Question

the motion of a lacrosse 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 - 4s: 16 m, right 4s - 12s: m, right how far away and in what direction from their starting position does she end up? unit, direction describe the lacrosse players speed during the following time periods? 0s - 4s: increasing speed 4s - 12s: decreasing speed

Explanation:

Step1: Recall distance - velocity - time relationship

The distance traveled is the area under the velocity - time graph. For the time interval \(4s - 12s\), the velocity - time graph is a trapezoid. The formula for the area of a trapezoid is \(A=\frac{(a + b)h}{2}\), where \(a\) and \(b\) are the lengths of the parallel sides and \(h\) is the height. Here, \(a = 8\) (velocity at \(t = 4s\)), \(b=0\) (velocity at \(t = 12s\)), and \(h=12 - 4=8s\).

Step2: Calculate the area of the trapezoid

\[A=\frac{(8 + 0)\times8}{2}=32\]
The area under the velocity - time graph gives the distance traveled. Since the velocity is positive in this interval, the direction is right.
For the total displacement, we need to find the net area under the graph from \(t = 0s\) to \(t=12s\). The area from \(0 - 4s\) is a triangle with base \(b = 4s\) and height \(h = 8m/s\), so its area \(A_1=\frac{1}{2}\times4\times8 = 16m\). The area from \(4 - 12s\) is \(32m\). The total displacement is \(16+32=48m\) to the right.

Answer:

4s - 12s: 32 m, right
Total: 48 m, right