Question

A velocity-time graph for a moving object is shown below. What would be the total displacement during time \(t = 0\) to \(t = 6~\text{s}\)?

Answer 1

The total displacement travelled by the object is the area under the \(v-t\) graph. Total displacement during \(t = 0\) to \(t = 6 ~\text{s}\) is equal to area under the graph \(= A_1 + A_2 + A_3\).

Since area \(A_3\) is on the negative side of the graph, the area will be negative hence, the net displacement.
Displacement \(= A_1 + A_2 + A_3\)
\(A_1= 5 \times 2\) = 10m
\(A_2= \dfrac{1}{2} \times 5 \times (4-2)\) = 5m
\(A_3= \dfrac{1}{2} \times (-5) \times (6-4)\)= -5m
Displacement \(= 10 + 5 – 5 = 10 ~\text{m}\)
Therefore, option A is correct.