# 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= ???

Jul 4, 2018

$8$ 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.

$\therefore a = \frac{\mathrm{dv}}{\mathrm{dt}}$

$= \frac{d}{\mathrm{dt}} \left(16 - {t}^{2}\right)$

$= - 2 t$

$\therefore - a = 2 t$

So, the deceleration would be $2 t$.

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

${t}^{2} = 16$

$t = \sqrt{16} = \pm 4$

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

The deceleration at this time is $2 \cdot 4 = 8$ units.

Jul 4, 2018

Acceleration $a$ is defined as the rate of change of velocity, and is given by derivative of expression for velocity with respect to time.

$a \equiv \frac{\mathrm{dv}}{\mathrm{dt}}$

From the given expression we get

$a = \frac{d}{\mathrm{dt}} \left(16 - {t}^{2}\right)$
$\implies a = - 2 t$ .....(1)

The point is it is momentarily at rest when $v = 0$. Inserting this in the given expression we get

$0 = 16 - {t}^{2}$
$\implies t = \sqrt{16} = \pm 4$

Inserting in (1) we get at $t = 4 \mathmr{and} t = - 4 \setminus u n i t s$

$a \left(4\right) = - 2 \times 4 = - 8 \setminus u n i t s$
$a \left(- 4\right) = - 2 \times \left(- 4\right) = 8 \setminus u n i t s$

Ignoring $- v e$ value of time and lower value of acceleration, as it does not give deceleration, we get
Deceleration when the point is momentarily at rest $= 8 \setminus u n i t s$