Distance traveled and displacement Suppose an object moves along a line with velocity (in ft/s) v(t)=6−2t, for 0≤t≤6, where t is measured in seconds.

c. Find the distance traveled by the object on the interval 0≤t≤6.

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Understand the difference between displacement and distance traveled: displacement is the net change in position, while distance traveled is the total length of the path covered, regardless of direction.

Identify the velocity function given: \(v(t) = 6 - 2t\) for \(0 \leq t \leq 6\). To find distance traveled, we need to consider when the velocity changes sign because the object might change direction.

Find the time when the velocity is zero by solving \(6 - 2t = 0\). This gives the critical point where the object changes direction.

Split the interval \([0,6]\) into subintervals based on the critical point found. On each subinterval, the velocity has a consistent sign, so integrate the absolute value of velocity over each subinterval to find the distance traveled in that segment.

Calculate the total distance traveled by summing the absolute values of the integrals of velocity over each subinterval: \(\text{Distance} = \int_0^{t_0} |v(t)| \, dt + \int_{t_0}^6 |v(t)| \, dt\), where \(t_0\) is the time when velocity is zero.

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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Velocity and Displacement

Velocity is the rate of change of position with respect to time and can be positive or negative, indicating direction. Displacement is the net change in position over a time interval and is found by integrating velocity. Understanding the difference between velocity and displacement is crucial for analyzing motion.

Distance Traveled vs. Displacement

Distance traveled is the total length of the path covered, regardless of direction, while displacement is the straight-line change in position. To find distance traveled from velocity, one must integrate the absolute value of velocity over the time interval, accounting for any changes in direction.

Definite Integrals and Absolute Value

Definite integrals calculate the accumulation of quantities over an interval. When velocity changes sign, the integral of velocity gives displacement, but to find total distance, the integral of the absolute value of velocity is needed. This requires identifying intervals where velocity is positive or negative and integrating accordingly.

Distance traveled and displacement Suppose an object moves along a line with velocity (in ft/s) v(t)=6−2t, for 0≤t≤6, where t is measured in seconds.c. Find the distance traveled by the object on the interval 0≤t≤6.