**Whenever we have to find total displacement from a velocity time graph we find the area under the curve.**

In the figure we clearly see

• Two triangles ABF & DCE having area A_{1} & A_{2} respectively

• A rectangles BEIF having area A_{3} .

A_{1} =\(\frac{(2-0)\times(5-0)}{2}\) = 5

A_{2} =\(\frac{(10-5)\times(10-6)}{2}\) = 10

A_{3} = (5-0)x(10-2) = 40

Total area under the curve = A_{1 }+ A_{2} + A_{3}

= 5 + 10 + 40 = 55

**Hence total displacement = 55 m**