A particle moves along a straight line such that its displacement(s) at any time(t) is given by $S = t^{3} - 3 t^{2} + 2$ . The velocity of the particle when its acceleration zero is?

Question

A particle moves along a straight line such that its displacement any time t is given by s - t-3 - 6t-2 - 3t - 4 metres- The velocity when the acceleration is zero is-3  m-s-12  m-s42  m-s-9  m-s

Toppr Admin · Accepted Answer

S-t3-x2212-62-3t-4v-dsdtSo -3t2-x2212-12-3-vagain differentiate -a-dvdt6t-x2212-12-0t-2sv-3-2-2-x2212-12-xD7-2-3-x2212-9ms-x2212-1Hence -D- is correct answer-

A particle moves along a straight line such that its displacement at any time $t$ is given by $s = t^{3} - 6 t^{2} + 3 t + 4$ metres. The velocity when the acceleration is zero is:-
$3 m / s$
$- 12 m / s$
$42 m / s$
$- 9 m / s$

$S = t^{3} - 6^{2} + 3 t + 4$
$v = \frac{d s}{d t}$

So $= 3 t^{2} - 12 + 3 = v$
again differentiate , $a = \frac{d v}{d t}$
$6 t - 12 = 0$
$t = 2 s$
$v = 3 (2)^{2} - 12 \times 2 + 3 = - 9 m s^{- 1}$

Hence (D) is correct answer.

A particle moves along a straight line such that its displacement at any time t is given by s=t3−6t2+3t+4 metres. The velocity when the acceleration is zero is:-3 m/s−12 m/s42 m/s−9 m/s

S=t3−62+3t+4v=dsdtSo =3t2−12+3=vagain differentiate ,a=dvdt6t−12=0t=2sv=3(2)2−12×2+3=−9ms−1Hence (D) is correct answer.

A particle moves along a straight line such that its displacement at any time $t$ is given by $s = t^{3} - 6 t^{2} + 3 t + 4$ metres. The velocity when the acceleration is zero is:-
$3 m / s$
$- 12 m / s$
$42 m / s$
$- 9 m / s$

$S = t^{3} - 6^{2} + 3 t + 4$
$v = \frac{d s}{d t}$

So $= 3 t^{2} - 12 + 3 = v$
again differentiate , $a = \frac{d v}{d t}$
$6 t - 12 = 0$
$t = 2 s$
$v = 3 (2)^{2} - 12 \times 2 + 3 = - 9 m s^{- 1}$

Hence (D) is correct answer.