Question

the motion of a soccer player is captured using a motion detector and displayed on the graph shown. use the graph to answer the following questions. right is defined as positive. answers have a 5% tolerance to allow for estimation. what is the distance and direction traveled during the following time periods? 0s - 6s: 15 m left; 6s - 18s: m right how far away and in what direction from their starting position does she end up? m direction describe the soccer players speed during the following time periods? 0s - 6s: decreasing speed; 6s - 18s: increasing speed

Explanation:

Step1: Recall distance - velocity relationship

The distance traveled is the area under the velocity - time graph. For \(0 - 6s\), the area of the trapezoid formed by the velocity - time graph and the time - axis is \(A_1=\frac{1}{2}(5 + 0)\times6= 15\) m. Since the velocity is negative in \(0 - 6s\), the direction is left.

Step2: Calculate distance for \(6 - 18s\)

For \(6 - 18s\), the area of the trapezoid is \(A_2=\frac{1}{2}(0 + 10)\times(18 - 6)=60\) m. Since the velocity is positive, the direction is right.

Step3: Find net displacement

The net displacement \(d=A_2 - A_1=60 - 15 = 45\) m, and the direction is right.

Answer:

For \(6 - 18s\): 60 m, right
For net displacement: 45 m, right