13–16. Displacement from velocity Consider an object moving along a line with the given velocity v. Assume time t is measured in seconds and velocities have units of m/s.

b. Find the displacement over the given interval.

v(t) = 3t²−6t on [0, 3]

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Identify the given velocity function: $v(t) = 3t^{2} - 6t$ and the time interval $[0, 3]$ seconds.

Recall that displacement over a time interval $[a, b]$ is found by integrating the velocity function over that interval: $\text{Displacement} = \int_{a}^{b} v(t) \, dt$.

Set up the definite integral for displacement: $\int_{0}^{3} (3t^{2} - 6t) \, dt$.

Integrate the function term-by-term: the integral of \$3t^{2}$ is $t^{3}$, and the integral of $-6t$ is $-3t^{2}$, so the antiderivative is $t^{3} - 3t^{2}$.

Evaluate the antiderivative at the bounds and subtract: calculate $\left[ t^{3} - 3t^{2} \right]_{0}^{3} = (3^{3} - 3 \times 3^{2}) - (0 - 0)$ to find the displacement.

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주요 개념

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Velocity and Displacement Relationship

Velocity represents the rate of change of position with respect to time. Displacement over a time interval is the net change in position, which can be found by integrating the velocity function over that interval. This integral sums the instantaneous velocities to give the overall change in position.

Definite Integral as Net Area

The definite integral of a velocity function over a time interval calculates the net area between the velocity curve and the time axis. Positive areas correspond to movement in one direction, while negative areas represent movement in the opposite direction. The sum of these areas gives the total displacement.

Evaluating Definite Integrals

To find displacement, compute the definite integral of v(t) from the start to end time. This involves finding the antiderivative of v(t), then applying the Fundamental Theorem of Calculus by subtracting the antiderivative evaluated at the lower limit from that at the upper limit.

13–16. Displacement from velocity Consider an object moving along a line with the given velocity v. Assume time t is measured in seconds and velocities have units of m/s.b. Find the displacement over the given interval. v(t) = 3t²−6t on [0, 3]