Position of a particle is given as: x = 4 t 4 − 2 t 3 + 2 11 t 2 − 6 t + 5 Choose the CORRECT option(s):

x = 4 t 4 − 2 t 3 + 2 11 t 2 − 6 t + 5 ⇒ Particle move on straight line d t d x = v = t 3 − 6 t 2 + 11 t − 6 For turning point v = 0 ⇒ t 3 − 6 t 2 + 11 t − 6 = 0 ⇒ t = 1 , 2 \& 3 . dt 2 d 2 x = a , 3 t 2 − 12 t + 11 Thus minima of acceleration occurs at 6 12 = 2 sec Since from 0-2 sec, particle changes its position ⇒ distance > |displacement|

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Physics - Kinematics Question with Solution | TestHub

PhysicsKinematics1-DMedium2 minQB

PhysicsMediummultiple choice

Position of a particle is given as:

$x = \frac{t ^{4}}{4} - 2 t^{3} + \frac{11 t ^{2}}{2} - 6 t + 5$

Choose the CORRECT option(s):

Options:(select one or more)

Answer:

A, B, C, D

Solution:

$x = \frac{t ^{4}}{4} - 2 t^{3} + \frac{11 t ^{2}}{2} - 6 t + 5 \Rightarrow$ Particle move on straight line $\frac{d x}{d t} = v = t^{3} - 6 t^{2} + 11 t - 6$ For turning point $v = 0 \Rightarrow t^{3} - 6 t^{2} + 11 t - 6 = 0 \Rightarrow t = 1, 2$ \& 3 . $\frac{d ^{2} x}{dt ^{2}} = a, 3 t^{2} - 12 t + 11$ Thus minima of acceleration occurs at $\frac{12}{6} = 2 sec$ Since from 0-2 sec, particle changes its position ⇒ distance > |displacement|

Stream:JEESubject:PhysicsTopic:KinematicsSubtopic:1-D

⏱ 2mℹ️ Source: QB

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