The force-time (F-t) graph of a 10\,kg moving object is shown...

The area under the force-time ($F$-$t$) graph gives the impulse, which is equal to the change in momentum of the object.

$\text{Impulse} = \text{Area of the trapezium}$

$\text{Area} = \dfrac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}$

$\text{Area} = \dfrac{1}{2} \times (25 + 15) \times 20$

$\text{Area} = \dfrac{1}{2} \times 40 \times 20 = 400 \, \text{N}\cdot\text{s}$

By the impulse-momentum theorem:

$\Delta p = m(v - u)$

Assuming the object starts from rest ($u = 0$):

$400 = 10 \times (v - 0)$

$v = 40 \, \text{m/s}$

Answer: $40\,\text{m/s}$