Area under a-t graph gives the velocity.

Velocity:

Part AB on a-t graph represents linearly decreasing acceleration.

Give $v_0 = 2 \ m/s$

$v_4 = v_0 + A(\Delta OAB)= 2 + \dfrac {1}{2} \times 4 \times 2 = 6 \ m/s$

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

$v_8 = v_4 + A(\Delta BEC)$

$= 6 + \dfrac {1}{2} \times (8-4)\times 5$

$=16 \ m/s$

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

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

The v-t curve is as shown.

Displacement:

Area under v-t graph gives the displacement

Area under curve GH=A (OGHK)=A(OGLK)+ A(GHL)

$= 4 \times 2 + \dfrac {2}{3} \times 4 \times (6-2) = \dfrac {56}{3} = 18.6667 \ m$

Assuming zero initially displacement,

Displacement after 4 sec (S4)=S_0 + A(OGHK)

$= 0+ 18.6667 = 18.6667 \ m$

Area under HI = A(HKMI)= A(HKMN)+ A(HNI)

$=(8-4) \times 6 + \dfrac {1}{3} \times (8-4) \times (16-6) = \dfrac {112}{3} = 37.3333 \ m$

Displacement after $8 \sec (S_8)=S_4+ A (HKMI)$

$=18.6667 \ m + 37.3333 \ m = 56 m$

Area under IJ=A (Trapezium IMPJ)

$= \dfrac {1}{2} \times (26+16) \times (10-8) = 42 m$

Displacement after $10 \sec(S_{10})= S_8 + A(IMPJ)$

$=56 m + 42m = 98 m$

The s-t curve is as shown.