The area under \(a-t\) graph gives change in velocity.

Given, \(u=-5 \,m / s\)

\(\Rightarrow\) Area on positive side \(=\frac{1}{2} \times 6 \times 10=30 \,ms ^{-1}\)

\(\Rightarrow\) Area on negative side \(=\frac{1}{2} \times 2 \times 10=10 \,ms ^{-1}\)

Net area \(=30-10=20 \,ms ^{-1}\)

\(\Delta v=\) Area

\(v-(-5)=20\)

\(\Rightarrow v=15 \,ms ^{-1}\)