Given x=0.5t^2+5t where t is in sec., what is the starting velocity and acceleration?

The expression for distance travelled in terms of time is
x = 0.5t^2 + 5t.

The derivative of displacement x with
respect to time, dx/dt, gives the instantaneous
velocity.

Here dx/dt = v = 2*0.5*t +
5

=> t + 5

When the
body starts, t = 0. The instantaneous velocity is 0 + 5 = 5
units/sec

The derivative of velocity with respect to time
gives the instantaneous acceleration.

Here dv/dt =
1

The acceleration is constant at 1
units/sec^2.

The required value for velocity
when the body starts is 5 units/sec and the acceleration when the body starts is 1
units/sec^2.