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Differentiation is a fundamental concept in calculus that deals with how a function changes as its input changes. Essentially, it finds the rate of change of one quantity with respect to another. In the context of motion, differentiation helps us derive the velocity and acceleration from the position function.

When you have a position function, such as \( s(t) = t^3 - t^2 \), differentiating this function with respect to \( t \), the time variable, gives you the velocity function \( v(t) \). This is because velocity is the rate of change of position over time.

• Step 1: Identify the position function: \( s(t) = t^3 - t^2 \)
• Step 2: Differentiate the position function: \( v(t) = \frac{\mathrm{d}s}{\mathrm{d}t} = 3t^2 - 2t \)

Differentiation is essentially used to extract more detailed information about motion, like how fast and in what direction something is moving at a given instant.

Velocity refers to the rate of change of an object's position with respect to time and is a vector quantity involving both speed and direction. From the position function \( s(t) = t^3 - t^2 \), we derived the velocity function, \( v(t) = 3t^2 - 2t \).

Velocity is dynamic and can change over time as a result of acceleration. In calculus, it's what you get when you differentiate the position function once. By substituting time values into the velocity function, you can discover how fast the object moves at any given time.

• To understand if the velocity changes over time, look at the next derivative, which is the acceleration.
• For instance, substituting \( t=1 \) gives us a velocity \( v(1) = 1 \mathop{\mathrm{ft/s}} \).

Therefore, velocity gives invaluable data on how quickly a position is changing, but to know how velocity itself evolves, we explore the acceleration.

Acceleration is the rate at which the velocity of an object changes with time. This rate of change gives insights into whether an object is speeding up, slowing down, or changing direction. To find acceleration, you differentiate the velocity function.

For the velocity function \( v(t) = 3t^2 - 2t \), we find the acceleration function by taking another derivative: \( a(t) = 6t - 2 \).

• This function tells us how much the velocity is changing over time.
• Evaluating acceleration at \( t = 1 \), we find \( a(1) = 4 \mathop{\mathrm{ft/s^2}} \).

If acceleration is positive, as it is here, the velocity is increasing at that point in time. If it were negative, velocity would be decreasing. In this problem, because \( a(1) > 0 \), we confirm that the velocity is indeed increasing when \( t = 1 \), meaning the particle is accelerating forward.