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Chapter 2: Problem 45

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

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

Step by step solution

01

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.
02

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: \(x(t) = 2.0 \cdot \frac{1}{3} t^3 + 1.0 \cdot \frac{1}{2} t^2\)
03

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.

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