Question

A particle moves along a straight line so that its distance s in time t sec is $s = t +6 t^{2} -t^{3}$. After what time is the acceleration zero
a) $2$ sec
b) $3$ sec
c) $4$ sec
d) $6$ sec

Answer 1

Hint: Zero acceleration is the acceleration with zero magnitudes. Motion with fixed velocity is just a particular case of motion with constant acceleration. For the body's acceleration to be zero, the difference in velocity of a body must be equivalent to zero.

Complete answer:
Given $s = t +6 t^{2} - t^{3}$
s is the distance
t is the time.
In the question, we have to find the time for which acceleration is zero.
On doing double differentiation on distance, we get the acceleration.
First, differentiate s with respect to t,
$\dfrac{ds}{dt} = \dfrac{d}{dt}( t + 6 t^{2} -t^{3})$
It gives velocity.
$v = 1+ 12 t – 3t^{2}$
Again, differentiate v with respect to time,
$\dfrac{dv}{dt} = \dfrac{d}{dt}( 1+ 12 t – 3t^{2})$
It gives acceleration.
$a = 12 -6t$
And we know, $a = 0$ .
$0 = 12 -6t$
It gives $t = 2$ sec.
Option (a) is correct.

Note: Zero acceleration means no difference in acceleration. That is no increment or reduction of acceleration throughout the track. On the other hand, steady acceleration means a uniform gain or drop of acceleration.