When `t = 2`, the particle is at `(8, 0.6)`.

`x=(20t)/(2t+1)`

so

`(dx)/(dt)=((2t+1)20-20t(2))/(2t+1)^2`

`=20/((2t+1)^2)`

At `t = 2`,

`dx/dt = v_x= 20/25 = 0.8\ "ms"^-1`.

Also, y = 0.1(t2 + t) so

`(dy)/(dt)=0.1(2t+1)`

When `t = 2`,

`dy/dt = v_y= 0.5\ "ms"^-1`.

So

`v=sqrt((v_x)^2+(v_y)^2)`

`=sqrt(0.8^2+0.5^2)`

`=0.943\ text(ms)^-1`

Graph of x against y

Now for the direction:

`tan\ theta_v=(v_y)/(v_x)=0.5/0.8 = 0.625`

So `θ_v= arctan(0.625) = 0.558` radians

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