# The deceleration of a point when it is momentarily at rest which is moving along a straight line with a velocity 16-t^2 is= ???

##### 2 Answers

#### Explanation:

I assume here that deceleration would be just negative acceleration, i.e. in the reverse direction.

Acceleration is the **rate of change of velocity**, or the derivative of the velocity.

So, the deceleration would be

When the object is at rest, its velocity is

Since time cannot be negative, we take

The deceleration at this time is

Acceleration **rate of change of velocity**, and is given by derivative of expression for velocity with respect to time.

#a-=(dv)/dt#

From the given expression we get

#a=d/dt(16-t^2)#

#=>a=-2t# .....(1)

The point is it is momentarily at rest when

#0=16-t^2#

#=>t=sqrt16=+-4#

Inserting in (1) we get at

#a(4)=-2xx4=-8\ units#

#a(-4)=-2xx(-4)=8\ units#

Ignoring

Deceleration when the point is momentarily at rest