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The position of a particle moving along X axis given by X = (-2t³ + 3t² + 5)m . The acceleration of the particle at the instant its velocity becomes zero is

The position of a particle moving along X axis given by X = (-2t³ + 3t² + 5)m . The acceleration of the particle at the instant its velocity becomes zero is

Grade:11

5 Answers

Aman gohel
71 Points
6 years ago
Given,X=-2t^3+3t^2+5.....(i) Differentiating x w.r.t t dx/dt V=d (-2t^3+3t^2+5)/dt V=-6t^2+9t.....(ii) Again differentiating b w.r.t t dv/dt a=d (-6t^2+9t)/dt =-12t+9....(iii)At v=0,From eqa.. ( ii)-6t^2+9t=0Therefore,t=1.5sPutting t=1.5s in equa...(iii)a=-12×1.5+9=-9ms^-2 (Ans)
Swadesh
14 Points
6 years ago
X=(-2t³+3t²+5)V=(dx/dt)=(-6t²+6t)Than v=0So (-6t²+6t)=0. We get t=1A=(dv/dt)=(-12t+6)A=12×1+6=-6m/s²
Nitin
11 Points
6 years ago
X = -2t^3+3t^2+5V=dx/dt=-6t^2+6tAt t= 1 , v =0 ( -6(1)^2-6(1)=0)At t=1 velocity =0 then acceleration = dv/dt = -12t+6 at t= 1 , a= -12(1)+6=-6m/s^2
Nithish
14 Points
5 years ago
x=-2t^3+3t^2+5
v=dx/dt=-6t^2+6t
v will be zero when t=1s
a=dv/dt=-12t+6
they have asked us to find the acceleration when velocity is zero
therefore put t=1s
a=-12(1)+6
a=-12+6
a=-6m/s^2
soni kumari
16 Points
2 years ago
  1. dx|dt = – 2t3 + 3t2 +5 
  2.            = -6t2 +6t
  3. v=0   A|Q -6t2 + 6t =0
  4. t=1s 
  5. dv|dt =-6t2+6t =0
  6.         =-12t +6 = -12*1 +6= -6m|s2

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