function is given by \(a_x(t) = At - Bt^2\). Acceleration often serves as the starting point for deriving other motion-related functions. By integrating the acceleration function, we were able to find the velocity. Interestingly, to find specific characteristics of the motion like the maximum velocity, one needs to differentiate the velocity function \(v_x(t)\). Setting its derivative to zero helps find the critical points where the acceleration briefly becomes zero, indicating a potential maximum or minimum for velocity.For this problem, the solution for the maximum velocity involves finding when the derivative of the velocity function equals zero:

• \( \frac{dv_x}{dt} = At - Bt^2 = 0 \)

The time when this occurs is \( t = \frac{A}{B} \). Calculate the maximum velocity by substituting back the time value into \(v_x(t)\). Such applications of differentiation highlight how closely knit acceleration, velocity, and position functions are in kinematics.