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Suppose you are driving a car along a straight road. Your velocity at any time t (in seconds) is given by the function v(t)=3t^(2)-2t+5 meters per second. Find the total distance traveled by the car from t=0 to t=3 seconds.

Suppose you are driving a car along a straight road. Your velocity at any time $t$ (in seconds) is given by the function $v(t)=3 t^{2}-2 t+5$ meters per second. Find the total distance traveled by the car from $t=0$ to $t=3$ seconds.

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Relate position, velocity, speed, and acceleration using derivatives

Full solution

Q. Suppose you are driving a car along a straight road. Your velocity at any time

t

(in seconds) is given by the function

v(t)=3 t^{2}-2 t+5

meters per second. Find the total distance traveled by the car from

t=0

to

t=3

seconds.

Understand the problem: Understand the problem and set up the integral. We are given the velocity function $v(t) = 3t^2 - 2t + 5$ , and we need to find the total distance traveled from $t=0$ to $t=3$ seconds. To find the distance, we need to integrate the velocity function over the given time interval.
Write integral: Write down the integral that represents the total distance traveled. $\newline$ The total distance traveled, $D$ , is the integral of the velocity function from $t=0$ to $t=3$ : $\newline$ $D = \int_{0}^{3} (3t^2 - 2t + 5) \, dt$
Calculate integral: Calculate the integral. $\newline$ To find the distance, we need to find the antiderivative of the velocity function and evaluate it from $t=0$ to $t=3$ . $\newline$ The antiderivative of $3t^2$ is $t^3$ , the antiderivative of $-2t$ is $-t^2$ , and the antiderivative of $5$ is $5t$ . So the integral becomes: $\newline$ $D = [t^3 - t^2 + 5t]$ from $0$ to $t=3$ $0$
Evaluate limits: Evaluate the antiderivative at the upper and lower limits of the integral. $\newline$ $D = [3^3 - 3^2 + 5(3)] - [0^3 - 0^2 + 5(0)]$ $\newline$ $D = [27 - 9 + 15] - [0 - 0 + 0]$ $\newline$ $D = [27 - 9 + 15] - 0$ $\newline$ $D = 27 - 9 + 15$ $\newline$ $D = 33$ meters

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