Question

Displacement and distance from velocity Consider the velocity function shown below of an object moving along a line. Assume time is measured in seconds and distance is measured in meters. The areas of four regions bounded by the velocity curve and the t-axis are also given.

e. Describe the position of the object relative to its initial position after 8 seconds.

Accepted Answer

Identify that the position of the object relative to its initial position after a certain time is given by the displacement, which is the integral of the velocity function over that time interval. Recall that displacement over the interval from $$t = 0 to$$ $$t = 8$$ seconds is calculated by integrating the velocity function $$v(t)$$ with respect to time: $$\displaystyle 	ext{Displacement} = \$$int_0$$^8 v(t) \, dt. $$Use the given areas of the regions bounded by the velocity curve and the $$t-$$axis to evaluate this integral. Remember that areas above the $$t-$$axis correspond to positive velocity (positive contribution to displacement), and areas below correspond to negative velocity (negative contribution to displacement). Sum the signed areas (positive areas minus negative areas) to find the net displacement after 8 seconds, which tells you how far and in which direction the object is from its initial position. Interpret the result: if the net displacement is positive, the object is to the right (or forward) of its initial position; if negative, it is to the left (or backward); if zero, it has returned to its starting point.

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Key Concepts

Velocity and its Relation to Position

Displacement vs. Distance

Area Under the Velocity Curve