JEE Main20199 Apr 2019Evening ShiftPhysicsMotion in One DimensionActual
The position of a particle as a function of time t , is given by x t = a t + b t 2 - c t 3 where a , b and c are constants. When the particles zero acceleration, then its velocity will be:
Options
- Aa + b 2 3 c
- Ba + b 2 2 c
- Ca + b 2 c
- Da + b 2 4 c
Correct answer
A. a + b 2 3 c
Step-by-step solution
x ( t ) = a t + b t 2 - c t 3 velocity v ( t ) = d d t ( x ) = d d t a t + b t 2 - c t 3 = a + 2 b t - 3 c t 2 Acceleration = d d t v ( t ) = 2 b - 6 t c acceleration = 0 ⇒ 2 b - 6 t c = 0 t = b 3 c ∴ velocity when t = b 3 c , v t = b 3 c = a + 2 b b 3 c - 3 b 3 c 2 = a + 2 b 2 3 c - b 2 3 c = a + b 2 3 c