Area under a-t graph gives the velocity.

Velocity:-

Part AB on a-t graph represents linearly decreasing acceleration

Given $v_0 =2m/s$

$v_4=v_0+A(\triangle OAB)= 2+\dfrac 12\times 4\times 2 =6m/s$

Part BC on a-t graph represents linearly increasing acceleration.

$v_8=v_4+A(\triangle BEC)\\ =6+\dfrac 12 \times (8-4) \times 5 \\ =16 m/s $

Part CD on a-t graph is parallel to X-axis. It represents constant acceleration.

$v_{10} =v_8 +A(CDFE)=16+(10-8)\times 5=26 m/s$

The v-t curve is as shown.

Displacement:

Area under curve $GH=A(OGHK) = A(OGLK) + A(GHL) \\ =4 \times 2 + \dfrac 23 \times 4 \times (6-2) = \dfrac {56}3 =18.6667 m$

Assuming zero initially displacement.

Displacement after 4 sec $(s_4 ) = s_0 + A (OGHK) \\ =0 + 18.6667 – 18.6667m $

Area under $H = A (HKMI) = A (HKMN) + A(HNI) \\ (8-4) \times 6 + \dfrac13 \times (8-4)\times (16-6)= \dfrac {112}3 =37.3333m $

Displacement after 4 sec $(s_3 ) = s_4 + A (HKMI) \\ =18.6667m + 37.3333m=56m $

Area under I = A (Trapezium IMPJ)

$=\dfrac 12 \times (26+16) \times (10-8)=42m $

Displacement after 10 sec $(s_{10} ) = s_8 + A (IMPJ) \\ = 56m +42m = 98m$

The s-t curve is shown.