Question

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= ???

Answer 1

`8` units

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.

`:.a=(dv)/dt`

`=d/dt(16-t^2)`

`=-2t`

`:.-a=2t`

So, the deceleration would be `2t`.

When the object is at rest, its velocity is `0`, and therefore `16-t^2=0`.

`t^2=16`

`t=sqrt16=+-4`

Since time cannot be negative, we take `t=4`.

The deceleration at this time is `2*4=8` units.

Answer 2

Acceleration `a` is defined as the 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 `v=0`. Inserting this in the given expression we get

`0=16-t^2`
`=>t=sqrt16=+-4`

Inserting in (1) we get at `t=4and t=-4\ units`

`a(4)=-2xx4=-8\ units`
`a(-4)=-2xx(-4)=8\ units`

Ignoring `-ve` value of time and lower value of acceleration, as it does not give deceleration, we get
Deceleration when the point is momentarily at rest `=8\ units`