Verified Solution • 2025–26 Session moderate

QJEEM19Phbfb2: JEE Main 2019 Physics - Class 11 JEEM19Phbfb2 Solution: The position of a particle as a function of time t, is given by …

The position of a particle as a function of time t, is given by
x(t) = at + bt2 – ct3
where a, b and c are constants. When the particle attains zero acceleration, then its velocity will be :

Multiple Choice Options

  • A a+b2ca + {{{b^2}} \over {c}}
  • B a+b24ca + {{{b^2}} \over {4c}}
  • C a+b23ca + {{{b^2}} \over {3c}}
  • D a+b22ca + {{{b^2}} \over {2c}}

Step-by-Step Verified Solution

x = at + bt2 – ct3

V=dxdt=a+2bt−3ct2V = {{dx} \over {dt}} = a + 2bt - 3c{t^2}

a=dvdt=2b−6cta = {{dv} \over {dt}} = 2b - 6ct

Put acceleration = 0

⇒t=b3c\Rightarrow t = {b \over {3c}}

Now V at t = b3c{b \over {3c}}

V=a+b23cV = a + {{{b^2}} \over {3c}}

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring