Question

retake velocity assessment
what is the displacement of the object from t = 0 to t = 6 s?
what is the displacement of the object from t = 6 to t = 16 s?
what is the displacement of the object from t = 0 to t = 16 s?

Explanation:

Step1: Recall displacement - area under v - t graph

Displacement from \(t = a\) to \(t = b\) is the area under the velocity - time graph from \(t=a\) to \(t = b\).

Step2: Calculate displacement from \(t = 0\) to \(t=6\ s\)

The area from \(t = 0\) to \(t = 6\ s\) is a triangle. The base \(b = 6\ s\) and height \(h=4\ m/s\). Using the formula for the area of a triangle \(A=\frac{1}{2}bh\), we have \(A_1=\frac{1}{2}\times6\times4 = 12\ m\).

Step3: Calculate displacement from \(t = 6\) to \(t = 16\ s\)

The area from \(t = 6\) to \(t=16\ s\) is composed of a triangle above the \(t\) - axis and two triangles below the \(t\) - axis.
The first triangle (from \(t = 6\) to \(t = 8\)) below the \(t\) - axis has base \(b_1 = 2\ s\) and height \(h_1=4\ m/s\), area \(A_{21}=\frac{1}{2}\times2\times4=4\ m\) (negative as below \(t\) - axis).
The second triangle (from \(t = 8\) to \(t = 12\)) below the \(t\) - axis has base \(b_2 = 4\ s\) and height \(h_2 = 4\ m/s\), area \(A_{22}=\frac{1}{2}\times4\times4 = 8\ m\) (negative as below \(t\) - axis).
The third triangle (from \(t = 12\) to \(t = 16\)) above the \(t\) - axis has base \(b_3=4\ s\) and height \(h_3 = 4\ m/s\), area \(A_{23}=\frac{1}{2}\times4\times4=8\ m\).
The total area \(A_2=-4 - 8+8=-4\ m\).

Step4: Calculate displacement from \(t = 0\) to \(t = 16\ s\)

We add the displacements from \(t = 0\) to \(t = 6\ s\) and from \(t = 6\) to \(t = 16\ s\). \(A=A_1 + A_2=12-4 = 8\ m\).

Answer:

Displacement from \(t = 0\) to \(t = 6\ s\): \(12\ m\)
Displacement from \(t = 6\) to \(t = 16\ s\): \(-4\ m\)
Displacement from \(t = 0\) to \(t = 16\ s\): \(8\ m\)