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.
Let the starting velocity be v and acceleration be a.
initially t has to be zero i.e. time= 0 sec
Hence initial velocity is 0+5 m/s=5 m/s
Hence the body moves with a constant acceleration of 1 m/s^2.
This question should have been posted in physics.
Anyways, Hope it helps..