Question

The velocity function of a moving particle on a coordinate line is \(v(t)\) \(=2 t+1\) for \(0 \leq t \leq 8 .\) At \(t=1,\) its position is \(-4 .\) Find the position of the particle at \(t=5\).

Short answer

Answer: The position of the particle at t = 5 is 10.

Step-by-step solution

Find the displacement function

To find the displacement function, integrate the velocity function with respect to time: $$s(t) = \int v(t) \, dt = \int (2t + 1) \, dt.$$

Apply the Fundamental Theorem of Calculus

Solve the integral using the FTC: $$s(t) = \int (2t + 1) \, dt = t^2 + t + C$$ We'll need to figure out the constant of integration (C).

Determine the constant of integration

We know that when t = 1, the position is -4. Plug in these values to solve for C: $$-4 = (1)^2 + (1) + C \Rightarrow C = -6$$ So, the displacement function is: $$s(t) = t^2 + t - 6$$

Find the displacement between t = 1 and t = 5

Now, we want to find the displacement between t = 1 and t = 5. To do this, subtract s(1) from s(5): $$ s(5) - s(1) = [(5)^2 + (5) - 6] - [(1)^2 + (1) - 6]$$ Calculating the difference, we get: $$s(5) - s(1) = 14$$

Calculate the position at t = 5

Finally, we need to find the position of the particle at t = 5. Add the displacement from step 4 to the initial position at t = 1: $$Position \: at \: t = 5 = Initial \: position + Displacement$$ $$Position \: at \: t = 5 = -4 + 14 = 10$$ So, the position of the particle at t = 5 is 10.

Key concepts

Velocity Function

The velocity function, denoted as \( v(t) \), represents how a particle's speed and direction change over time. It's essential in understanding motion, as it provides the rate of change of the particle's position. In this exercise, the velocity function is given by \( v(t) = 2t + 1 \). This expression indicates that the velocity increases linearly as time progresses.

- The term \( 2t \) suggests that for every unit increase in time, the velocity will increase by 2 units.- The \(+1\) signifies the initial velocity when \( t = 0 \).

This function will be integrated to determine the particle's displacement over a specified time interval. Understanding velocity functions is crucial because they describe how fast and in what direction an object is moving at any time \( t \).

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus (FTC) is a bridge between differentiation and integration. It simplifies the calculation of a definite integral by linking it to the antiderivative of a function.
When integrating the velocity function of \( v(t) = 2t + 1 \), the FTC is applied to determine the displacement function. In mathematical terms, the FTC states:

• If \( F(t) \) is an antiderivative of \( f(t) \) on an interval \( [a, b] \), then:
\[ \int_{a}^{b} f(t) \, dt = F(b) - F(a) \]

Thus, by finding the antiderivative of \( v(t) \), one can easily evaluate the integral over any interval. This powerful theorem allows us to not only compute areas under curves but also find quantities like displacement, which is essential for solving kinematic problems.

Displacement Function

The displacement function tells us how far a particle has moved from its initial position over a given time period. In this context, after integrating the velocity function, we obtain a displacement formula: \( s(t) = t^2 + t + C \). This represents the total change in position from the starting point.

- The expression \( t^2 + t \) refers to the combined areas under the curve of the velocity function over time, efficiently capturing the total displacement.- The constant \( C \) is crucial, as it adjusts the formula based on known initial conditions.

Once the integration is complete, the displacement function allows you to calculate the particle's position at any time by considering its initial position and the subsequent changes. This function is pivotal in translating velocity information into actual physical movement.

Integration

Integration is a fundamental mathematical process used to find the antiderivative of a function, such as a velocity function, to determine quantities like displacement. In this exercise, integrating the velocity function \( v(t) = 2t + 1 \) yields the displacement function.
Intuitively, integration calculates the accumulation of area under a curve, turning information about rates (like speed) into total change (like distance). It helps transition from understanding motion terms in rates to tangible distances.
Integration involves several steps:

• First, identify the function to be integrated, \( 2t + 1 \).
• Find the antiderivative: \( t^2 + t + C \).
• Evaluate the antiderivative over the interval of interest.

This process unveils the underlying movement dictated by velocity, essential for solving many physics and calculus problems.

Constant of Integration

When determining the indefinite integral of a function, such as the velocity function \( v(t) \), we encounter a constant term, often denoted as \( C \). This constant of integration arises because when differentiating a constant, it vanishes, so we must account for any initial conditions not captured by standard integration.
In this problem, after integrating \( v(t) = 2t + 1 \), we arrived at \( s(t) = t^2 + t + C \). To find \( C \), one utilizes known values. Here, when the particle is positioned at \( t = 1 \), it's known to be at position \(-4\).
By substituting into the equation:

• \(-4 = 1^2 + 1 + C\)
• We solve for \( C \) to find \( C = -6 \).

This constant ensures the displacement function reflects the true initial condition, allowing for accurate computation of the position at any subsequent time.