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

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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.

v=dx/dt

= d(0.5t^2+5t)/dt

=2*0.5t+5

=t+5

initially t has to be zero i.e. time= 0 sec

Hence initial velocity is 0+5 m/s=5 m/s

Now, a=dv/dt

=> a=d(t+5)/dt

=(1+0) m/s^2

=1 m/s^2

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..

Thanx.