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
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.

:.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.

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-=(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