Firstly we need to calculate time at which velocity becomes 0

Slope of line from t=10 to t=15 is equal to acceleration

$a = -60/5 = -12 m/s^2 \\ Now, 0 = 30 - (12)t \\ T = 2.5 sec $

Hence, velocity is zero at $t=12.5 sec $

Position calculation for x-t graph:

From v-t curve, position can be calculated using

$X_F = X_I +$ area under v-t curve

Now, for $t=0 \space \space to\space \space t=10, X_I = -25 \\ X_{10} = -25 + (1/2)*10*30 \\ X_{10} = 125 m \\ For, t=10 \space \space to \space \space t=12.5 \\ X_{12.5} = 125 + (1/2)*5*30 \\ X_{12.5} = 162.5 m \\ For, t=12.5\space \space to\space \space t=15 \\ X_{15} = 162.5 - (1/2)*5*30...........[\text { -ve sign indicates that area is below x axis }] \\ X_{12.5} = 125 m \\ For, t=15\space \space to\space \space t=24 \\ X_{24} = 125 - 30*9 \\ X_{24} = -145 m $

x-t curve will be as follows

Acceleration calculations for a t graph:

$For t=0\space \space to\space \space t=10 $

a= slope of line $= (30-0)/(10-0) = 3 m/s^2 \\ For t=10 \space \space to\space \space t=15 $

a= slope of line =$ (30-0)/(10-0) = 3 m/s^2 \\ For \space \space t=15\space \space to\space \space t=24 $

a= slope of line = 0

a-t curve will be as follows: