Question

The velocity as a function of time for a car on an amusement park ride is given as \(v=A t^{2}+B t\) with constants \(A=2.0 \mathrm{~m} / \mathrm{s}^{3}\) and \(B=1.0 \mathrm{~m} / \mathrm{s}^{2} .\) If the car starts at the origin, what is its position at \(t=3.0\) s?

Short answer

Answer: The position of the car at t = 3.0 seconds is 22.5 meters from the origin.

Step-by-step solution

Integrate Velocity Function

In order to find the position of the car at any given time, we are required to integrate the velocity function with respect to time. So let's integrate the given velocity function \(v=A t^{2}+B t\): \(\int (2.0t^2 + 1.0t) dt = 2.0 \int t^2 dt + 1.0 \int t dt\) Now doing the integrations: \(\Rightarrow 2.0 \cdot \frac{1}{3} t^3 + 1.0 \cdot \frac{1}{2} t^2 + C\) where C is the constant of integration.

Determine the Constant of Integration

We know that the car starts at the origin, i.e. its position is 0 when \(t=0\). Using this information, we can find the value of the constant of integration: \(0 = 2.0 \cdot \frac{1}{3} (0)^3 + 1.0 \cdot \frac{1}{2} (0)^2 + C\) which gives us \(C = 0\). Our position function now becomes: \(x(t) = 2.0 \cdot \frac{1}{3} t^3 + 1.0 \cdot \frac{1}{2} t^2\)

Find the Position at \(t = 3.0\) s

Now we have the position function x(t), and we can find the car's position at \(t = 3.0\) s by plugging in the value of t: \(x(3) = 2.0 \cdot \frac{1}{3} (3)^3 + 1.0 \cdot \frac{1}{2} (3)^2\) Evaluating this expression: \(x(3) = 2.0 \cdot \frac{1}{3}(27) + 1.0 \cdot \frac{1}{2}(9)\) \(x(3) = 18 + 4.5 = 22.5\,\mathrm{m}\) Thus, the car's position at \(t = 3.0\) s is 22.5 meters from the origin.

Key concepts

Velocity Function

To calculate the position of an object in physics, you often start by analyzing its velocity function. In this exercise, the velocity is given as a function of time:

• \(v(t) = A t^2 + B t\)

Here, "\(A\)" and "\(B\)" are constants that tell us how velocity changes over time. Specifically:

• The term "\(A t^2\)" represents the part of velocity that changes with the square of time, indicating an acceleration that increases over time.
• The "\(B t\)" term is a linear term, depicting a constant rate of change in velocity, similar to having a steady acceleration.

Understanding these components helps you see how an object speeds up at different rates. To find how far something has moved, you first consider its velocity, which serves as the rate of change of its position.
Integrating the velocity function over time can then yield the total distance traveled.

Position Calculation

Once we have the velocity function, the next step is to calculate the position of the object. To do this, we integrate the velocity function with respect to time:

• \(v(t) = 2.0 t^2 + 1.0 t\)

Integration reverses differentiation, which means we're finding an anti-derivative. When we integrate the given function:

• \(\int (2.0t^2 + 1.0t) \, dt = 2.0 \cdot \frac{1}{3} t^3 + 1.0 \cdot \frac{1}{2} t^2 + C\)

The resulting expression represents the car's position as a function of time. By integrating, you add up all the little changes in position (represented by the velocity) to find out where the car ends up over an interval of time.

Constant of Integration

During the integration process, we encounter a constant of integration denoted by "\(C\)". This constant is crucial because it accounts for the initial conditions of the problem. In our particular scenario:

• The car starts from the origin, meaning its initial position at time \(t = 0\) is \(0\).

To determine "\(C\)", you set the position function equal to the initial position:

• \(0 = 2.0 \cdot \frac{1}{3} (0)^3 + 1.0 \cdot \frac{1}{2} (0)^2 + C\)

Solving for \(C\), you find \(C = 0\).
This value is crucial because it adjusts our position function to align with the car's starting point. With "\(C\)" known, we finalize our position function as:

• \(x(t) = 2.0 \cdot \frac{1}{3} t^3 + 1.0 \cdot \frac{1}{2} t^2\)

Finally, to find the position at any specific time, like \(t = 3.0\) seconds, you simply substitute \(t\) with the desired value and solve for \(x(t)\), confirming the car's position at the given instant.