When `t = 10`, the particle is at `(5000, 400)`.

Here is the graph of the motion.

**Note**:

- The axes are
*x*and*y*(and do not involve*t*). - The particle is accelerating as time goes on (the red dots are at one second intervals)

We are told that

x= 5t^{3}

So

`dx/dt=15t^2`

At *t* = 10, the velocity in the *x*-direction is given by:

`dx/dt = v_x=15(10)^2=1500\ "ms"^-1`

Also, *y* = 4*t*^{2}** ** so the velocity in the *y*-direction is:

`dy/dt=8t`

When *t* = 10, the velocity in the *y*-direction is:

`dy/dt=v_y=8(10)=80\ "ms"^-1`

So the magnitude of the velocity will be:

`v=sqrt((v_x)^2+(v_y)^2)=sqrt(1500^2+80^2)=1502.1\ "ms"^-1`

Now for the direction of the velocity (it is an angle, relative to the positive *x*-axis):

`tan\ theta_v=v_y/v_x=80/1500`

So `θ_v= 0.053` radians `= 3.05^@`.