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Chapter 11: Problem 66

The velocity of a car (in feet/second) \(t\) sec after starting from rest is given by the function $$ f(t)=2 \sqrt{t} \quad(0 \leq t \leq 30) $$ Find the car's position, \(s(t)\), at any time \(t\). Assume \(s(0)=0\).

Short Answer

Expert verified
The car's position at any time \(t\) is given by the function \(s(t) = \frac{4}{3} t^{(3/2)}\), with \(0 \leq t \leq 30\).

Step by step solution

01

Find the antiderivative of f(t)

First, we need to find the antiderivative of the velocity function, f(t) = 2鈭歵. We rewrite 鈭歵 as t^(1/2) to make it easier to work with: \(f(t) = 2t^{(1/2)}\) Now, applying the power rule for integration, we get: \(s(t) = \int 2t^{(1/2)} dt\) Which can be evaluated as follows: \(s(t) = \frac{2}{(1/2) + 1} t^{(1/2) + 1} + C\)
02

Simplify the antiderivative equation

We can simplify the antiderivative expression: \(s(t) = \frac{2}{(3/2)} t^{(3/2)} + C = \frac{4}{3} t^{(3/2)} + C\)
03

Solve for the constant of integration, C

We are given that the initial position of the car, s(0), is 0. We can use this information to solve for the constant of integration, C: \(s(0) = \frac{4}{3}(0)^{(3/2)} + C\) \(0 = C\) Therefore, we don't need to include the constant C in our final expression of the position function.
04

Write the position function, s(t)

Now that we have found the antiderivative of the velocity function and have solved for the constant of integration, we can write the position function, s(t): \(s(t) = \frac{4}{3} t^{(3/2)}\) So, the car's position at any time t is given by this function, with \(0 \leq t \leq 30\).

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